Let $p$ be a prime integer. Show that for each $a ∈ GF(p)$ there exist elements $b$ and $c$ of $GF(p)$ satisfying $a = b^2 + c^2$. Let $p$ be a prime integer. Show that for each $a ∈ GF(p)$ there exist elements $b$ and $c$ of $GF(p)$ satisfying $a = b^2 + c^2$.
I got some ideas like to show the elements of the form $b^2 + c^2$ is asubfield so they are all the field but I dont know how to prove that too, so I ask for some help ;)
 A: Consider the numbers, 
$$ 0^2, 1^2, 2^2, ... , \left( \frac{p-1}{2} \right)^2, c - 0^2, c - 1^2, c-2^2, ... , c - \left( \frac{p-1}{2} \right)^2 $$
The first $(p+1)/2$ are different mod $p$. So are the next $(p+1)/2$. 
By pigieoning, two are equal mod $p$. 
A: This follows from Chevalley-Warning theorem. If $f(X,Y,Z) = X^2+Y^2-aZ^2$, then Chevalley-Warning says that the number of solutions to $f(X,Y,Z)=0$ in $\mathbb F_p^3$ is divisible by $p$. Since there is the trivial solution $f(0,0,0)=0$, this implies that there are at least $p$ solutions, hence at least $p-1$ nontrivial solutions. Argue that one of those solutions must have $Z\neq 0$, so that we can divide by $Z^2$ to get $(X/Z)^2 + (Y/Z)^2 = a$.
Alternatively, you can argue directly that one of the numbers $a-c^2$ is a square mod $p$, as there are $(p+1)/2$ of them but only $(p-1)/2$ quadratic nonresidues.
A: For the quadratic residues, of course, we can just chose $b$ as one of the roots of $a$ and choose $c=0$. In fact we can recast the problem to say that for all $a \in GF(p)$, there exist quadratic residues $q_i,q_j \bmod p$ such that $q_i+q_j\equiv a \bmod p$, and not worry about which root we take. 
The numbers chosen for the remaining $a$ depend on $n :=$ the smallest quadratic non-residue $\bmod p$.  If $n=2$, we can obtain all quadratic non-residues by setting $q_i=q_j$ for each of the quadratic residues $\neq 0$. 
For other values of $n$, we can get $n$ itself by choosing $q_i=1$ and $q_j=n-1$. Then - in the cases I've looked at - we can get all non-residues as sums by successive multiplying $(q_i,q_j)$ by $n-1$. I'm working on showing that $n-1$ effectively generates the quadratic residues, which would allow me to complete the proof.
