Solution for December 2011

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

The solution:

We received correct solutions from

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 $\large \frac{2^{p-1} -1}{p} = \frac{1}{2}$. Therefore $p$ must be an odd prime. Suppose that $\large \frac{2^{p-1} -1}{p} = a^2$. Then $2^{p-1} -1 = pa^2$. Since $p$ is odd, $\frac{p-1}{2}$ is an integer, and the left side factors as

$$2^{p-1} -1 = (2^{\frac{p-1}{2}} + 1)(2^{\frac{p-1}{2}} – 1).$$

$\large 2^{\frac{p-1}{2}} + 1$ and $\large 2^{\frac{p-1}{2}} – 1$ are odd integers with a difference of two, they do not have a common factor other than 1. Hence $2^{p-1} -1 = pa^2$ implies that there exist integers $x$, $y$ such that $\large 2^{\frac{p-1}{2}} – 1 = x^2$ and $\large 2^{\frac{p-1}{2}} + 1 = py^2$ or $\large 2^{\frac{p-1}{2}} – 1 = py^2$ and $\large 2^{\frac{p-1}{2}} + 1 = x^2$.

Case 1: $\large 2^{\frac{p-1}{2}} – 1 = x^2$, where $x$ is an odd integer. The square of an odd integer is always congruent to 1 mod 4, while $\large 2^{\frac{p-1}{2}} – 1$ 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: $\large 2^{\frac{p-1}{2}} + 1 = x^2$, implies $\large 2^{\frac{p-1}{2}} = x^2 – 1 = (x + 1)(x – 1)$. Hence $x – 1$ and $x+1$ are consecutive even numbers whose product is a power of 2, namely 2 and 4. This implies that $x = 3$, $\large 2^{\frac{p-1}{2}} = 8$ and $p = 7$. The only solution is $p = 7$.