BxMO/EGMO Team Selection Test

The only divisor $p-q$ and $p+q$ can have in common is $2$, because $p$ and $q$ are different prime numbers. Indeed, a divisor $d$ of $p+q$ and $p-q$ is also a divisor of $(p+q)+(p-q)=2p$ and of $(p+q)-(p-q)=2q$. And we know that $\text{gcd}(2p,2q)=2$, so $d$ must be a divisor of $2$.

Since each prime divisor of $|p-q|$ or $p+q$ must also be a prime divisor of the other, because of the equation, we now know that $|p-q|$ and $p+q$ are both powers of $2$ (unequal to $1$). But the greatest common divisor is $2$, so the smallest of the two (and this must be $|p-q|$) is equal to $2$. So $|p-q|=2$ and $p+q$ is a power of $2$. So there is exactly one number between $p$ and $q$, namely $\frac{p+q}{2}$. Since $p+q$ is a power of two, $p+q$ is not divisible by $3$. So $\frac{p+q}{2}$ is also not divisible by $3$. Since of three consecutive numbers, one must be divisible by $3$, it follows that $p$ or $q$ is divisible by $3$. Since $1$ is not a prime number, the minimum of $p$ and $q$ must equal $3$ and the other must equal $5$. We find two solutions, $(p,q)=(3,5)$ and $(p,q)=(5,3)$, both of which satisfy the equation. $\square$