Slovenia — National Math Olympiad

Let $x_1$ and $x_2$ be the integer solutions of this quadratic equation. We may assume that $x_1 \le x_2$. By Viete's formulas we have $x_1 + x_2 = -n$ and $x_1x_2 = n + 5$. Adding both identities together we get $x_1x_2 + x_1 + x_2 = 5$, so

$$(x_1 + 1)(x_2 + 1) = 6.$$

Since $6 = 1 \cdot 6 = 2 \cdot 3 = (-3) \cdot (-2) = (-6) \cdot (-1)$, we have four possible cases to consider. In the first case we have $x_1 = 0$, $x_2 = 5$, in the second case $x_1 = 1$, $x_2 = 2$, in the third case $x_1 = -4$, $x_2 = -3$ and in the final case $x_1 = -7$, $x_2 = -2$. Using the identity $x_1 + x_2 = -n$ we see that $n$ is equal to $-5, -3, 7$ or $9$.