# Find the sum of all quadratic residues modulo $p$ where $p \equiv 1 \pmod{4}$

I read one theorem in the book, they said there will be exactly $\dfrac{p-1}{2}$ quadratic residues of $p$. So for each $i$, $$x^2 \equiv a_i \pmod{p} \text{ where } 1 \leq i \leq p - 1$$

But if we sum all $a_i$, then what does this sum equal to? $$\sum_{i=1}^{\frac{p-1}{2}}a_i = ?$$

I haven't used the condition that $p \equiv 1 \pmod{4}$, so I think I missed one important point here. Any idea?

Update
The original problem
Let $p$ be a prime such that $p \equiv 1 \pmod{4}$.
Prove that the sum of those numbers $1 \leq r \leq p - 1$ that are quadratic residues modulo $p$ is $\dfrac{p(p-1)}{4}$.

Thanks,

-

Because $p\equiv 1\bmod 4$, we have that $-1$ is a quadratic residue modulo $p$. The product of two quadratic residues is a quadratic residue. This means that for any quadratic residue $a$, we have that $-a$ is also a quadratic residue. Thus the sum cancels to $0\bmod p$.
Because there are $\frac{p-1}{2}$ quadratic residues mod $p$, and they all occur in pairs $a$ and $p-a$, there are $\frac{p-1}{4}$ pairs each of whose sum is $p$, hence the sum of them all is $\frac{p(p-1)}{4}$.
@Zev Chonoles: Thank you. However, the question asked to prove it equals to $\dfrac{p(p-1)}{4}$. I'm confused now. – Chan Apr 4 '11 at 2:23
@Chan - my apologies then, I had incorrectly assumed the sum was to be evaluated modulo p. However, the stated answer is not in contradiction with my answer - note that $\frac{p(p-1)}{4}\equiv 0\bmod p$. – Zev Chonoles Apr 4 '11 at 2:27