29^3-7^3

enzocreti · 2020-02-18, 09:32

let be p and q primes

the Wieferich prime 1093 divides 29^3-7^3.

Are there other p and q prime such that

p^3-q^3 is a multiple of 1093?

axn · 2020-02-18, 11:10

Every prime q>=2 will have infinitely many p's that give the solution (except 1093, of course).

For example, for q=3, we have p=5009,8747,11383, etc...
q=5, p=6563,13121,15307, etc...

EDIT:- The next smallest solution is 131^3-107^3

Dr Sardonicus · 2020-02-18, 13:53

Of course, there are three classes of pairs (p,q) mod 1093 (p and q not 1093) for which p^3 == q^3 (mod 1093).

One of these classes is p == q (mod 1093). We can take any pairs p < q

Another class is that of (p,q) for which q == p*29/7 (mod 1093).

And, obviously, a third class consists of pairs (p, q) for which q == 7/29*p (mod 1093).

Each of these classes has 1092 pairs of congruence classes (mod 1093), each of which contains infinitely many primes.

But this is all very elementary, and there's no challenge. Anybody who had any real interest in number theory would already know this.

The real challenge is to do the totally mindless computation to produce all pairs of primes p < q < 1093 for which q^3 - p^3 is divisible by 1093.

I have sunk to the challenge! There are thirty such pairs.

[[7, 29], [107, 131], [103, 251], [277, 293], [61, 467], [191, 479], [349, 509], [277, 523], [293, 523], [607, 641], [233, 653], [19, 683], [439, 709], [103, 739], [251, 739], [613, 751], [179, 797], [541, 809], [599, 823], [577, 829], [353, 839], [433, 857], [223, 883], [151, 941], [907, 947], [571, 967], [383, 997], [137, 1013], [557, 1039], [919, 1051]]

2020-02-18, 09:32	#1
enzocreti Mar 2018 1000010010₂ Posts	29^3-7^3 let be p and q primes the Wieferich prime 1093 divides 29^3-7^3. Are there other p and q prime such that p^3-q^3 is a multiple of 1093?

2020-02-18, 11:10	#2
axn Jun 2003 13BC₁₆ Posts	Every prime q>=2 will have infinitely many p's that give the solution (except 1093, of course). For example, for q=3, we have p=5009,8747,11383, etc... q=5, p=6563,13121,15307, etc... EDIT:- The next smallest solution is 131^3-107^3 Last fiddled with by axn on 2020-02-18 at 11:20

2020-02-18, 13:53	#3
Dr Sardonicus Feb 2017 Nowhere 4,643 Posts	Of course, there are three classes of pairs (p,q) mod 1093 (p and q not 1093) for which p^3 == q^3 (mod 1093). One of these classes is p == q (mod 1093). We can take any pairs p < q Another class is that of (p,q) for which q == p29/7 (mod 1093). And, obviously, a third class consists of pairs (p, q) for which q == 7/29p (mod 1093). Each of these classes has 1092 pairs of congruence classes (mod 1093), each of which contains infinitely many primes. But this is all very elementary, and there's no challenge. Anybody who had any real interest in number theory would already know this. The real challenge is to do the totally mindless computation to produce all pairs of primes p < q < 1093 for which q^3 - p^3 is divisible by 1093. I have sunk to the challenge! There are thirty such pairs. [[7, 29], [107, 131], [103, 251], [277, 293], [61, 467], [191, 479], [349, 509], [277, 523], [293, 523], [607, 641], [233, 653], [19, 683], [439, 709], [103, 739], [251, 739], [613, 751], [179, 797], [541, 809], [599, 823], [577, 829], [353, 839], [433, 857], [223, 883], [151, 941], [907, 947], [571, 967], [383, 997], [137, 1013], [557, 1039], [919, 1051]]