Mathematical competitions in Croatia

For arbitrary $a$ and $b$ of different parity define
$$c = \frac{1 + a^2 + b^2 - 2a - 2b - 2ab}{4}.$$
Note that $c$ is an integer, since $a$ and $b$ are of different parity (for example, if $a$ is even and $b$ is odd, then $1 - 2b + b^2 = (1 - b)^2$ and $a^2 - 2a - 2ab$ are divisible by $4$).
Then
$$\begin{aligned} a + c &= \frac{1 + a^2 + b^2 + 2a - 2b - 2ab}{4} = \left( \frac{1 + a - b}{2} \right)^2, \\ b + c &= \frac{1 + a^2 + b^2 - 2a + 2b - 2ab}{4} = \left( \frac{1 - a + b}{2} \right)^2, \\ ab + c &= \frac{1 + a^2 + b^2 - 2a - 2b + 2ab}{4} = \left( \frac{1 - a - b}{2} \right)^2. \end{aligned}$$
Since $a$ and $b$ are of different parity, the numbers $1+a-b$, $1-a+b$ and $1-a-b$ are even, which completes the proof.