XV Junior Macedonian Mathematical Olympiad

From $m^3+m^2+7=(m^2-m+1)(m+2)+(m+5)$ it follows that $m^2-m+1$ is a divisor of $m+5$. Obviously, one solution is $m=-5$.

Let $m\ne -5$. Then $|m^2-m+1|\le |m+5|$. From $m^2-m+1=(m-\frac{1}{2}{)}^2+\frac{3}{4}>0$ it follows that $m^2-m+1\le m+5$ or $m^2-m+1\le -m-5$. The case $m^2-m+1\le -m-5$ is not possible ($m^2+6\le 0$). From $m^2-m+1\le m+5$ it follows that $m^2-2m-4\le 0$ i.e. $(m-1{)}^2\le 5$. The last inequality is satisfied for $m\in \{-1,0,1,2,3\}$. Checking each value, we conclude that $m=0$ and $m=1$ satisfy the initial condition.

Finally $m\in \{-5,0,1\}$.