If $a, b, c, d$ are non-zero integers such that $ab = cd$, then $a^2+b^2+c^2+d^2$ is composite. Proving the case with one odd and three evens.

Question

I am doing a number theory course from scratch right now, and this question here has put a hault on my journey.

Let $a, b, c, d$ be non-zero integers such that $ab = cd$, prove that $a^2 + b^2 + c^2 + d^2$ is composite.

I started off really well on this, checking the pairity. Solved all the cases except one.

Let N = $a^2 + b^2 + c^2 + d^2$.

If there are an even number of odd numbers in {$a, b, c, d$} we get $N$ to be even. And since $N \ge 4$, trivially, its composite.

If we have three odds and only one even in {$a, b, c, d$}, one of {$ab, cd$} would be even and the other would be odd, which does not satisfy the given $ab = cd$ condition.

So the only case left to do is when exactly one of {$a, b, c, d$} is odd and the rest are even. But I don't see the way out. Thanks in advance.