Number Theory: PrimesDate: 07/10/2003 at 04:23:10 From: Carlos Subject: Number theory Find all primes p,q such that ((7^p-2^p)(7^q-2^q))/(pq) is an integer. If p divides (7^p-2^p)(7^q-2^q), then p divides either (7^p-2^p) or (7^q-2^q) and vice versa. It can easily be deduced from Fermat's Little theorem that 7^p - 2^p = 5 (mod p). Thus p divides (7^p-2^p) iff p = 5. From this I have figured that (p=5,q=5),(p=5,q=11),(p=5,q=61) are all solutions. Where I get stuck is for the case 7^q=2^q mod p and 7^p=2^p mod q. I wonder if 7^q=2^q mod p and 7^p=2^p mod q have any solutions. I suspect that for p (not 5) to divide (7^q-2^q), p > q. This will prove that there are no other solutions. But I cannot prove this assertion. Date: 07/10/2003 at 04:43:59 From: Doctor Jacques Subject: Re: Number theory Hi Carlos, You are on the right track. Assume that 7^p = 2^p (mod q) with p and q primes, and q <> 5. As q is obviously odd, 2 has an inverse mod q, namely: a = (q+1)/2 and we may write: (7a)^p = 1 (mod q) Now, the order of (7a) mod q is either 1 (in which case q = 5) or p, and divides (q - 1), by Fermat's theorem. This should allow you to use your idea to conclude the proof. Does this help? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ |
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