Solution for December 2011
The Problem: 

Find all primes $p$ such that $\large \frac{2^{p1} 1}{p}$ is a perfect square.

The solution:
We received correct solutions from
Lamis Alsheikh
(Syria)

Diana Andrei (Sweden)

Lou Cairoli (USA)

Bojan Bašić (Serbia)

MeiHui Fang (Austria)

Ruben Victor Cohen (Argentina)

Gruian Cornel (Romania)

Bernard Collignon (France)

Allen Druze(USA)

Philippe Fondanaiche (France)

Kipp Johnson (USA)

HansChristian Jung (Germany)

Wolfgang Kais (Germany)

Patrick J LoPresti (USA)

Albert Stadler (Switzerland)

Hakan Summakoğlu (Turkey)

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 "Thιorie
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 $\large
\frac{2^{p1} 1}{p} = \frac{1}{2}$. Therefore $p$ must be
an odd prime. Suppose that $\large \frac{2^{p1} 1}{p} =
a^2$. Then $2^{p1} 1 = pa^2$. Since $p$ is odd,
$\frac{p1}{2}$ is an integer, and the left side factors as
$$2^{p1} 1 = (2^{\frac{p1}{2}} +
1)(2^{\frac{p1}{2}} 1).$$
$\large 2^{\frac{p1}{2}} + 1$ and $\large
2^{\frac{p1}{2}} 1$ are odd integers with a difference
of two, they do not have a common factor other than 1. Hence
$2^{p1} 1 = pa^2$ implies that there exist integers $x$,
$y$ such that $\large 2^{\frac{p1}{2}} 1 = x^2$ and
$\large 2^{\frac{p1}{2}} + 1 = py^2$ or $\large
2^{\frac{p1}{2}} 1 = py^2$ and $\large
2^{\frac{p1}{2}} + 1 = x^2$.
Case 1: $\large 2^{\frac{p1}{2}} 1 =
x^2$, where $x$ is an odd integer. The square of an odd
integer is always congruent to 1 mod 4, while $\large
2^{\frac{p1}{2}} 1$ 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: $\large 2^{\frac{p1}{2}} + 1 = x^2$,
implies $\large 2^{\frac{p1}{2}} = x^2 1 = (x + 1)(x
1)$. Hence $x 1$ and $x+1$ are consecutive even numbers
whose product is a power of 2, namely 2 and 4. This implies
that $x = 3$, $\large 2^{\frac{p1}{2}} = 8$ and $p = 7$.
The only solution is $p = 7$.
