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### Number Theory: Primes

```Date: 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.

some more, or if you have any other questions.

- Doctor Jacques, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory

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