Find those triples of integers, the product of which is four times their sum, and one of which is twice the sum of the other two.
Solution. For the triplet of numbers $a, b, c$, according to the conditions, we can write the following equations: $a b c=4(a+b+c)$, and for example, $c=2(a+b)$. Substituting this:

$$
a b \cdot 2(a+b)=4(a+b+2(a+b))
$$

Rearranging and factoring out $(a+b)$:

$$
2 a b(a+b)=12(a+b)
$$

If $a+b \neq 0$, we arrive at the equation $a b=6$. If $a= \pm 1$, then $b= \pm 6$ and $c= \pm 14$ (the signs of $a, b$, and $c$ are of course the same). In this case,

$$
a b c= \pm 84 \quad \text { and } \quad a+b+c= \pm 21
$$

thus the requirements of the problem are satisfied. If $a= \pm 2$, then $b= \pm 3$ and $c= \pm 10$. Now,

$$
a b c= \pm 60 \quad \text { and } \quad a+b+c= \pm 15
$$

and this also meets the requirements of the problem. If $a= \pm 3$ or $\pm 6$, we get the same solutions as above, only the roles of $a$ and $b$ are swapped.

Finally, if $a+b=0$, that is, $b=-a$, then $c=0$, and these triplets $(a,-a, 0)$ - for any integer $a$ - are also solutions to the problem.