Sum of four squares not a prime Let $ a, b, c, d $ be natural numbers such that $ ab=cd $. Prove that $ a^2+b^2+c^2+d^2 $ is not a prime.

I am clueless on this one. I tried contradiction, but didn't get anywhere. 

Can you help?

Edit: I understand natural numbers to be strictly positive, excluding $0 $.
Since $d=\frac{ab}{c}$, we obtain

$$
a^2+b^2+c^2+d^2=\frac{(a^2+c^2)(b^2+c^2)}{c^2}=\left(\left(\frac{a}{a'}\right)^2+b'^2\right)\left(\left(\frac{b}{b'}\right)^2+a'^2\right).
$$
where $c=a'b'$ such that $a' \mid a$ and $b' \mid b$.
A rather odd way to perhaps get there..

ab = cd 

no prime > 2 is even.

sum of any even number of odd numbers is even.

square of odd numbers is odd and of even numbers is even. 

so either 1 or 3 of a, b, c, d must be odd if we are possibly to get a prime.  

It cannot be that only one of them is odd or ab != cd.  So we need 3 of them odd.

If a,b,c are odd then ab/c = d which is a contradiction.  QED
:)