Primes and Squares

Date: 05/03/2001 at 19:32:49
From: ali
Subject: Number theory

For what values of prime number p is (2^(p-1)-1)/p a perfect square?

Date: 05/04/2001 at 17:44:58
From: Doctor Schwa
Subject: Re: Number theory

Hi Ali,

What a neat problem!

First I tried all the primes up to 47, and found that only 3 and 7 
worked, and I guessed that those were the only ones, and set about 
trying to prove it.

I started by supposing that

   (2^(p-1)-1)/p = a^2

so that multiplying by p, I get

   2^(p-1) - 1 = p*a^2

and, since p-1 is even (call it 2m), I can factor the left side
as difference of squares,

   (2^m - 1)(2^m + 1) = p * a^2

Since 2^m - 1 and 2^m + 1 are both odd and differ by 2, they can have 
no common factor, so one of them is p times a perfect square, and the 
other one is a perfect square by itself (since they can't have any 
common factors, they must make the squares all by themselves without 
help from the other factor).

So, case 1: 2^m - 1 = p * b^2, 2^m + 1 = c^2

Then 2^m = c^2 - 1,

   2^m = (c-1)(c+1)

so how can c-1 and c+1 both be powers of 2? Must be that c = 3, and 
hence m = 3, p = 7. So p = 7 is one of the possibilities.

The other case is somewhat trickier, but similar sorts of factoring
arguments eventually establish that p = 3 is the only number that 
works in the other case.

I'll leave it to you to try that (harder!) half of the proof, and 
please do write us back if you get stuck.

Enjoy,

- Doctor Schwa, The Math Forum
  http://mathforum.org/dr.math/