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 $x_1{x}_2=n+5$. Adding both identities together we get $x_1{x}_2+x_1+x_2=5$, 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 $9$.