National Math Olympiad in Slovenia

Let $p+q = x^2$ and $p+4q = y^2$ for some positive integers $x$ and $y$. Subtracting the equations we get $3q = y^2 - x^2 = (y-x)(y+x)$. Since $q$ is a prime and $x+y \ge 2$ we have the following possibilities:

1. $y - x = 1$ and $y + x = 3q$. From the first equation we get $y = x + 1$. Plugging this into the second equation we see that $2x + 1 = 3q$. This implies that $q$ is odd, so we may write $q = 2m + 1$ for some positive integer $m$. Now, $x = 3m + 1$ and $p = x^2 - q = 9m^2 + 4m = m(9m + 4)$. Since $p$ is a prime, we conclude that $m = 1$. Thus, $p = 13$ and $q = 3$, which are indeed prime.

2. $y - x = 3$ and $y + x = q$. From the first equation we get $y = x + 3$. Plugging this into the second equation we see that $2x + 3 = q$. The initial equality then implies $p = x^2 - p = x^2 - q = x^2 - 2x - 3 = (x + 1)(x - 3)$. Since $p$ is a prime we have $x = 4$. Thus, $p = 5$ and $q = 11$, which are indeed prime.

3. $y - x = q$ and $y + x = 3$. Since $x$ and $y$ are positive integers and $y > x$, we get $y = 2$ and $x = 1$, which is a contradiction since it immediately follows that $q = 1$.

The two solutions are $(5, 11)$ and $(13, 3)$.