If $a, b, c, d$ are natural numbers, such that, $ab = cd$, prove that $a^2 + b^2 + c^2 + d^2$ is a composite number. 
If $a, b, c, d$ are natural numbers, such that, $ab = cd$, prove that $a^2 + b^2 + c^2 + d^2$ is a composite number.

Came across this question while solving an exercise on prime numbers.
Now I found a pretty simple solution to this problem here.
(It is not the exact same problem but the method described there in most of the answers can be easily used on this one too.)
But when I was trying to solve this on my own, I took another path.
I noticed that $a^2 + b^2 + c^2 + d^2$ is even (and therefore composite) if:
(1) All of $a, b, c, d$ have the same parity (i.e. all are odd or all are even).
(2) Two of them are even and two are odd (say $a, c$ are even and $b, d$ are odd).
So the only case left to worry about is when $a, b, c$ are even and $d$ is odd (wlog).
Example: $26\times 28 = 56\times 13$ ...and... $13^2 + 26^2 + 28^2 + 56^2 = 4765$.
And this is where I'm stuck. Any ideas?
EDIT: As pointed out in some if the answers, my initial assumption of the given expression always being divisible by 5 was wrong (I apologise for that). But I'd still like to see if it can proved, in the manner I started out, that the given expression is composite.

It's not always divisible by $5$:
$6 \times 6 = 4 \times 9$ but $6^2+6^2+4^2+9^2 = 169$
