Prove that $4n+2=x^2+y^2+z^2$ for some odd $x,y$ and even $z$ Show that for all $n\in \mathbb{N}$, exists $x,y,z \in \mathbb{N}$, such that $x,y$ are odd and $z$ is even, such that $4n+2=x^2+y^2+z^2$.
I started by using the fact that every natural number has a decomposition to a sum of 4 square, and tried to prove that one of the numbers is zero, basically $4n+2=x^2+y^2+z^2+w^2$. 
Firstly, I falsely assumed that all of those numbers weren't 0, and then I proved that at least one of those numbers has to be odd, because if they were all even then $4n+2=x^2+y^2+z^2+w^2 \equiv 0 (mod 4)$, contradicting the fact that $4n+2\equiv 2 (mod 4)$. Then I showed that since $4n+2$ is even then there have to be 2 even numbers exactly and 2 odd numbers exactly among $x,y,z,w$. 
From here I assumed that $x,y$ are odd, and that $z,w$ are even and tried to prove that at least one of $z,w$ is 0, but I didn't really have ideas in this direction, so this didn't work out. 

If x and y are odd while z is even, then you can set $x = 2a+1, y = 2b+1, z = 2c$
$4n+2 = (2a+1)^2+(2b+1)^2+(2c)^2 = 4a^2 + 4a+1+4b^2+4b+1+4c^2 = 4(a^2+a+b^2+b+c^2)+2$
Now you're trying to prove that every possible $n$ can be described as $n = a^2+a+b^2+b+c^2 = a(a+1) + b(b+1) + c^2$
Note that $a(a+1)+b(b+1) \equiv 0 (mod 2)$
Does this help?
