1. Determine all integer values of the parameter \( a \) for which the equation

$$
(x+a)(x+2a)=3a
$$

has at least one integer root.
SOLUTION. After expanding the left side and moving the term $3a$ from the right side to the left, we obtain the quadratic equation

$$
x^{2}+3 a x+2 a^{2}-3 a=0 .
$$

Its roots (if they exist) have, according to a well-known formula, the form

$$
x_{1,2}=\frac{-3 a+\sqrt{a^{2}+12 a}}{2}
$$

The value of such an expression is an integer only when the number $a^{2}+12 a$ is the square of some integer $b$, which we can assume is non-negative. The equality $b=\sqrt{a^{2}+12 a}$ can be rearranged by squaring and completing the square to the form

$$
(a+6)^{2}=b^{2}+36 \text {, or }(a+6+b)(a+6-b)=36 \text {. }
$$

We have obtained the factorization of the number 36 into the product of two integer factors, which must therefore have the same sign. Since their difference

$$
(a+6+b)-(a+6-b)=2 b
$$

is an even non-negative number (we remind that $b \geqq 0$), both factors have the same parity (they are either both even or both odd) and the second factor is not greater than the first factor. All this together means that there are only four possibilities:

(1) $a+6+b=18$ and $a+6-b=2$. This system of equations has a unique solution $a=4$ and $b=8$. Check: the equation $(x+4)(x+8)=12$ has roots $-10$ and $-2$.

(2) $a+6+b=6$ and $a+6-b=6$. In this case, $a=0$ and $b=0$. Check: the equation $(x+0)(x+0)=0$ has a double root 0.

(3) $a+6+b=-2$ and $a+6-b=-18$. In this case, $a=-16$ and $b=8$. Check: the equation $(x-16)(x-32)=-48$ has roots 20 and 28.

(4) $a+6+b=-6$ and $a+6-b=-6$. In this case, $a=-12$ and $b=0$. Check: the equation $(x-12)(x-24)=-36$ has a double root 18.

Answer: The sought values of the parameter $a$ are four, namely the numbers 4, 0, -16, and -12.