Primes and SquaresDate: 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/ |
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