Most even numbers is a sum $a+b+c+d$ where $a^2+b^2+c^2=d^2$ I have no idea how hard this conjecture is to prove:
Any even number $n\ge 36$ can be written as $n=a+b+c+d$ where $a^2+b^2+c^2=d^2$ and $a,b,c,d\in\mathbb Z^+$.
Small exceptions are $n=2, 4, 6, 10, 12, 14, 20, 26, 34.\,$ Tested for $n\le 10,000$.
A counter-example would be as interesting as a proof.

$n$ as above must be even
$d-c,c+d|a^2+b^2$

$$a^2+b^2+c^2=d^2$$
$$a=2ps$$
$$b=2ks$$ 
$$c=s^2-p^2-k^2$$ 
$$d=s^2+p^2+k^2$$
$$2n=2ps+2ks+s^2-p^2-k^2+s^2+p^2+k^2$$ 
$$qt=n=s(p+k+s)$$ 
Lay on multipliers and pick the right.
Or other item.
$$a=2s(p-k)$$
$$b=2s(p+k)$$
$$c=p^2+k^2-2s^2$$
$$d=p^2+k^2+2s^2$$
$$2n=2p^2+2k^2+4ps$$
$$p(p+2s)=n-k^2$$
It remains to try all possible  $k$  . That expression was greater than $0$.
