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
$begingroup$
$84537841287167$ 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
edited Apr 4 at 10:11
tarit goswami
2,2361422
$endgroup$
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
|
show 3 more comments
$begingroup$
$84537841287167$ 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
edited Apr 4 at 10:11
tarit goswami
2,2361422
$endgroup$
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
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 maximafor 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
|
show 3 more comments
$begingroup$
$84537841287167$ 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
edited Apr 4 at 10:11
tarit goswami
2,2361422
$endgroup$
$84537841287167$ 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
edited Apr 4 at 10:11
tarit goswami
2,2361422
edited Apr 4 at 10:11
tarit goswami
2,2361422
edited Apr 4 at 10:11
tarit goswami
2,2361422
edited Apr 4 at 10:11
tarit goswami
2,2361422
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
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
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 maximafor 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
|
show 3 more comments
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 maximafor 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
|
show 3 more comments
5 Answers
5
active
oldest
votes
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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]
edited Apr 5 at 9:45
Community♦
1
answered Apr 4 at 12:20
robjohn♦robjohn
271k27315643
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1
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"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?
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– spyr03
Apr 4 at 15:46
3
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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 $nequiv 3pmod5$.
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– OppoInfinity
Apr 4 at 15:54
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@spyr03: I have expanded the section about the values skipped
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– robjohn♦
Apr 4 at 17:27
add a comment |
$begingroup$
$3$ 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.
answered Apr 4 at 9:31
ArthurArthur
123k7122211
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some bigger one
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– user660792
Apr 4 at 9:32
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@DietrichBurde It's a cube, see the question body.
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– Arthur
Apr 4 at 9:34
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@DietrichBurde I'm sure we can. $17+19 = 6^2$, so $6$ is really prolific here.
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– Arthur
Apr 4 at 9:37
1
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with Maple can you find other examples?
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– user660792
Apr 4 at 9:41
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@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.
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– Arthur
Apr 4 at 9:42
|
show 4 more comments
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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.
answered Apr 4 at 11:34
PeterPeter
49.2k1240138
$endgroup$
add a comment |
$begingroup$
$(3,5),(107,109)$ 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)
answered Apr 4 at 10:02
tarit goswamitarit goswami
2,2361422
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$begingroup$
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.
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– Mees de Vries
Apr 4 at 14:32
$begingroup$
@MeesdeVries Updated it, please have a look.
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– tarit goswami
Apr 4 at 19:20
add a comment |
$begingroup$
$p + (p + 2) = 2p + 2 = 2(p + 1)$
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 $4m^3$.
answered Apr 4 at 9:36
11235813211123581321
352210
$endgroup$
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
add a comment |
5 Answers
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$begingroup$
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]
edited Apr 5 at 9:45
Community♦
1
answered Apr 4 at 12:20
robjohn♦robjohn
271k27315643
$endgroup$
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
add a comment |
$begingroup$
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]
edited Apr 5 at 9:45
Community♦
1
answered Apr 4 at 12:20
robjohn♦robjohn
271k27315643
$endgroup$
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
add a comment |
$begingroup$
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]
edited Apr 5 at 9:45
Community♦
1
answered Apr 4 at 12:20
robjohn♦robjohn
271k27315643
$endgroup$
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]
edited Apr 5 at 9:45
Community♦
1
answered Apr 4 at 12:20
robjohn♦robjohn
271k27315643
edited Apr 5 at 9:45
Community♦
1
edited Apr 5 at 9:45
Community♦
1
edited Apr 5 at 9:45
Community♦
1
1
answered Apr 4 at 12:20
robjohn♦robjohn
271k27315643
answered Apr 4 at 12:20
robjohn♦robjohn
271k27315643
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
add a comment |
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
add a comment |
$begingroup$
$3$ 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.
answered Apr 4 at 9:31
ArthurArthur
123k7122211
$endgroup$
$begingroup$
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
|
show 4 more comments
$begingroup$
$3$ 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.
answered Apr 4 at 9:31
ArthurArthur
123k7122211
$endgroup$
$begingroup$
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
|
show 4 more comments
$begingroup$
$3$ 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.
answered Apr 4 at 9:31
ArthurArthur
123k7122211
$endgroup$
$3$ 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.
answered Apr 4 at 9:31
ArthurArthur
123k7122211
edited Apr 4 at 11:12
answered Apr 4 at 9:31
ArthurArthur
123k7122211
answered Apr 4 at 9:31
ArthurArthur
123k7122211
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
|
show 4 more comments
$begingroup$
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
|
show 4 more comments
$begingroup$
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.
answered Apr 4 at 11:34
PeterPeter
49.2k1240138
$endgroup$
add a comment |
$begingroup$
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.
answered Apr 4 at 11:34
PeterPeter
49.2k1240138
$endgroup$
add a comment |
$begingroup$
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.
answered Apr 4 at 11:34
PeterPeter
49.2k1240138
$endgroup$
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.
answered Apr 4 at 11:34
PeterPeter
49.2k1240138
answered Apr 4 at 11:34
PeterPeter
49.2k1240138
answered Apr 4 at 11:34
PeterPeter
49.2k1240138
answered Apr 4 at 11:34
PeterPeter
49.2k1240138
49.2k1240138
add a comment |
add a comment |
$begingroup$
$(3,5),(107,109)$ 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)
answered Apr 4 at 10:02
tarit goswamitarit goswami
2,2361422
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$(3,5),(107,109)$ 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)
answered Apr 4 at 10:02
tarit goswamitarit goswami
2,2361422
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$(3,5),(107,109)$ 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)
answered Apr 4 at 10:02
tarit goswamitarit goswami
2,2361422
$endgroup$
$(3,5),(107,109)$ 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)
answered Apr 4 at 10:02
tarit goswamitarit goswami
2,2361422
edited Apr 5 at 9:06
answered Apr 4 at 10:02
tarit goswamitarit goswami
2,2361422
answered Apr 4 at 10:02
tarit goswamitarit goswami
2,2361422
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
$p + (p + 2) = 2p + 2 = 2(p + 1)$
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 $4m^3$.
answered Apr 4 at 9:36
11235813211123581321
352210
$endgroup$
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
add a comment |
$begingroup$
$p + (p + 2) = 2p + 2 = 2(p + 1)$
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 $4m^3$.
answered Apr 4 at 9:36
11235813211123581321
352210
$endgroup$
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
add a comment |
$begingroup$
$p + (p + 2) = 2p + 2 = 2(p + 1)$
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 $4m^3$.
answered Apr 4 at 9:36
11235813211123581321
352210
$endgroup$
$p + (p + 2) = 2p + 2 = 2(p + 1)$
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 $4m^3$.
answered Apr 4 at 9:36
11235813211123581321
352210
answered Apr 4 at 9:36
11235813211123581321
352210
answered Apr 4 at 9:36
11235813211123581321
352210
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
add a comment |
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
add a comment |
2
$begingroup$
What do you want to do with other examples?
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– 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.. $107,2634011,29659499,57395627,104792291,271669247,485149499,cdots\ 83797007573051, colorred84537841287167, 88875400155587,89731703685707,89846291353499,cdots$
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– achille hui
Apr 4 at 10:23
1
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@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
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oeis.org/A061308 is worth a look.
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– Barry Cipra
Apr 4 at 12:15
2
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similarly oeis.org/A240169
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– KBusc
Apr 4 at 16:00