Slovenian Mathematical Olympiad, 2016

From the second equation we express $d = 2a + 4b + 6c - 26$ and insert it into the first equation to get
$$a^2 + b^2 + c^2 = 2a + 4b + 6c - 13.$$
Moving all the terms to the left we have $a^2 + b^2 + c^2 - 2a - 4b - 6c + 13 = 0$, which we now rewrite as the sum of perfect squares
$$(a - 1)^2 + (b - 2)^2 + (c - 3)^2 = 1.$$
All three perfect squares on the left are non-negative integers, so one of them is equal to 1 and the other two are 0. If $(a - 1)^2 = 1$, we have $a = 0$ or $a = 2$ as well as $b = 2$ and $c = 3$. If $(b - 2)^2 = 1$, we have $b = 1$ or $b = 3$ as well as $a = 1$ and $c = 3$. If $(c - 3)^2 = 1$, we have $c = 2$ or $c = 4$ as well as $a = 1$ and $b = 2$. In each case we can determine $d$ from the equation given above. The integer solutions $(a, b, c, d)$ of the given system of equations are therefore $(0, 2, 3, 0)$, $(2, 2, 3, 4)$, $(1, 1, 3, -2)$, $(1, 3, 3, 6)$, $(1, 2, 2, -4)$ and $(1, 2, 4, 8)$.