Mathematical Reflections 2017, Issue 2 - Problem S403

Problem:
Find all primes $p$ and $q$ such that $$\dfrac{2^{p^2-q^2}-1}{pq}$$ is a product of two primes.

Proposed by Adrian Andreescu, Dallas, Texas, USA

Solution:
Since $2^{p^2-q^2}-1$ is odd and $$\dfrac{2^{p^2-q^2}-1}{pq}$$ is an integer, then $p$ and $q$ are odd primes and clearly $p>q$. So, $p^2-q^2 \equiv 0 \pmod{8}$. If $q>3$, then $p^2-q^2 \equiv 0 \pmod{3}$, so $p^2-q^2$ is divisible by $24$. Then
$$2^{p^2-q^2}-1=2^{24k}-1,$$ where $k \in \mathbb{N}^*$. Since $2^{24k}-1$ is divisible by $2^{24}-1=16777215=3^2\cdot 5\cdot 7 \cdot 13 \cdot 17 \cdot 241$, then $\dfrac{2^{p^2-q^2}-1}{pq}$ cannot be the product of two primes. So, $q=3$ and
$$\dfrac{2^{p^2-q^2}-1}{pq}=\dfrac{2^{p^2-9}-1}{3p}.$$ Since $p^2-9$ is divisible by $8$, then $p^2-9=8k$ for some $k \in \mathbb{N}^*$, so $$\dfrac{2^{p^2-9}-1}{3p}=\dfrac{2^{8k}-1}{3p}=\dfrac{(2^k-1)(2^k+1)(2^{2k}+1)(2^{4k}+1)}{3p}.$$ An easy check shows that it must be $k \geq 2$. If $k$ is even, then $3 \ | \ (2^k-1)$ and if $k>2$ we have $2^k-1=3n$ for some $n \in \mathbb{N}$ and $n>1$. But then
$$\dfrac{n(2^k+1)(2^{2k}+1)(2^{4k}+1)}{p}$$ is the product of at least three primes, contradiction. So, $k=2$ and we get $p=5$, which satisfies the condition. If $k$ is odd, then $3 \ | \ (2^k+1)$ and since $k \geq 3$, then $2^k+1=3n$ for some $n \in \mathbb{N}$ and $n>1$. But then, we have that $$\dfrac{n(2^k-1)(2^{2k}+1)(2^{4k}+1)}{p}$$ is the product of at least three primes, contradiction.
In conclusion, $(p,q)=(5,3)$.