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 $.
 A: 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: 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.
A: Suppose not. Note $p:=a^2+b^2+c^2+d^2=(a+b)^2+(c-d)^2=(a-b)^2+(c+d)^2$. That is, we expressed $p$ in two ways as a sum of two squares.
But since $p$ is prime, it can be expressed as the sum of two squares in at most one way, up to interchanging the numbers. This corresponds to the fact $p$ has a unique prime factorization in the ring of Gaussian integers $\mathbb{Z}[i]$: $p=(a+bi)(a-bi)=a^2+b^2$.
Therefore either $a+b=a-b$ and $c-d=c+d$, which is impossible, or $a+b=c+d$ and $c-d=a-b$ which implies $a=c$ and $b=d$ and therefore $2|p$. Contradiction because $p>2$.
A: 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
:)
A: Quick and dirty proof:
We start with the information given.
a^2 + b^2 + c^2 + d^2 = k such that k is prime (we will contradict this)
ab = cd
We rewrite the terms of k in the following way:
(cd/b)^2 + b^2 + (ab/d)^2 + (ab/c)^2
= (c^2 * d^2)/b^2 + b^2 + (b^2)(a/d)^2 + (b^2)(a/c)^2 [1]
The last three terms in [1] are multiples of b^2.
We have already shown, in those last three terms, that 
c^2 and d^2 are both multiples of b^2.
Thus the first term of [1] is a multiple of b^4, since we the product of c^2 and d^2 in the numerator.
Dividing by b^2 leaves it as a multiple of b^2.
Hence all terms in k are a multiple of b^2 and k cannot be a prime.
PS. I have no idea how to format equations on here. Truly your forgiveness I implore.
PSS. If anyone notices a mistake please let me know!!!
