Twin primes whose sum is a cube [closed] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)$84537841287167$ and $84537841287169$ a particular pair of twin primes!Conjecture on twin primesAbout a paper by Gold & Tucker (characterizing twin primes)Twin Primes Problem: Need HelpNumber Theory : Primes not in Twin PrimesIs there more than one occurrence of a power of two between twin primes?Twin Primes, their Arithmetic Means and some properties.Twin Primes and Complete Residue SystemsWhat is wrong with this reasoning regarding twin primes?A sieve for twin primes; does it imply there are infinite many twin primes?A twin prime theorem, and a reformulation of the twin prime conjecture

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Twin primes whose sum is a cube [closed]



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)$84537841287167$ and $84537841287169$ a particular pair of twin primes!Conjecture on twin primesAbout a paper by Gold & Tucker (characterizing twin primes)Twin Primes Problem: Need HelpNumber Theory : Primes not in Twin PrimesIs there more than one occurrence of a power of two between twin primes?Twin Primes, their Arithmetic Means and some properties.Twin Primes and Complete Residue SystemsWhat is wrong with this reasoning regarding twin primes?A sieve for twin primes; does it imply there are infinite many twin primes?A twin prime theorem, and a reformulation of the twin prime conjecture










6












$begingroup$


$84537841287167$ and $84537841287169$ are a pair of twin primes.



The sum of $84537841287167+84537841287169$ is a cube.



Are there other examples of twin pairs $p$, $p+2$ whose sum is a cube?



Have they to have a particular form?






elementary-number-theory







share|cite|improve this question











$endgroup$



closed as off-topic by user21820, Adrian Keister, Saad, José Carlos Santos, YiFan Apr 5 at 23:39


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Adrian Keister, Saad, José Carlos Santos, YiFan
If this question can be reworded to fit the rules in the help center, please edit the question.











  • 2




    $begingroup$
    What do you want to do with other examples?
    $endgroup$
    – Dietrich Burde
    Apr 4 at 9:50






  • 6




    $begingroup$
    For $k le 10000$, there are $185$ values of $k$ where $p = 108k^3-1$ and $p+2$ are twin primes that sum to a cube. Following are some examples.. $107,2634011,29659499,57395627,104792291,271669247,485149499,cdots\ 83797007573051, colorred84537841287167, 88875400155587,89731703685707,89846291353499,cdots$
    $endgroup$
    – achille hui
    Apr 4 at 10:23






  • 1




    $begingroup$
    @taritgoswami it is a single line command using the CAS maxima for k : 1 thru 10000 do if (primep(108*k^3-1) and primep(108*k^3+1)) then print(k," ",108*k^3-1);
    $endgroup$
    – achille hui
    Apr 4 at 12:10






  • 3




    $begingroup$
    oeis.org/A061308 is worth a look.
    $endgroup$
    – Barry Cipra
    Apr 4 at 12:15






  • 2




    $begingroup$
    similarly oeis.org/A240169
    $endgroup$
    – KBusc
    Apr 4 at 16:00















6












$begingroup$


$84537841287167$ and $84537841287169$ are a pair of twin primes.



The sum of $84537841287167+84537841287169$ is a cube.



Are there other examples of twin pairs $p$, $p+2$ whose sum is a cube?



Have they to have a particular form?






elementary-number-theory







share|cite|improve this question











$endgroup$



closed as off-topic by user21820, Adrian Keister, Saad, José Carlos Santos, YiFan Apr 5 at 23:39


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Adrian Keister, Saad, José Carlos Santos, YiFan
If this question can be reworded to fit the rules in the help center, please edit the question.











  • 2




    $begingroup$
    What do you want to do with other examples?
    $endgroup$
    – Dietrich Burde
    Apr 4 at 9:50






  • 6




    $begingroup$
    For $k le 10000$, there are $185$ values of $k$ where $p = 108k^3-1$ and $p+2$ are twin primes that sum to a cube. Following are some examples.. $107,2634011,29659499,57395627,104792291,271669247,485149499,cdots\ 83797007573051, colorred84537841287167, 88875400155587,89731703685707,89846291353499,cdots$
    $endgroup$
    – achille hui
    Apr 4 at 10:23






  • 1




    $begingroup$
    @taritgoswami it is a single line command using the CAS maxima for k : 1 thru 10000 do if (primep(108*k^3-1) and primep(108*k^3+1)) then print(k," ",108*k^3-1);
    $endgroup$
    – achille hui
    Apr 4 at 12:10






  • 3




    $begingroup$
    oeis.org/A061308 is worth a look.
    $endgroup$
    – Barry Cipra
    Apr 4 at 12:15






  • 2




    $begingroup$
    similarly oeis.org/A240169
    $endgroup$
    – KBusc
    Apr 4 at 16:00













6












6








6


2



$begingroup$


$84537841287167$ and $84537841287169$ are a pair of twin primes.



The sum of $84537841287167+84537841287169$ is a cube.



Are there other examples of twin pairs $p$, $p+2$ whose sum is a cube?



Have they to have a particular form?






elementary-number-theory







share|cite|improve this question











$endgroup$




$84537841287167$ and $84537841287169$ are a pair of twin primes.



The sum of $84537841287167+84537841287169$ is a cube.



Are there other examples of twin pairs $p$, $p+2$ whose sum is a cube?



Have they to have a particular form?






elementary-number-theory




elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 4 at 10:11









tarit goswami

2,2361422




2,2361422










asked Apr 4 at 9:29







user660792











closed as off-topic by user21820, Adrian Keister, Saad, José Carlos Santos, YiFan Apr 5 at 23:39


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Adrian Keister, Saad, José Carlos Santos, YiFan
If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by user21820, Adrian Keister, Saad, José Carlos Santos, YiFan Apr 5 at 23:39


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Adrian Keister, Saad, José Carlos Santos, YiFan
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 2




    $begingroup$
    What do you want to do with other examples?
    $endgroup$
    – Dietrich Burde
    Apr 4 at 9:50






  • 6




    $begingroup$
    For $k le 10000$, there are $185$ values of $k$ where $p = 108k^3-1$ and $p+2$ are twin primes that sum to a cube. Following are some examples.. $107,2634011,29659499,57395627,104792291,271669247,485149499,cdots\ 83797007573051, colorred84537841287167, 88875400155587,89731703685707,89846291353499,cdots$
    $endgroup$
    – achille hui
    Apr 4 at 10:23






  • 1




    $begingroup$
    @taritgoswami it is a single line command using the CAS maxima for k : 1 thru 10000 do if (primep(108*k^3-1) and primep(108*k^3+1)) then print(k," ",108*k^3-1);
    $endgroup$
    – achille hui
    Apr 4 at 12:10






  • 3




    $begingroup$
    oeis.org/A061308 is worth a look.
    $endgroup$
    – Barry Cipra
    Apr 4 at 12:15






  • 2




    $begingroup$
    similarly oeis.org/A240169
    $endgroup$
    – KBusc
    Apr 4 at 16:00












  • 2




    $begingroup$
    What do you want to do with other examples?
    $endgroup$
    – Dietrich Burde
    Apr 4 at 9:50






  • 6




    $begingroup$
    For $k le 10000$, there are $185$ values of $k$ where $p = 108k^3-1$ and $p+2$ are twin primes that sum to a cube. Following are some examples.. $107,2634011,29659499,57395627,104792291,271669247,485149499,cdots\ 83797007573051, colorred84537841287167, 88875400155587,89731703685707,89846291353499,cdots$
    $endgroup$
    – achille hui
    Apr 4 at 10:23






  • 1




    $begingroup$
    @taritgoswami it is a single line command using the CAS maxima for k : 1 thru 10000 do if (primep(108*k^3-1) and primep(108*k^3+1)) then print(k," ",108*k^3-1);
    $endgroup$
    – achille hui
    Apr 4 at 12:10






  • 3




    $begingroup$
    oeis.org/A061308 is worth a look.
    $endgroup$
    – Barry Cipra
    Apr 4 at 12:15






  • 2




    $begingroup$
    similarly oeis.org/A240169
    $endgroup$
    – KBusc
    Apr 4 at 16:00







2




2




$begingroup$
What do you want to do with other examples?
$endgroup$
– Dietrich Burde
Apr 4 at 9:50




$begingroup$
What do you want to do with other examples?
$endgroup$
– Dietrich Burde
Apr 4 at 9:50




6




6




$begingroup$
For $k le 10000$, there are $185$ values of $k$ where $p = 108k^3-1$ and $p+2$ are twin primes that sum to a cube. Following are some examples.. $107,2634011,29659499,57395627,104792291,271669247,485149499,cdots\ 83797007573051, colorred84537841287167, 88875400155587,89731703685707,89846291353499,cdots$
$endgroup$
– achille hui
Apr 4 at 10:23




$begingroup$
For $k le 10000$, there are $185$ values of $k$ where $p = 108k^3-1$ and $p+2$ are twin primes that sum to a cube. Following are some examples.. $107,2634011,29659499,57395627,104792291,271669247,485149499,cdots\ 83797007573051, colorred84537841287167, 88875400155587,89731703685707,89846291353499,cdots$
$endgroup$
– achille hui
Apr 4 at 10:23




1




1




$begingroup$
@taritgoswami it is a single line command using the CAS maxima for k : 1 thru 10000 do if (primep(108*k^3-1) and primep(108*k^3+1)) then print(k," ",108*k^3-1);
$endgroup$
– achille hui
Apr 4 at 12:10




$begingroup$
@taritgoswami it is a single line command using the CAS maxima for k : 1 thru 10000 do if (primep(108*k^3-1) and primep(108*k^3+1)) then print(k," ",108*k^3-1);
$endgroup$
– achille hui
Apr 4 at 12:10




3




3




$begingroup$
oeis.org/A061308 is worth a look.
$endgroup$
– Barry Cipra
Apr 4 at 12:15




$begingroup$
oeis.org/A061308 is worth a look.
$endgroup$
– Barry Cipra
Apr 4 at 12:15




2




2




$begingroup$
similarly oeis.org/A240169
$endgroup$
– KBusc
Apr 4 at 16:00




$begingroup$
similarly oeis.org/A240169
$endgroup$
– KBusc
Apr 4 at 16:00










5 Answers
5






active

oldest

votes


















21












$begingroup$

List of Some Twin Primes Greater than $bf3$ whose Sum is a Cube



The sum of twin primes $6k-1$ and $6k+1$ is $12k$ and $12k=j^3implies6mid j$. So to look for such twin primes, look for twin primes of the form
$$
(108n^3-1,108n^3+1)
$$
We can skip $nequiv2pmod5$ since
$$
beginalign
108cdot n^3+1
&equiv3cdot2^3+1&pmod5\
&equiv0&pmod5
endalign
$$
and we can skip $nequiv3pmod5$ since
$$
beginalign
108cdot n^3-1
&equiv3cdot3^3-1&pmod5\
&equiv0&pmod5
endalign
$$
Checking $nequiv0,1,4pmod5$, we get
$$
beginarrayl
n&108n^3-1&108n^3+1\hline
1&107&109\
29&2634011&2634013\
65&29659499&29659501\
81&57395627&57395629\
99&104792291&104792293\
136&271669247&271669249\
165&485149499&485149501\
174&568946591&568946593\
176&588791807&588791809\
191&752530067&752530069\
200&863999999&864000001\
266&2032678367&2032678369\
295&2772616499&2772616501\
301&2945257307&2945257309\
319&3505869971&3505869973\
346&4473547487&4473547489\
351&4670303507&4670303509\
370&5470523999&5470524001\
400&6911999999&6912000001\
411&7498065347&7498065349\
431&8646803027&8646803029\
434&8828622431&8828622433\
436&8951240447&8951240449\
456&10240432127&10240432129\
491&12784043267&12784043269\
494&13019808671&13019808673\
526&15717410207&15717410209\
541&17100765467&17100765469\
599&23211554291&23211554293\
651&29796600707&29796600709\
676&33362903807&33362903809\
714&39311389151&39311389153\
746&44837381087&44837381089\
790&53248211999&53248212001\
924&85200014591&85200014593\
956&94362064127&94362064129\
991&105110165267&105110165269\
endarray
$$
Note that the sum of the primes is $(6n)^3$.



This list does not include $3+5=2^3$ because $(3,5)$ is the only twin prime that is not of the form $(6k-1,6k+1)$; that is, all primes greater than $3$ must be relatively prime to $6$.



Mathematica Code



Module[s = "", m, 
 For[n = 1, n < 1000, ++n, 
 If[MemberQ[0, 1, 4, Mod[n, 5]] && PrimeQ[(m = 108 n^3) - 1] && 
 PrimeQ[m + 1], 
 s = StringJoin[s, 
 ToString[n] <> "&" <> ToString[m - 1] <> "&" <> ToString[m + 1] <>
 "\n"]]]; s]





share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    "We can immediately skip $nequiv2,3pmod5$ since $5mid108cdot2^3+1$ and $5mid108cdot3^3-1$" I don't understand this line, could you explain this a little further?
    $endgroup$
    – spyr03
    Apr 4 at 15:46






  • 3




    $begingroup$
    As $5$ divides $108cdot 2^3+1$, for any $k$ of the form $5t+2$ we have $108cdot (5t+2)^3+ 1$. Now using binomial expansion you can see that it will left $5cdot m+108cdot 2^3$, which is divisible by $5$. Hence it can't be prime. So we are not checking for that $n$. Same argument goes for $nequiv 3pmod5$.
    $endgroup$
    – OppoInfinity
    Apr 4 at 15:54










  • $begingroup$
    @spyr03: I have expanded the section about the values skipped
    $endgroup$
    – robjohn♦
    Apr 4 at 17:27



















13












$begingroup$

$3$ and $5$ add to $2^3$, and $107 + 109 = 6^3$. So yes, there are other examples.



Clearly the sum has to be even, so if you want to systematically look for examples, then you could look through numbers of the form
$$
frac(2n)^32pm 1
$$
and check whether they are twin primes. That's how I found the $6^3$ one, at least. I'm sure there are clever ways to cut down the search space further. Or maybe it's better to look through all twin primes and see whether they add to cubes. I have no idea which would actually be faster.



Edit Apart from $3+5$, the primes cannot be divisible by $3$ (obviously), and since neither $p$ nor $p+2$ are divisible by $3$, that means their sum is divisible by $3$. So instead of just $frac(2n)^32pm1$, you can narrow it to $frac(6n)^32pm 1$.



Once can do a similar analysis on the basis of, say, $5$, and find that $nnotequiv 2, 3pmod 5$, meaning the numerator must be on one of the forms $(30n)^3, (30n + 6)^3$ or $(30n + 24)^3$.



New edit:



I used my above idea to search with the following Python program:



import math 

def is_prime(n):
 if(n%2==0):
 return False
 for i in range(3,int(math.sqrt(n))+1,2):
 if(n%i==0):
 return False
 return True

for i in range(1,300):
 for k in [30*i, 30*i + 6, 30*i + 24]: 
 if is_prime(k**3/2 - 1) & is_prime(k**3/2 + 1):
 print "%d & %d & %d\\" % (k**3/2 - 1, k**3/2 + 1, k)


(The speed of the program starts to go down at about 300.)



It found the following list:
$$
beginarray
hline
p&p+2&sqrt[3]textsum\
hline
2634011 & 2634013 & 174\
29659499 & 29659501 & 390\
57395627 & 57395629 & 486\
104792291 & 104792293 & 594\
271669247 & 271669249 & 816\
485149499 & 485149501 & 990\
568946591 & 568946593 & 1044\
588791807 & 588791809 & 1056\
752530067 & 752530069 & 1146\
863999999 & 864000001 & 1200\
2032678367 & 2032678369 & 1596\
2772616499 & 2772616501 & 1770\
2945257307 & 2945257309 & 1806\
3505869971 & 3505869973 & 1914\
4473547487 & 4473547489 & 2076\
4670303507 & 4670303509 & 2106\
5470523999 & 5470524001 & 2220\
6911999999 & 6912000001 & 2400\
7498065347 & 7498065349 & 2466\
8646803027 & 8646803029 & 2586\
8828622431 & 8828622433 & 2604\
8951240447 & 8951240449 & 2616\
10240432127 & 10240432129 & 2736\
12784043267 & 12784043269 & 2946\
13019808671 & 13019808673 & 2964\
15717410207 & 15717410209 & 3156\
17100765467 & 17100765469 & 3246\
23211554291 & 23211554293 & 3594\
29796600707 & 29796600709 & 3906\
33362903807 & 33362903809 & 4056\
39311389151 & 39311389153 & 4284\
44837381087 & 44837381089 & 4476\
53248211999 & 53248212001 & 4740\
85200014591 & 85200014593 & 5544\
94362064127 & 94362064129 & 5736\
105110165267 & 105110165269 & 5946\
111603347747 & 111603347749 & 6066\
156246957827 & 156246957829 & 6786\
169013118347 & 169013118349 & 6966\
183838613471 & 183838613473 & 7164\
215526633731 & 215526633733 & 7554\
223322272991 & 223322272993 & 7644\
226492415999 & 226492416001 & 7680\
239472986111 & 239472986113 & 7824\
280145695391 & 280145695393 & 8244\
hline
endarray
$$
This skips $2^3 = 3+5$ and $6^3 = 107+109$, and $24^3 = 6911 + 6913$ (although $6913 = 31cdot 223$, so that's irrelevant), because of some hiccup with the range in the primality test that I didn't bother fixing. Thanks to @taritgoswami for the prime test I copy-pasted from him and trimmed down a little. It still thinks $2$ is non-prime, but that's OK in this case. In fact, I could theoretically have taken away the whole parity check there, since I know I'm only feeding it odd numbers.



The print statement is formatted like that so that I could copy-paste it directly from the Python output into a MathJaX array without having to manually type all the formatting necessary.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    some bigger one
    $endgroup$
    – user660792
    Apr 4 at 9:32










  • $begingroup$
    @DietrichBurde It's a cube, see the question body.
    $endgroup$
    – Arthur
    Apr 4 at 9:34










  • $begingroup$
    @DietrichBurde I'm sure we can. $17+19 = 6^2$, so $6$ is really prolific here.
    $endgroup$
    – Arthur
    Apr 4 at 9:37






  • 1




    $begingroup$
    with Maple can you find other examples?
    $endgroup$
    – user660792
    Apr 4 at 9:41










  • $begingroup$
    @pantareis I'm sure I could. If I knew how to use Maple. Or you could do it: I've already told you one possible algorithm.
    $endgroup$
    – Arthur
    Apr 4 at 9:42



















5












$begingroup$

To demonstrate that there are also huge solutions :



Define $$k=10^100+303593$$ $$s=4k^3-1$$ $$t=4k^3+1$$ then $(s,t)$ is a twin prime pair of the desired form which can be searched with this PARI/GP - routine 



? z=prod(j=1,3*10^4,prime(j));k=10^100-1;gef=0;while(gef==0,k=k+1;s=4*k^3-1;t=4*
k^3+1;if(gcd(s*t,z)==1,if(ispseudoprime(s)==1,print(k-10^100);if(ispseudoprime(t)
==1,gef=1))))


$ s $and $ t $ are proven primes with $ 301 $ digits. Assuming the generalized bunyakovsky conjecture, there are infinite many pairs of the desired form.






share|cite|improve this answer









$endgroup$




















    4












    $begingroup$

    $(3,5),(107,109)$ and $(2634011,2634013)$ are only such twin prime pairs below $10^7$. You can check for more by increasing the range using this small Python code below. I just run it upto $10^8$, $(29659499,29659501)$ and $(57395627,57395629)$ are only such pairs with $10^7<p<10^8$.



    Edit:
    For optimization purpose, I will use the fact that all primes more than $3$ can be represented in the form $6kpm 1$.



    First let check manually for $p=2,3$. For $p=2$ it's clearly not possible. For $p=3$, we have $3+5=8=2^3$, so, $(3,5)$ is such pair.



    Suppose, $p=6k-1$, with $p>5$, then we have $12k=n^3$. That means, $12|n^3$. But, as $n^3$ is a perfect cube, and we have $12=3cdot 2^2$, at least $3^3cdot 2^3=216$ will divide $n^3$. So, the pair $(108m^3-1,108m^3+1)$ will be such pair if both of them are prime.



    While dealing with cubes, usually prefer to work in $mathbbZ_7$, as any cube is either of $0,1,6$ in this field,i.e; $n^3equiv 0,1,6pmod7$ . So, we can have $2p+2equiv 0pmod7$, which gives $pequiv 6pmod7$, or, $2p+2equiv 1pmod7$ which implies $pequiv 3pmod7$ or $2p+2equiv 6pmod7$ which implies $pequiv 2pmod7$. Hence, only $3$ pssibilities. Here is the updated program: 



    import math
    import time

    start_time=time.time()
    def is_prime(n):
     flag=0
     if(n==2):
     return True
     if(n%2==0):
     return False
     else:
     for i in range(3,int(math.sqrt(n))+1,2):
     if(n%i==0):
     flag=1
     break
     if(flag==0):
     return True
     return False 

    for i in range(1,1000):#change the number inside this braket to check for larger numbers
     c=(6*i)**3 
     p=c//2-1 
     if(p%7==2 or p%7==4 or p%7==6):
     if(is_prime(p) & is_prime(p+2)):
     print(p, "is such twin prime with sum",c) 

    print(time.time()-start_time)





    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      I am not the downvoter, but my guess is that it is because your script, the substance of your answer, is quite inefficient compared to some others that were posted.
      $endgroup$
      – Mees de Vries
      Apr 4 at 14:32










    • $begingroup$
      @MeesdeVries Updated it, please have a look.
      $endgroup$
      – tarit goswami
      Apr 4 at 19:20


















    1












    $begingroup$

    $p + (p + 2) = 2p + 2 = 2(p + 1)$



    For the sum to be a cube, $p + 1$ must be divisible by 4, so that the sum becomes $2 times 4m^3$ for some integer $m$. So $p + 1$ is of the form $4m^3$.






    share|cite|improve this answer









    $endgroup$








    • 2




      $begingroup$
      Also, with the exception of the twin prime pair 3,5, every pair of twin primes must be of the form 6n-1,6n+1. Using your notation, you can further refine based on this to $p+1=108k^3$
      $endgroup$
      – Moko19
      Apr 4 at 9:52
















    5 Answers
    5






    active

    oldest

    votes








    5 Answers
    5






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    21












    $begingroup$

    List of Some Twin Primes Greater than $bf3$ whose Sum is a Cube



    The sum of twin primes $6k-1$ and $6k+1$ is $12k$ and $12k=j^3implies6mid j$. So to look for such twin primes, look for twin primes of the form
    $$
    (108n^3-1,108n^3+1)
    $$
    We can skip $nequiv2pmod5$ since
    $$
    beginalign
    108cdot n^3+1
    &equiv3cdot2^3+1&pmod5\
    &equiv0&pmod5
    endalign
    $$
    and we can skip $nequiv3pmod5$ since
    $$
    beginalign
    108cdot n^3-1
    &equiv3cdot3^3-1&pmod5\
    &equiv0&pmod5
    endalign
    $$
    Checking $nequiv0,1,4pmod5$, we get
    $$
    beginarrayl
    n&108n^3-1&108n^3+1\hline
    1&107&109\
    29&2634011&2634013\
    65&29659499&29659501\
    81&57395627&57395629\
    99&104792291&104792293\
    136&271669247&271669249\
    165&485149499&485149501\
    174&568946591&568946593\
    176&588791807&588791809\
    191&752530067&752530069\
    200&863999999&864000001\
    266&2032678367&2032678369\
    295&2772616499&2772616501\
    301&2945257307&2945257309\
    319&3505869971&3505869973\
    346&4473547487&4473547489\
    351&4670303507&4670303509\
    370&5470523999&5470524001\
    400&6911999999&6912000001\
    411&7498065347&7498065349\
    431&8646803027&8646803029\
    434&8828622431&8828622433\
    436&8951240447&8951240449\
    456&10240432127&10240432129\
    491&12784043267&12784043269\
    494&13019808671&13019808673\
    526&15717410207&15717410209\
    541&17100765467&17100765469\
    599&23211554291&23211554293\
    651&29796600707&29796600709\
    676&33362903807&33362903809\
    714&39311389151&39311389153\
    746&44837381087&44837381089\
    790&53248211999&53248212001\
    924&85200014591&85200014593\
    956&94362064127&94362064129\
    991&105110165267&105110165269\
    endarray
    $$
    Note that the sum of the primes is $(6n)^3$.



    This list does not include $3+5=2^3$ because $(3,5)$ is the only twin prime that is not of the form $(6k-1,6k+1)$; that is, all primes greater than $3$ must be relatively prime to $6$.



    Mathematica Code



    Module[s = "", m, 
     For[n = 1, n < 1000, ++n, 
     If[MemberQ[0, 1, 4, Mod[n, 5]] && PrimeQ[(m = 108 n^3) - 1] && 
     PrimeQ[m + 1], 
     s = StringJoin[s, 
     ToString[n] <> "&" <> ToString[m - 1] <> "&" <> ToString[m + 1] <>
     "\n"]]]; s]





    share|cite|improve this answer











    $endgroup$








    • 1




      $begingroup$
      "We can immediately skip $nequiv2,3pmod5$ since $5mid108cdot2^3+1$ and $5mid108cdot3^3-1$" I don't understand this line, could you explain this a little further?
      $endgroup$
      – spyr03
      Apr 4 at 15:46






    • 3




      $begingroup$
      As $5$ divides $108cdot 2^3+1$, for any $k$ of the form $5t+2$ we have $108cdot (5t+2)^3+ 1$. Now using binomial expansion you can see that it will left $5cdot m+108cdot 2^3$, which is divisible by $5$. Hence it can't be prime. So we are not checking for that $n$. Same argument goes for $nequiv 3pmod5$.
      $endgroup$
      – OppoInfinity
      Apr 4 at 15:54










    • $begingroup$
      @spyr03: I have expanded the section about the values skipped
      $endgroup$
      – robjohn♦
      Apr 4 at 17:27
















    21












    $begingroup$

    List of Some Twin Primes Greater than $bf3$ whose Sum is a Cube



    The sum of twin primes $6k-1$ and $6k+1$ is $12k$ and $12k=j^3implies6mid j$. So to look for such twin primes, look for twin primes of the form
    $$
    (108n^3-1,108n^3+1)
    $$
    We can skip $nequiv2pmod5$ since
    $$
    beginalign
    108cdot n^3+1
    &equiv3cdot2^3+1&pmod5\
    &equiv0&pmod5
    endalign
    $$
    and we can skip $nequiv3pmod5$ since
    $$
    beginalign
    108cdot n^3-1
    &equiv3cdot3^3-1&pmod5\
    &equiv0&pmod5
    endalign
    $$
    Checking $nequiv0,1,4pmod5$, we get
    $$
    beginarrayl
    n&108n^3-1&108n^3+1\hline
    1&107&109\
    29&2634011&2634013\
    65&29659499&29659501\
    81&57395627&57395629\
    99&104792291&104792293\
    136&271669247&271669249\
    165&485149499&485149501\
    174&568946591&568946593\
    176&588791807&588791809\
    191&752530067&752530069\
    200&863999999&864000001\
    266&2032678367&2032678369\
    295&2772616499&2772616501\
    301&2945257307&2945257309\
    319&3505869971&3505869973\
    346&4473547487&4473547489\
    351&4670303507&4670303509\
    370&5470523999&5470524001\
    400&6911999999&6912000001\
    411&7498065347&7498065349\
    431&8646803027&8646803029\
    434&8828622431&8828622433\
    436&8951240447&8951240449\
    456&10240432127&10240432129\
    491&12784043267&12784043269\
    494&13019808671&13019808673\
    526&15717410207&15717410209\
    541&17100765467&17100765469\
    599&23211554291&23211554293\
    651&29796600707&29796600709\
    676&33362903807&33362903809\
    714&39311389151&39311389153\
    746&44837381087&44837381089\
    790&53248211999&53248212001\
    924&85200014591&85200014593\
    956&94362064127&94362064129\
    991&105110165267&105110165269\
    endarray
    $$
    Note that the sum of the primes is $(6n)^3$.



    This list does not include $3+5=2^3$ because $(3,5)$ is the only twin prime that is not of the form $(6k-1,6k+1)$; that is, all primes greater than $3$ must be relatively prime to $6$.



    Mathematica Code



    Module[s = "", m, 
     For[n = 1, n < 1000, ++n, 
     If[MemberQ[0, 1, 4, Mod[n, 5]] && PrimeQ[(m = 108 n^3) - 1] && 
     PrimeQ[m + 1], 
     s = StringJoin[s, 
     ToString[n] <> "&" <> ToString[m - 1] <> "&" <> ToString[m + 1] <>
     "\n"]]]; s]





    share|cite|improve this answer











    $endgroup$








    • 1




      $begingroup$
      "We can immediately skip $nequiv2,3pmod5$ since $5mid108cdot2^3+1$ and $5mid108cdot3^3-1$" I don't understand this line, could you explain this a little further?
      $endgroup$
      – spyr03
      Apr 4 at 15:46






    • 3




      $begingroup$
      As $5$ divides $108cdot 2^3+1$, for any $k$ of the form $5t+2$ we have $108cdot (5t+2)^3+ 1$. Now using binomial expansion you can see that it will left $5cdot m+108cdot 2^3$, which is divisible by $5$. Hence it can't be prime. So we are not checking for that $n$. Same argument goes for $nequiv 3pmod5$.
      $endgroup$
      – OppoInfinity
      Apr 4 at 15:54










    • $begingroup$
      @spyr03: I have expanded the section about the values skipped
      $endgroup$
      – robjohn♦
      Apr 4 at 17:27














    21












    21








    21





    $begingroup$

    List of Some Twin Primes Greater than $bf3$ whose Sum is a Cube



    The sum of twin primes $6k-1$ and $6k+1$ is $12k$ and $12k=j^3implies6mid j$. So to look for such twin primes, look for twin primes of the form
    $$
    (108n^3-1,108n^3+1)
    $$
    We can skip $nequiv2pmod5$ since
    $$
    beginalign
    108cdot n^3+1
    &equiv3cdot2^3+1&pmod5\
    &equiv0&pmod5
    endalign
    $$
    and we can skip $nequiv3pmod5$ since
    $$
    beginalign
    108cdot n^3-1
    &equiv3cdot3^3-1&pmod5\
    &equiv0&pmod5
    endalign
    $$
    Checking $nequiv0,1,4pmod5$, we get
    $$
    beginarrayl
    n&108n^3-1&108n^3+1\hline
    1&107&109\
    29&2634011&2634013\
    65&29659499&29659501\
    81&57395627&57395629\
    99&104792291&104792293\
    136&271669247&271669249\
    165&485149499&485149501\
    174&568946591&568946593\
    176&588791807&588791809\
    191&752530067&752530069\
    200&863999999&864000001\
    266&2032678367&2032678369\
    295&2772616499&2772616501\
    301&2945257307&2945257309\
    319&3505869971&3505869973\
    346&4473547487&4473547489\
    351&4670303507&4670303509\
    370&5470523999&5470524001\
    400&6911999999&6912000001\
    411&7498065347&7498065349\
    431&8646803027&8646803029\
    434&8828622431&8828622433\
    436&8951240447&8951240449\
    456&10240432127&10240432129\
    491&12784043267&12784043269\
    494&13019808671&13019808673\
    526&15717410207&15717410209\
    541&17100765467&17100765469\
    599&23211554291&23211554293\
    651&29796600707&29796600709\
    676&33362903807&33362903809\
    714&39311389151&39311389153\
    746&44837381087&44837381089\
    790&53248211999&53248212001\
    924&85200014591&85200014593\
    956&94362064127&94362064129\
    991&105110165267&105110165269\
    endarray
    $$
    Note that the sum of the primes is $(6n)^3$.



    This list does not include $3+5=2^3$ because $(3,5)$ is the only twin prime that is not of the form $(6k-1,6k+1)$; that is, all primes greater than $3$ must be relatively prime to $6$.



    Mathematica Code



    Module[s = "", m, 
     For[n = 1, n < 1000, ++n, 
     If[MemberQ[0, 1, 4, Mod[n, 5]] && PrimeQ[(m = 108 n^3) - 1] && 
     PrimeQ[m + 1], 
     s = StringJoin[s, 
     ToString[n] <> "&" <> ToString[m - 1] <> "&" <> ToString[m + 1] <>
     "\n"]]]; s]





    share|cite|improve this answer











    $endgroup$



    List of Some Twin Primes Greater than $bf3$ whose Sum is a Cube



    The sum of twin primes $6k-1$ and $6k+1$ is $12k$ and $12k=j^3implies6mid j$. So to look for such twin primes, look for twin primes of the form
    $$
    (108n^3-1,108n^3+1)
    $$
    We can skip $nequiv2pmod5$ since
    $$
    beginalign
    108cdot n^3+1
    &equiv3cdot2^3+1&pmod5\
    &equiv0&pmod5
    endalign
    $$
    and we can skip $nequiv3pmod5$ since
    $$
    beginalign
    108cdot n^3-1
    &equiv3cdot3^3-1&pmod5\
    &equiv0&pmod5
    endalign
    $$
    Checking $nequiv0,1,4pmod5$, we get
    $$
    beginarrayl
    n&108n^3-1&108n^3+1\hline
    1&107&109\
    29&2634011&2634013\
    65&29659499&29659501\
    81&57395627&57395629\
    99&104792291&104792293\
    136&271669247&271669249\
    165&485149499&485149501\
    174&568946591&568946593\
    176&588791807&588791809\
    191&752530067&752530069\
    200&863999999&864000001\
    266&2032678367&2032678369\
    295&2772616499&2772616501\
    301&2945257307&2945257309\
    319&3505869971&3505869973\
    346&4473547487&4473547489\
    351&4670303507&4670303509\
    370&5470523999&5470524001\
    400&6911999999&6912000001\
    411&7498065347&7498065349\
    431&8646803027&8646803029\
    434&8828622431&8828622433\
    436&8951240447&8951240449\
    456&10240432127&10240432129\
    491&12784043267&12784043269\
    494&13019808671&13019808673\
    526&15717410207&15717410209\
    541&17100765467&17100765469\
    599&23211554291&23211554293\
    651&29796600707&29796600709\
    676&33362903807&33362903809\
    714&39311389151&39311389153\
    746&44837381087&44837381089\
    790&53248211999&53248212001\
    924&85200014591&85200014593\
    956&94362064127&94362064129\
    991&105110165267&105110165269\
    endarray
    $$
    Note that the sum of the primes is $(6n)^3$.



    This list does not include $3+5=2^3$ because $(3,5)$ is the only twin prime that is not of the form $(6k-1,6k+1)$; that is, all primes greater than $3$ must be relatively prime to $6$.



    Mathematica Code



    Module[s = "", m, 
     For[n = 1, n < 1000, ++n, 
     If[MemberQ[0, 1, 4, Mod[n, 5]] && PrimeQ[(m = 108 n^3) - 1] && 
     PrimeQ[m + 1], 
     s = StringJoin[s, 
     ToString[n] <> "&" <> ToString[m - 1] <> "&" <> ToString[m + 1] <>
     "\n"]]]; s]






    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Apr 5 at 9:45









    Community♦

    1




    1










    answered Apr 4 at 12:20









    robjohn♦robjohn

    271k27315643




    271k27315643







    • 1




      $begingroup$
      "We can immediately skip $nequiv2,3pmod5$ since $5mid108cdot2^3+1$ and $5mid108cdot3^3-1$" I don't understand this line, could you explain this a little further?
      $endgroup$
      – spyr03
      Apr 4 at 15:46






    • 3




      $begingroup$
      As $5$ divides $108cdot 2^3+1$, for any $k$ of the form $5t+2$ we have $108cdot (5t+2)^3+ 1$. Now using binomial expansion you can see that it will left $5cdot m+108cdot 2^3$, which is divisible by $5$. Hence it can't be prime. So we are not checking for that $n$. Same argument goes for $nequiv 3pmod5$.
      $endgroup$
      – OppoInfinity
      Apr 4 at 15:54










    • $begingroup$
      @spyr03: I have expanded the section about the values skipped
      $endgroup$
      – robjohn♦
      Apr 4 at 17:27













    • 1




      $begingroup$
      "We can immediately skip $nequiv2,3pmod5$ since $5mid108cdot2^3+1$ and $5mid108cdot3^3-1$" I don't understand this line, could you explain this a little further?
      $endgroup$
      – spyr03
      Apr 4 at 15:46






    • 3




      $begingroup$
      As $5$ divides $108cdot 2^3+1$, for any $k$ of the form $5t+2$ we have $108cdot (5t+2)^3+ 1$. Now using binomial expansion you can see that it will left $5cdot m+108cdot 2^3$, which is divisible by $5$. Hence it can't be prime. So we are not checking for that $n$. Same argument goes for $nequiv 3pmod5$.
      $endgroup$
      – OppoInfinity
      Apr 4 at 15:54










    • $begingroup$
      @spyr03: I have expanded the section about the values skipped
      $endgroup$
      – robjohn♦
      Apr 4 at 17:27








    1




    1




    $begingroup$
    "We can immediately skip $nequiv2,3pmod5$ since $5mid108cdot2^3+1$ and $5mid108cdot3^3-1$" I don't understand this line, could you explain this a little further?
    $endgroup$
    – spyr03
    Apr 4 at 15:46




    $begingroup$
    "We can immediately skip $nequiv2,3pmod5$ since $5mid108cdot2^3+1$ and $5mid108cdot3^3-1$" I don't understand this line, could you explain this a little further?
    $endgroup$
    – spyr03
    Apr 4 at 15:46




    3




    3




    $begingroup$
    As $5$ divides $108cdot 2^3+1$, for any $k$ of the form $5t+2$ we have $108cdot (5t+2)^3+ 1$. Now using binomial expansion you can see that it will left $5cdot m+108cdot 2^3$, which is divisible by $5$. Hence it can't be prime. So we are not checking for that $n$. Same argument goes for $nequiv 3pmod5$.
    $endgroup$
    – OppoInfinity
    Apr 4 at 15:54




    $begingroup$
    As $5$ divides $108cdot 2^3+1$, for any $k$ of the form $5t+2$ we have $108cdot (5t+2)^3+ 1$. Now using binomial expansion you can see that it will left $5cdot m+108cdot 2^3$, which is divisible by $5$. Hence it can't be prime. So we are not checking for that $n$. Same argument goes for $nequiv 3pmod5$.
    $endgroup$
    – OppoInfinity
    Apr 4 at 15:54












    $begingroup$
    @spyr03: I have expanded the section about the values skipped
    $endgroup$
    – robjohn♦
    Apr 4 at 17:27





    $begingroup$
    @spyr03: I have expanded the section about the values skipped
    $endgroup$
    – robjohn♦
    Apr 4 at 17:27












    13












    $begingroup$

    $3$ and $5$ add to $2^3$, and $107 + 109 = 6^3$. So yes, there are other examples.



    Clearly the sum has to be even, so if you want to systematically look for examples, then you could look through numbers of the form
    $$
    frac(2n)^32pm 1
    $$
    and check whether they are twin primes. That's how I found the $6^3$ one, at least. I'm sure there are clever ways to cut down the search space further. Or maybe it's better to look through all twin primes and see whether they add to cubes. I have no idea which would actually be faster.



    Edit Apart from $3+5$, the primes cannot be divisible by $3$ (obviously), and since neither $p$ nor $p+2$ are divisible by $3$, that means their sum is divisible by $3$. So instead of just $frac(2n)^32pm1$, you can narrow it to $frac(6n)^32pm 1$.



    Once can do a similar analysis on the basis of, say, $5$, and find that $nnotequiv 2, 3pmod 5$, meaning the numerator must be on one of the forms $(30n)^3, (30n + 6)^3$ or $(30n + 24)^3$.



    New edit:



    I used my above idea to search with the following Python program:



    import math 

    def is_prime(n):
     if(n%2==0):
     return False
     for i in range(3,int(math.sqrt(n))+1,2):
     if(n%i==0):
     return False
     return True

    for i in range(1,300):
     for k in [30*i, 30*i + 6, 30*i + 24]: 
     if is_prime(k**3/2 - 1) & is_prime(k**3/2 + 1):
     print "%d & %d & %d\\" % (k**3/2 - 1, k**3/2 + 1, k)


    (The speed of the program starts to go down at about 300.)



    It found the following list:
    $$
    beginarray
    hline
    p&p+2&sqrt[3]textsum\
    hline
    2634011 & 2634013 & 174\
    29659499 & 29659501 & 390\
    57395627 & 57395629 & 486\
    104792291 & 104792293 & 594\
    271669247 & 271669249 & 816\
    485149499 & 485149501 & 990\
    568946591 & 568946593 & 1044\
    588791807 & 588791809 & 1056\
    752530067 & 752530069 & 1146\
    863999999 & 864000001 & 1200\
    2032678367 & 2032678369 & 1596\
    2772616499 & 2772616501 & 1770\
    2945257307 & 2945257309 & 1806\
    3505869971 & 3505869973 & 1914\
    4473547487 & 4473547489 & 2076\
    4670303507 & 4670303509 & 2106\
    5470523999 & 5470524001 & 2220\
    6911999999 & 6912000001 & 2400\
    7498065347 & 7498065349 & 2466\
    8646803027 & 8646803029 & 2586\
    8828622431 & 8828622433 & 2604\
    8951240447 & 8951240449 & 2616\
    10240432127 & 10240432129 & 2736\
    12784043267 & 12784043269 & 2946\
    13019808671 & 13019808673 & 2964\
    15717410207 & 15717410209 & 3156\
    17100765467 & 17100765469 & 3246\
    23211554291 & 23211554293 & 3594\
    29796600707 & 29796600709 & 3906\
    33362903807 & 33362903809 & 4056\
    39311389151 & 39311389153 & 4284\
    44837381087 & 44837381089 & 4476\
    53248211999 & 53248212001 & 4740\
    85200014591 & 85200014593 & 5544\
    94362064127 & 94362064129 & 5736\
    105110165267 & 105110165269 & 5946\
    111603347747 & 111603347749 & 6066\
    156246957827 & 156246957829 & 6786\
    169013118347 & 169013118349 & 6966\
    183838613471 & 183838613473 & 7164\
    215526633731 & 215526633733 & 7554\
    223322272991 & 223322272993 & 7644\
    226492415999 & 226492416001 & 7680\
    239472986111 & 239472986113 & 7824\
    280145695391 & 280145695393 & 8244\
    hline
    endarray
    $$
    This skips $2^3 = 3+5$ and $6^3 = 107+109$, and $24^3 = 6911 + 6913$ (although $6913 = 31cdot 223$, so that's irrelevant), because of some hiccup with the range in the primality test that I didn't bother fixing. Thanks to @taritgoswami for the prime test I copy-pasted from him and trimmed down a little. It still thinks $2$ is non-prime, but that's OK in this case. In fact, I could theoretically have taken away the whole parity check there, since I know I'm only feeding it odd numbers.



    The print statement is formatted like that so that I could copy-paste it directly from the Python output into a MathJaX array without having to manually type all the formatting necessary.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      some bigger one
      $endgroup$
      – user660792
      Apr 4 at 9:32










    • $begingroup$
      @DietrichBurde It's a cube, see the question body.
      $endgroup$
      – Arthur
      Apr 4 at 9:34










    • $begingroup$
      @DietrichBurde I'm sure we can. $17+19 = 6^2$, so $6$ is really prolific here.
      $endgroup$
      – Arthur
      Apr 4 at 9:37






    • 1




      $begingroup$
      with Maple can you find other examples?
      $endgroup$
      – user660792
      Apr 4 at 9:41










    • $begingroup$
      @pantareis I'm sure I could. If I knew how to use Maple. Or you could do it: I've already told you one possible algorithm.
      $endgroup$
      – Arthur
      Apr 4 at 9:42
















    13












    $begingroup$

    $3$ and $5$ add to $2^3$, and $107 + 109 = 6^3$. So yes, there are other examples.



    Clearly the sum has to be even, so if you want to systematically look for examples, then you could look through numbers of the form
    $$
    frac(2n)^32pm 1
    $$
    and check whether they are twin primes. That's how I found the $6^3$ one, at least. I'm sure there are clever ways to cut down the search space further. Or maybe it's better to look through all twin primes and see whether they add to cubes. I have no idea which would actually be faster.



    Edit Apart from $3+5$, the primes cannot be divisible by $3$ (obviously), and since neither $p$ nor $p+2$ are divisible by $3$, that means their sum is divisible by $3$. So instead of just $frac(2n)^32pm1$, you can narrow it to $frac(6n)^32pm 1$.



    Once can do a similar analysis on the basis of, say, $5$, and find that $nnotequiv 2, 3pmod 5$, meaning the numerator must be on one of the forms $(30n)^3, (30n + 6)^3$ or $(30n + 24)^3$.



    New edit:



    I used my above idea to search with the following Python program:



    import math 

    def is_prime(n):
     if(n%2==0):
     return False
     for i in range(3,int(math.sqrt(n))+1,2):
     if(n%i==0):
     return False
     return True

    for i in range(1,300):
     for k in [30*i, 30*i + 6, 30*i + 24]: 
     if is_prime(k**3/2 - 1) & is_prime(k**3/2 + 1):
     print "%d & %d & %d\\" % (k**3/2 - 1, k**3/2 + 1, k)


    (The speed of the program starts to go down at about 300.)



    It found the following list:
    $$
    beginarray
    hline
    p&p+2&sqrt[3]textsum\
    hline
    2634011 & 2634013 & 174\
    29659499 & 29659501 & 390\
    57395627 & 57395629 & 486\
    104792291 & 104792293 & 594\
    271669247 & 271669249 & 816\
    485149499 & 485149501 & 990\
    568946591 & 568946593 & 1044\
    588791807 & 588791809 & 1056\
    752530067 & 752530069 & 1146\
    863999999 & 864000001 & 1200\
    2032678367 & 2032678369 & 1596\
    2772616499 & 2772616501 & 1770\
    2945257307 & 2945257309 & 1806\
    3505869971 & 3505869973 & 1914\
    4473547487 & 4473547489 & 2076\
    4670303507 & 4670303509 & 2106\
    5470523999 & 5470524001 & 2220\
    6911999999 & 6912000001 & 2400\
    7498065347 & 7498065349 & 2466\
    8646803027 & 8646803029 & 2586\
    8828622431 & 8828622433 & 2604\
    8951240447 & 8951240449 & 2616\
    10240432127 & 10240432129 & 2736\
    12784043267 & 12784043269 & 2946\
    13019808671 & 13019808673 & 2964\
    15717410207 & 15717410209 & 3156\
    17100765467 & 17100765469 & 3246\
    23211554291 & 23211554293 & 3594\
    29796600707 & 29796600709 & 3906\
    33362903807 & 33362903809 & 4056\
    39311389151 & 39311389153 & 4284\
    44837381087 & 44837381089 & 4476\
    53248211999 & 53248212001 & 4740\
    85200014591 & 85200014593 & 5544\
    94362064127 & 94362064129 & 5736\
    105110165267 & 105110165269 & 5946\
    111603347747 & 111603347749 & 6066\
    156246957827 & 156246957829 & 6786\
    169013118347 & 169013118349 & 6966\
    183838613471 & 183838613473 & 7164\
    215526633731 & 215526633733 & 7554\
    223322272991 & 223322272993 & 7644\
    226492415999 & 226492416001 & 7680\
    239472986111 & 239472986113 & 7824\
    280145695391 & 280145695393 & 8244\
    hline
    endarray
    $$
    This skips $2^3 = 3+5$ and $6^3 = 107+109$, and $24^3 = 6911 + 6913$ (although $6913 = 31cdot 223$, so that's irrelevant), because of some hiccup with the range in the primality test that I didn't bother fixing. Thanks to @taritgoswami for the prime test I copy-pasted from him and trimmed down a little. It still thinks $2$ is non-prime, but that's OK in this case. In fact, I could theoretically have taken away the whole parity check there, since I know I'm only feeding it odd numbers.



    The print statement is formatted like that so that I could copy-paste it directly from the Python output into a MathJaX array without having to manually type all the formatting necessary.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      some bigger one
      $endgroup$
      – user660792
      Apr 4 at 9:32










    • $begingroup$
      @DietrichBurde It's a cube, see the question body.
      $endgroup$
      – Arthur
      Apr 4 at 9:34










    • $begingroup$
      @DietrichBurde I'm sure we can. $17+19 = 6^2$, so $6$ is really prolific here.
      $endgroup$
      – Arthur
      Apr 4 at 9:37






    • 1




      $begingroup$
      with Maple can you find other examples?
      $endgroup$
      – user660792
      Apr 4 at 9:41










    • $begingroup$
      @pantareis I'm sure I could. If I knew how to use Maple. Or you could do it: I've already told you one possible algorithm.
      $endgroup$
      – Arthur
      Apr 4 at 9:42














    13












    13








    13





    $begingroup$

    $3$ and $5$ add to $2^3$, and $107 + 109 = 6^3$. So yes, there are other examples.



    Clearly the sum has to be even, so if you want to systematically look for examples, then you could look through numbers of the form
    $$
    frac(2n)^32pm 1
    $$
    and check whether they are twin primes. That's how I found the $6^3$ one, at least. I'm sure there are clever ways to cut down the search space further. Or maybe it's better to look through all twin primes and see whether they add to cubes. I have no idea which would actually be faster.



    Edit Apart from $3+5$, the primes cannot be divisible by $3$ (obviously), and since neither $p$ nor $p+2$ are divisible by $3$, that means their sum is divisible by $3$. So instead of just $frac(2n)^32pm1$, you can narrow it to $frac(6n)^32pm 1$.



    Once can do a similar analysis on the basis of, say, $5$, and find that $nnotequiv 2, 3pmod 5$, meaning the numerator must be on one of the forms $(30n)^3, (30n + 6)^3$ or $(30n + 24)^3$.



    New edit:



    I used my above idea to search with the following Python program:



    import math 

    def is_prime(n):
     if(n%2==0):
     return False
     for i in range(3,int(math.sqrt(n))+1,2):
     if(n%i==0):
     return False
     return True

    for i in range(1,300):
     for k in [30*i, 30*i + 6, 30*i + 24]: 
     if is_prime(k**3/2 - 1) & is_prime(k**3/2 + 1):
     print "%d & %d & %d\\" % (k**3/2 - 1, k**3/2 + 1, k)


    (The speed of the program starts to go down at about 300.)



    It found the following list:
    $$
    beginarray
    hline
    p&p+2&sqrt[3]textsum\
    hline
    2634011 & 2634013 & 174\
    29659499 & 29659501 & 390\
    57395627 & 57395629 & 486\
    104792291 & 104792293 & 594\
    271669247 & 271669249 & 816\
    485149499 & 485149501 & 990\
    568946591 & 568946593 & 1044\
    588791807 & 588791809 & 1056\
    752530067 & 752530069 & 1146\
    863999999 & 864000001 & 1200\
    2032678367 & 2032678369 & 1596\
    2772616499 & 2772616501 & 1770\
    2945257307 & 2945257309 & 1806\
    3505869971 & 3505869973 & 1914\
    4473547487 & 4473547489 & 2076\
    4670303507 & 4670303509 & 2106\
    5470523999 & 5470524001 & 2220\
    6911999999 & 6912000001 & 2400\
    7498065347 & 7498065349 & 2466\
    8646803027 & 8646803029 & 2586\
    8828622431 & 8828622433 & 2604\
    8951240447 & 8951240449 & 2616\
    10240432127 & 10240432129 & 2736\
    12784043267 & 12784043269 & 2946\
    13019808671 & 13019808673 & 2964\
    15717410207 & 15717410209 & 3156\
    17100765467 & 17100765469 & 3246\
    23211554291 & 23211554293 & 3594\
    29796600707 & 29796600709 & 3906\
    33362903807 & 33362903809 & 4056\
    39311389151 & 39311389153 & 4284\
    44837381087 & 44837381089 & 4476\
    53248211999 & 53248212001 & 4740\
    85200014591 & 85200014593 & 5544\
    94362064127 & 94362064129 & 5736\
    105110165267 & 105110165269 & 5946\
    111603347747 & 111603347749 & 6066\
    156246957827 & 156246957829 & 6786\
    169013118347 & 169013118349 & 6966\
    183838613471 & 183838613473 & 7164\
    215526633731 & 215526633733 & 7554\
    223322272991 & 223322272993 & 7644\
    226492415999 & 226492416001 & 7680\
    239472986111 & 239472986113 & 7824\
    280145695391 & 280145695393 & 8244\
    hline
    endarray
    $$
    This skips $2^3 = 3+5$ and $6^3 = 107+109$, and $24^3 = 6911 + 6913$ (although $6913 = 31cdot 223$, so that's irrelevant), because of some hiccup with the range in the primality test that I didn't bother fixing. Thanks to @taritgoswami for the prime test I copy-pasted from him and trimmed down a little. It still thinks $2$ is non-prime, but that's OK in this case. In fact, I could theoretically have taken away the whole parity check there, since I know I'm only feeding it odd numbers.



    The print statement is formatted like that so that I could copy-paste it directly from the Python output into a MathJaX array without having to manually type all the formatting necessary.






    share|cite|improve this answer











    $endgroup$



    $3$ and $5$ add to $2^3$, and $107 + 109 = 6^3$. So yes, there are other examples.



    Clearly the sum has to be even, so if you want to systematically look for examples, then you could look through numbers of the form
    $$
    frac(2n)^32pm 1
    $$
    and check whether they are twin primes. That's how I found the $6^3$ one, at least. I'm sure there are clever ways to cut down the search space further. Or maybe it's better to look through all twin primes and see whether they add to cubes. I have no idea which would actually be faster.



    Edit Apart from $3+5$, the primes cannot be divisible by $3$ (obviously), and since neither $p$ nor $p+2$ are divisible by $3$, that means their sum is divisible by $3$. So instead of just $frac(2n)^32pm1$, you can narrow it to $frac(6n)^32pm 1$.



    Once can do a similar analysis on the basis of, say, $5$, and find that $nnotequiv 2, 3pmod 5$, meaning the numerator must be on one of the forms $(30n)^3, (30n + 6)^3$ or $(30n + 24)^3$.



    New edit:



    I used my above idea to search with the following Python program:



    import math 

    def is_prime(n):
     if(n%2==0):
     return False
     for i in range(3,int(math.sqrt(n))+1,2):
     if(n%i==0):
     return False
     return True

    for i in range(1,300):
     for k in [30*i, 30*i + 6, 30*i + 24]: 
     if is_prime(k**3/2 - 1) & is_prime(k**3/2 + 1):
     print "%d & %d & %d\\" % (k**3/2 - 1, k**3/2 + 1, k)


    (The speed of the program starts to go down at about 300.)



    It found the following list:
    $$
    beginarray
    hline
    p&p+2&sqrt[3]textsum\
    hline
    2634011 & 2634013 & 174\
    29659499 & 29659501 & 390\
    57395627 & 57395629 & 486\
    104792291 & 104792293 & 594\
    271669247 & 271669249 & 816\
    485149499 & 485149501 & 990\
    568946591 & 568946593 & 1044\
    588791807 & 588791809 & 1056\
    752530067 & 752530069 & 1146\
    863999999 & 864000001 & 1200\
    2032678367 & 2032678369 & 1596\
    2772616499 & 2772616501 & 1770\
    2945257307 & 2945257309 & 1806\
    3505869971 & 3505869973 & 1914\
    4473547487 & 4473547489 & 2076\
    4670303507 & 4670303509 & 2106\
    5470523999 & 5470524001 & 2220\
    6911999999 & 6912000001 & 2400\
    7498065347 & 7498065349 & 2466\
    8646803027 & 8646803029 & 2586\
    8828622431 & 8828622433 & 2604\
    8951240447 & 8951240449 & 2616\
    10240432127 & 10240432129 & 2736\
    12784043267 & 12784043269 & 2946\
    13019808671 & 13019808673 & 2964\
    15717410207 & 15717410209 & 3156\
    17100765467 & 17100765469 & 3246\
    23211554291 & 23211554293 & 3594\
    29796600707 & 29796600709 & 3906\
    33362903807 & 33362903809 & 4056\
    39311389151 & 39311389153 & 4284\
    44837381087 & 44837381089 & 4476\
    53248211999 & 53248212001 & 4740\
    85200014591 & 85200014593 & 5544\
    94362064127 & 94362064129 & 5736\
    105110165267 & 105110165269 & 5946\
    111603347747 & 111603347749 & 6066\
    156246957827 & 156246957829 & 6786\
    169013118347 & 169013118349 & 6966\
    183838613471 & 183838613473 & 7164\
    215526633731 & 215526633733 & 7554\
    223322272991 & 223322272993 & 7644\
    226492415999 & 226492416001 & 7680\
    239472986111 & 239472986113 & 7824\
    280145695391 & 280145695393 & 8244\
    hline
    endarray
    $$
    This skips $2^3 = 3+5$ and $6^3 = 107+109$, and $24^3 = 6911 + 6913$ (although $6913 = 31cdot 223$, so that's irrelevant), because of some hiccup with the range in the primality test that I didn't bother fixing. Thanks to @taritgoswami for the prime test I copy-pasted from him and trimmed down a little. It still thinks $2$ is non-prime, but that's OK in this case. In fact, I could theoretically have taken away the whole parity check there, since I know I'm only feeding it odd numbers.



    The print statement is formatted like that so that I could copy-paste it directly from the Python output into a MathJaX array without having to manually type all the formatting necessary.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Apr 4 at 11:12

























    answered Apr 4 at 9:31









    ArthurArthur

    123k7122211




    123k7122211











    • $begingroup$
      some bigger one
      $endgroup$
      – user660792
      Apr 4 at 9:32










    • $begingroup$
      @DietrichBurde It's a cube, see the question body.
      $endgroup$
      – Arthur
      Apr 4 at 9:34










    • $begingroup$
      @DietrichBurde I'm sure we can. $17+19 = 6^2$, so $6$ is really prolific here.
      $endgroup$
      – Arthur
      Apr 4 at 9:37






    • 1




      $begingroup$
      with Maple can you find other examples?
      $endgroup$
      – user660792
      Apr 4 at 9:41










    • $begingroup$
      @pantareis I'm sure I could. If I knew how to use Maple. Or you could do it: I've already told you one possible algorithm.
      $endgroup$
      – Arthur
      Apr 4 at 9:42

















    • $begingroup$
      some bigger one
      $endgroup$
      – user660792
      Apr 4 at 9:32










    • $begingroup$
      @DietrichBurde It's a cube, see the question body.
      $endgroup$
      – Arthur
      Apr 4 at 9:34










    • $begingroup$
      @DietrichBurde I'm sure we can. $17+19 = 6^2$, so $6$ is really prolific here.
      $endgroup$
      – Arthur
      Apr 4 at 9:37






    • 1




      $begingroup$
      with Maple can you find other examples?
      $endgroup$
      – user660792
      Apr 4 at 9:41










    • $begingroup$
      @pantareis I'm sure I could. If I knew how to use Maple. Or you could do it: I've already told you one possible algorithm.
      $endgroup$
      – Arthur
      Apr 4 at 9:42
















    $begingroup$
    some bigger one
    $endgroup$
    – user660792
    Apr 4 at 9:32




    $begingroup$
    some bigger one
    $endgroup$
    – user660792
    Apr 4 at 9:32












    $begingroup$
    @DietrichBurde It's a cube, see the question body.
    $endgroup$
    – Arthur
    Apr 4 at 9:34




    $begingroup$
    @DietrichBurde It's a cube, see the question body.
    $endgroup$
    – Arthur
    Apr 4 at 9:34












    $begingroup$
    @DietrichBurde I'm sure we can. $17+19 = 6^2$, so $6$ is really prolific here.
    $endgroup$
    – Arthur
    Apr 4 at 9:37




    $begingroup$
    @DietrichBurde I'm sure we can. $17+19 = 6^2$, so $6$ is really prolific here.
    $endgroup$
    – Arthur
    Apr 4 at 9:37




    1




    1




    $begingroup$
    with Maple can you find other examples?
    $endgroup$
    – user660792
    Apr 4 at 9:41




    $begingroup$
    with Maple can you find other examples?
    $endgroup$
    – user660792
    Apr 4 at 9:41












    $begingroup$
    @pantareis I'm sure I could. If I knew how to use Maple. Or you could do it: I've already told you one possible algorithm.
    $endgroup$
    – Arthur
    Apr 4 at 9:42





    $begingroup$
    @pantareis I'm sure I could. If I knew how to use Maple. Or you could do it: I've already told you one possible algorithm.
    $endgroup$
    – Arthur
    Apr 4 at 9:42












    5












    $begingroup$

    To demonstrate that there are also huge solutions :



    Define $$k=10^100+303593$$ $$s=4k^3-1$$ $$t=4k^3+1$$ then $(s,t)$ is a twin prime pair of the desired form which can be searched with this PARI/GP - routine 



    ? z=prod(j=1,3*10^4,prime(j));k=10^100-1;gef=0;while(gef==0,k=k+1;s=4*k^3-1;t=4*
    k^3+1;if(gcd(s*t,z)==1,if(ispseudoprime(s)==1,print(k-10^100);if(ispseudoprime(t)
    ==1,gef=1))))


    $ s $and $ t $ are proven primes with $ 301 $ digits. Assuming the generalized bunyakovsky conjecture, there are infinite many pairs of the desired form.






    share|cite|improve this answer









    $endgroup$

















      5












      $begingroup$

      To demonstrate that there are also huge solutions :



      Define $$k=10^100+303593$$ $$s=4k^3-1$$ $$t=4k^3+1$$ then $(s,t)$ is a twin prime pair of the desired form which can be searched with this PARI/GP - routine 



      ? z=prod(j=1,3*10^4,prime(j));k=10^100-1;gef=0;while(gef==0,k=k+1;s=4*k^3-1;t=4*
      k^3+1;if(gcd(s*t,z)==1,if(ispseudoprime(s)==1,print(k-10^100);if(ispseudoprime(t)
      ==1,gef=1))))


      $ s $and $ t $ are proven primes with $ 301 $ digits. Assuming the generalized bunyakovsky conjecture, there are infinite many pairs of the desired form.






      share|cite|improve this answer









      $endgroup$















        5












        5








        5





        $begingroup$

        To demonstrate that there are also huge solutions :



        Define $$k=10^100+303593$$ $$s=4k^3-1$$ $$t=4k^3+1$$ then $(s,t)$ is a twin prime pair of the desired form which can be searched with this PARI/GP - routine 



        ? z=prod(j=1,3*10^4,prime(j));k=10^100-1;gef=0;while(gef==0,k=k+1;s=4*k^3-1;t=4*
        k^3+1;if(gcd(s*t,z)==1,if(ispseudoprime(s)==1,print(k-10^100);if(ispseudoprime(t)
        ==1,gef=1))))


        $ s $and $ t $ are proven primes with $ 301 $ digits. Assuming the generalized bunyakovsky conjecture, there are infinite many pairs of the desired form.






        share|cite|improve this answer









        $endgroup$



        To demonstrate that there are also huge solutions :



        Define $$k=10^100+303593$$ $$s=4k^3-1$$ $$t=4k^3+1$$ then $(s,t)$ is a twin prime pair of the desired form which can be searched with this PARI/GP - routine 



        ? z=prod(j=1,3*10^4,prime(j));k=10^100-1;gef=0;while(gef==0,k=k+1;s=4*k^3-1;t=4*
        k^3+1;if(gcd(s*t,z)==1,if(ispseudoprime(s)==1,print(k-10^100);if(ispseudoprime(t)
        ==1,gef=1))))


        $ s $and $ t $ are proven primes with $ 301 $ digits. Assuming the generalized bunyakovsky conjecture, there are infinite many pairs of the desired form.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 4 at 11:34









        PeterPeter

        49.2k1240138




        49.2k1240138





















            4












            $begingroup$

            $(3,5),(107,109)$ and $(2634011,2634013)$ are only such twin prime pairs below $10^7$. You can check for more by increasing the range using this small Python code below. I just run it upto $10^8$, $(29659499,29659501)$ and $(57395627,57395629)$ are only such pairs with $10^7<p<10^8$.



            Edit:
            For optimization purpose, I will use the fact that all primes more than $3$ can be represented in the form $6kpm 1$.



            First let check manually for $p=2,3$. For $p=2$ it's clearly not possible. For $p=3$, we have $3+5=8=2^3$, so, $(3,5)$ is such pair.



            Suppose, $p=6k-1$, with $p>5$, then we have $12k=n^3$. That means, $12|n^3$. But, as $n^3$ is a perfect cube, and we have $12=3cdot 2^2$, at least $3^3cdot 2^3=216$ will divide $n^3$. So, the pair $(108m^3-1,108m^3+1)$ will be such pair if both of them are prime.



            While dealing with cubes, usually prefer to work in $mathbbZ_7$, as any cube is either of $0,1,6$ in this field,i.e; $n^3equiv 0,1,6pmod7$ . So, we can have $2p+2equiv 0pmod7$, which gives $pequiv 6pmod7$, or, $2p+2equiv 1pmod7$ which implies $pequiv 3pmod7$ or $2p+2equiv 6pmod7$ which implies $pequiv 2pmod7$. Hence, only $3$ pssibilities. Here is the updated program: 



            import math
            import time

            start_time=time.time()
            def is_prime(n):
             flag=0
             if(n==2):
             return True
             if(n%2==0):
             return False
             else:
             for i in range(3,int(math.sqrt(n))+1,2):
             if(n%i==0):
             flag=1
             break
             if(flag==0):
             return True
             return False 

            for i in range(1,1000):#change the number inside this braket to check for larger numbers
             c=(6*i)**3 
             p=c//2-1 
             if(p%7==2 or p%7==4 or p%7==6):
             if(is_prime(p) & is_prime(p+2)):
             print(p, "is such twin prime with sum",c) 

            print(time.time()-start_time)





            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              I am not the downvoter, but my guess is that it is because your script, the substance of your answer, is quite inefficient compared to some others that were posted.
              $endgroup$
              – Mees de Vries
              Apr 4 at 14:32










            • $begingroup$
              @MeesdeVries Updated it, please have a look.
              $endgroup$
              – tarit goswami
              Apr 4 at 19:20















            4












            $begingroup$

            $(3,5),(107,109)$ and $(2634011,2634013)$ are only such twin prime pairs below $10^7$. You can check for more by increasing the range using this small Python code below. I just run it upto $10^8$, $(29659499,29659501)$ and $(57395627,57395629)$ are only such pairs with $10^7<p<10^8$.



            Edit:
            For optimization purpose, I will use the fact that all primes more than $3$ can be represented in the form $6kpm 1$.



            First let check manually for $p=2,3$. For $p=2$ it's clearly not possible. For $p=3$, we have $3+5=8=2^3$, so, $(3,5)$ is such pair.



            Suppose, $p=6k-1$, with $p>5$, then we have $12k=n^3$. That means, $12|n^3$. But, as $n^3$ is a perfect cube, and we have $12=3cdot 2^2$, at least $3^3cdot 2^3=216$ will divide $n^3$. So, the pair $(108m^3-1,108m^3+1)$ will be such pair if both of them are prime.



            While dealing with cubes, usually prefer to work in $mathbbZ_7$, as any cube is either of $0,1,6$ in this field,i.e; $n^3equiv 0,1,6pmod7$ . So, we can have $2p+2equiv 0pmod7$, which gives $pequiv 6pmod7$, or, $2p+2equiv 1pmod7$ which implies $pequiv 3pmod7$ or $2p+2equiv 6pmod7$ which implies $pequiv 2pmod7$. Hence, only $3$ pssibilities. Here is the updated program: 



            import math
            import time

            start_time=time.time()
            def is_prime(n):
             flag=0
             if(n==2):
             return True
             if(n%2==0):
             return False
             else:
             for i in range(3,int(math.sqrt(n))+1,2):
             if(n%i==0):
             flag=1
             break
             if(flag==0):
             return True
             return False 

            for i in range(1,1000):#change the number inside this braket to check for larger numbers
             c=(6*i)**3 
             p=c//2-1 
             if(p%7==2 or p%7==4 or p%7==6):
             if(is_prime(p) & is_prime(p+2)):
             print(p, "is such twin prime with sum",c) 

            print(time.time()-start_time)





            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              I am not the downvoter, but my guess is that it is because your script, the substance of your answer, is quite inefficient compared to some others that were posted.
              $endgroup$
              – Mees de Vries
              Apr 4 at 14:32










            • $begingroup$
              @MeesdeVries Updated it, please have a look.
              $endgroup$
              – tarit goswami
              Apr 4 at 19:20













            4












            4








            4





            $begingroup$

            $(3,5),(107,109)$ and $(2634011,2634013)$ are only such twin prime pairs below $10^7$. You can check for more by increasing the range using this small Python code below. I just run it upto $10^8$, $(29659499,29659501)$ and $(57395627,57395629)$ are only such pairs with $10^7<p<10^8$.



            Edit:
            For optimization purpose, I will use the fact that all primes more than $3$ can be represented in the form $6kpm 1$.



            First let check manually for $p=2,3$. For $p=2$ it's clearly not possible. For $p=3$, we have $3+5=8=2^3$, so, $(3,5)$ is such pair.



            Suppose, $p=6k-1$, with $p>5$, then we have $12k=n^3$. That means, $12|n^3$. But, as $n^3$ is a perfect cube, and we have $12=3cdot 2^2$, at least $3^3cdot 2^3=216$ will divide $n^3$. So, the pair $(108m^3-1,108m^3+1)$ will be such pair if both of them are prime.



            While dealing with cubes, usually prefer to work in $mathbbZ_7$, as any cube is either of $0,1,6$ in this field,i.e; $n^3equiv 0,1,6pmod7$ . So, we can have $2p+2equiv 0pmod7$, which gives $pequiv 6pmod7$, or, $2p+2equiv 1pmod7$ which implies $pequiv 3pmod7$ or $2p+2equiv 6pmod7$ which implies $pequiv 2pmod7$. Hence, only $3$ pssibilities. Here is the updated program: 



            import math
            import time

            start_time=time.time()
            def is_prime(n):
             flag=0
             if(n==2):
             return True
             if(n%2==0):
             return False
             else:
             for i in range(3,int(math.sqrt(n))+1,2):
             if(n%i==0):
             flag=1
             break
             if(flag==0):
             return True
             return False 

            for i in range(1,1000):#change the number inside this braket to check for larger numbers
             c=(6*i)**3 
             p=c//2-1 
             if(p%7==2 or p%7==4 or p%7==6):
             if(is_prime(p) & is_prime(p+2)):
             print(p, "is such twin prime with sum",c) 

            print(time.time()-start_time)





            share|cite|improve this answer











            $endgroup$



            $(3,5),(107,109)$ and $(2634011,2634013)$ are only such twin prime pairs below $10^7$. You can check for more by increasing the range using this small Python code below. I just run it upto $10^8$, $(29659499,29659501)$ and $(57395627,57395629)$ are only such pairs with $10^7<p<10^8$.



            Edit:
            For optimization purpose, I will use the fact that all primes more than $3$ can be represented in the form $6kpm 1$.



            First let check manually for $p=2,3$. For $p=2$ it's clearly not possible. For $p=3$, we have $3+5=8=2^3$, so, $(3,5)$ is such pair.



            Suppose, $p=6k-1$, with $p>5$, then we have $12k=n^3$. That means, $12|n^3$. But, as $n^3$ is a perfect cube, and we have $12=3cdot 2^2$, at least $3^3cdot 2^3=216$ will divide $n^3$. So, the pair $(108m^3-1,108m^3+1)$ will be such pair if both of them are prime.



            While dealing with cubes, usually prefer to work in $mathbbZ_7$, as any cube is either of $0,1,6$ in this field,i.e; $n^3equiv 0,1,6pmod7$ . So, we can have $2p+2equiv 0pmod7$, which gives $pequiv 6pmod7$, or, $2p+2equiv 1pmod7$ which implies $pequiv 3pmod7$ or $2p+2equiv 6pmod7$ which implies $pequiv 2pmod7$. Hence, only $3$ pssibilities. Here is the updated program: 



            import math
            import time

            start_time=time.time()
            def is_prime(n):
             flag=0
             if(n==2):
             return True
             if(n%2==0):
             return False
             else:
             for i in range(3,int(math.sqrt(n))+1,2):
             if(n%i==0):
             flag=1
             break
             if(flag==0):
             return True
             return False 

            for i in range(1,1000):#change the number inside this braket to check for larger numbers
             c=(6*i)**3 
             p=c//2-1 
             if(p%7==2 or p%7==4 or p%7==6):
             if(is_prime(p) & is_prime(p+2)):
             print(p, "is such twin prime with sum",c) 

            print(time.time()-start_time)






            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Apr 5 at 9:06

























            answered Apr 4 at 10:02









            tarit goswamitarit goswami

            2,2361422




            2,2361422











            • $begingroup$
              I am not the downvoter, but my guess is that it is because your script, the substance of your answer, is quite inefficient compared to some others that were posted.
              $endgroup$
              – Mees de Vries
              Apr 4 at 14:32










            • $begingroup$
              @MeesdeVries Updated it, please have a look.
              $endgroup$
              – tarit goswami
              Apr 4 at 19:20
















            • $begingroup$
              I am not the downvoter, but my guess is that it is because your script, the substance of your answer, is quite inefficient compared to some others that were posted.
              $endgroup$
              – Mees de Vries
              Apr 4 at 14:32










            • $begingroup$
              @MeesdeVries Updated it, please have a look.
              $endgroup$
              – tarit goswami
              Apr 4 at 19:20















            $begingroup$
            I am not the downvoter, but my guess is that it is because your script, the substance of your answer, is quite inefficient compared to some others that were posted.
            $endgroup$
            – Mees de Vries
            Apr 4 at 14:32




            $begingroup$
            I am not the downvoter, but my guess is that it is because your script, the substance of your answer, is quite inefficient compared to some others that were posted.
            $endgroup$
            – Mees de Vries
            Apr 4 at 14:32












            $begingroup$
            @MeesdeVries Updated it, please have a look.
            $endgroup$
            – tarit goswami
            Apr 4 at 19:20




            $begingroup$
            @MeesdeVries Updated it, please have a look.
            $endgroup$
            – tarit goswami
            Apr 4 at 19:20











            1












            $begingroup$

            $p + (p + 2) = 2p + 2 = 2(p + 1)$



            For the sum to be a cube, $p + 1$ must be divisible by 4, so that the sum becomes $2 times 4m^3$ for some integer $m$. So $p + 1$ is of the form $4m^3$.






            share|cite|improve this answer









            $endgroup$








            • 2




              $begingroup$
              Also, with the exception of the twin prime pair 3,5, every pair of twin primes must be of the form 6n-1,6n+1. Using your notation, you can further refine based on this to $p+1=108k^3$
              $endgroup$
              – Moko19
              Apr 4 at 9:52















            1












            $begingroup$

            $p + (p + 2) = 2p + 2 = 2(p + 1)$



            For the sum to be a cube, $p + 1$ must be divisible by 4, so that the sum becomes $2 times 4m^3$ for some integer $m$. So $p + 1$ is of the form $4m^3$.






            share|cite|improve this answer









            $endgroup$








            • 2




              $begingroup$
              Also, with the exception of the twin prime pair 3,5, every pair of twin primes must be of the form 6n-1,6n+1. Using your notation, you can further refine based on this to $p+1=108k^3$
              $endgroup$
              – Moko19
              Apr 4 at 9:52













            1












            1








            1





            $begingroup$

            $p + (p + 2) = 2p + 2 = 2(p + 1)$



            For the sum to be a cube, $p + 1$ must be divisible by 4, so that the sum becomes $2 times 4m^3$ for some integer $m$. So $p + 1$ is of the form $4m^3$.






            share|cite|improve this answer









            $endgroup$



            $p + (p + 2) = 2p + 2 = 2(p + 1)$



            For the sum to be a cube, $p + 1$ must be divisible by 4, so that the sum becomes $2 times 4m^3$ for some integer $m$. So $p + 1$ is of the form $4m^3$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Apr 4 at 9:36









            11235813211123581321

            352210




            352210







            • 2




              $begingroup$
              Also, with the exception of the twin prime pair 3,5, every pair of twin primes must be of the form 6n-1,6n+1. Using your notation, you can further refine based on this to $p+1=108k^3$
              $endgroup$
              – Moko19
              Apr 4 at 9:52












            • 2




              $begingroup$
              Also, with the exception of the twin prime pair 3,5, every pair of twin primes must be of the form 6n-1,6n+1. Using your notation, you can further refine based on this to $p+1=108k^3$
              $endgroup$
              – Moko19
              Apr 4 at 9:52







            2




            2




            $begingroup$
            Also, with the exception of the twin prime pair 3,5, every pair of twin primes must be of the form 6n-1,6n+1. Using your notation, you can further refine based on this to $p+1=108k^3$
            $endgroup$
            – Moko19
            Apr 4 at 9:52




            $begingroup$
            Also, with the exception of the twin prime pair 3,5, every pair of twin primes must be of the form 6n-1,6n+1. Using your notation, you can further refine based on this to $p+1=108k^3$
            $endgroup$
            – Moko19
            Apr 4 at 9:52



            -