1. Find all integers $a$ for which the equation

$$
a\left(x^{2}+x\right)=3\left(x^{2}+1\right)
$$

has at least one integer root.

(Jaromír Šimša)
Solution. If the equation has an integer root $m$, then the number $3\left(m^{2}+1\right)$ is equal to $a m(m+1)$, so it is divisible by the number $m$, which is clearly different from zero. Given the coprimality of the numbers $m$ and $m^{2}+1$, it follows that the number $m$ is a divisor of the number 3, so $m \in\{-3,-1,1,3\}$. Substituting all four values of $m$ into the original equation, we find that an integer $a$ results for exactly two of them. It is $x=-3$, which corresponds to $a=5$, and $x=1$, for which $a=3$. (For $x=-1$ we get $0 \cdot a=6$, and for $x=3$ the equation $12 a=30$ has a non-integer root $a$.)

Answer. The equation has an integer root for $a=3$ or $a=5$.

Note. The equality for the integer root $m$ could also be rearranged to the form

$$
\operatorname{am}(m+1)=3\left(m^{2}+1\right)=3(m-1)(m+1)+6,
$$

from which we see that the number $m+1$ divides the number 6. This consideration leads to eight possible values for the number $m$ with a similar discussion.