Find all primes p such that 2p−1−1p is a perfect square.

The Problem:
	Find all primes p such that 2p−1−1p is a perfect square.

The solution:

The solution: p=3 and p=7 .

We found this problem in the 2006 Thai Mathematical Olympiad. Jean Drabbe kindly informed us that it is also covered on page 423 of Lucas' book "Théorie des Nombres" (1891). Those who read french can download it from http://www.archive.org/details/thoriedesnombre00lucagoog. Our solvers responses were similar to Lucas' argument.

We first rule out p=2 , since 2p−1−1p=12 . Therefore p must be an odd prime. Suppose that 2p−1−1p=a2 . Then 2p−1−1=pa2 . Since p is odd, p−12 is an integer, and the left side factors as

2p−1−1=(2p−12+1)(2p−12–1).

2p−12+1 and 2p−12–1 are odd integers with a difference of two, they do not have a common factor other than 1. Hence 2p−1−1=pa2 implies that there exist integers x , y such that 2p−12–1=x2 and 2p−12+1=py2 or 2p−12–1=py2 and 2p−12+1=x2 .

Case 1: 2p−12–1=x2 , where x is an odd integer. The square of an odd integer is always congruent to 1 mod 4, while 2p−12–1 is congruent to 1 mod 4 when p=3 and and to 3 mod 4 when p>3 . Hence the only solution is p=3 .

Case 2: 2p−12+1=x2 , implies 2p−12=x2–1=(x+1)(x–1) . Hence x–1 and x+1 are consecutive even numbers whose product is a power of 2, namely 2 and 4. This implies that x=3 , 2p−12=8 and p=7 . The only solution is p=7 .

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