Mathematical Reflections 2014, Issue 4 - Problem J308

Problem:
Are there triples $(p, q, r)$ of primes for which $(p^2-7)(q^2-7)(r^2-7)$ is a perfect square?

Proposed by Titu Andreescu.

Solution:
The answer is no. Let $N=(p^2-7)(q^2-7)(r^2-7)$. If exactly one or exactly three among $p,q,r$ are equal to $2$, then $N<0$ and if exactly two numbers are equal to $2$, then $N \equiv 2 \pmod{4}$. In every case $N$ is not a perfect square. Assume that $p,q,r$ are all odd. Then there exist $a,b,c \in \mathbb{N}$ such that $p^2=4a+1,q^2=4b+1,r^2=4c+1$ and
$$N=2^3(2a-3)(2b-3)(2c-3).$$ Since in the factorization of $N$ the number $2$ has an odd exponent, we conclude that $N$ is not a perfect square. Therefore, $N$ is never a perfect square.