TITLE: Sum of four squares not a prime
QUESTION [23 upvotes]: 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 $.

REPLY [3 votes]: Let $N := a^2 + b^2 + c^2 + d^2$
Note that $bd(a^2 + c^2) = ac(b^2 + d^2)$, while $ac < a^2+c^2$, so that   $$lcm(a^2 + c^2, b^2 + d^2) < (a^2 + c^2)(b^2 + d^2)$$ This tells us that $a^2 + c^2$ and $b^2 + d^2$ are not coprime, and so $$ 1 < gcd(a^2 + c^2, b^2 + d^2)< N $$
Finally this $gcd|N$, which completes the proof.