$\sum_{x=0}^{p-1}(\frac{ax^2+bx+c}{p})=-(\frac{a}{p})$, for $p\geq5$ an odd prime with $p \nmid b^2-4ac$, where $(\frac{a}{p})$ is the Legendre symbol

Question

Let $\left(\dfrac{a}{p}\right)$ denote the Legendre symbol and let $a,b,c$ be integers integers and $p \geq 5$ an odd prime number. Prove that $$\sum_{x=0}^{p-1}\left(\dfrac{ax^2+bx+c}{p}\right) = -\left(\dfrac{a}{p}\right)$$ if $p \nmid b^2-4ac$.

I wasn't sure how to prove this using the properties of Legendre symbols. How do we deal with the sum of Legendre symbols?

Answer 1

If $p|a$, then $p\nmid b$ since $p\nmid b^{2}-4ac$. In this case, $\{bx+c\}_{x=0}^{p=1}$ forms complete residue system mod $p$, so both side became $0$.

For $p\nmid a$, Multiply both side by $\left(\frac{4a}{p}\right)$, then it is equivalent to $$ \sum_{x=0}^{p-1}\left(\frac{4a(ax^{2}+bx+c)}{p}\right)=\sum_{x=0}^{p-1}\left(\frac{(2ax+b)^{2}+(4ac-b^{2})}{p}\right)=-1$$

Since $p\nmid 2a$, $\{2ax+b\}_{x=0}^{p-1}$ forms complete residue system, so we can rewrite as $$ \sum_{X=0}^{p-1} \left(\frac{X^{2}+A}{p}\right)=-1$$ where $p\nmid A=4ac-b^{2}$. You can see proof here for this identity.

Answer 2

If $p\mid a$, then both the sides are $0$ as $\{bx+c\}_{x=0}^{p-1}$ forms a complete residue system mod $p$. So, let us assume $p\nmid a$.

Let $ f(x) = ax^2 + bx + c $ $ \Rightarrow \left( \frac{f(x)}{p} \right) \equiv f(x)^{\frac{p-1}{2}} \equiv a_0 + a_1x + \dots a_{p-1}x^{p-1} ( \mbox{ mod } p \ ) $

Thus\begin{align*} \sum_{x=0}^{p-1} \left( \frac{f(x)}{p} \right) & = \sum_{x=0}^{p-1} f(x)^{\frac{p-1}{2}} \\ & \equiv \sum_{x=0}^{p-1} \sum_{j=0}^{p-1} a_j x^j ( \mbox{ mod } p \ ) \\ & \equiv \sum_{j=0}^{p-1}a_j \sum_{x=0}^{p-1} x^j ( \mbox{ mod } p \ ) \\ & \equiv -a_{p-1} \equiv -a^{\frac{p-1}{2}} \equiv - \left( \frac{a}{p} \right) ( \mbox{ mod } p \ ) \end{align*}

The third last congruence follows from the fact that defining $ S_n = \sum_{x=0}^{p-1} x^n $, we know that $ S_n \equiv -1 ( \mbox{ mod } p \ ) $ if $ p-1 \ | \ n $ and $ S_n \equiv 0 ( \mbox{ mod } p \ ) $ otherwise.

Also, note that proving $$\sum_{x=0}^{p-1} \left( \frac{f(x)}{p} \right)\equiv - \left( \frac{a}{p} \right) ( \mbox{ mod } p \ )$$ is enough as the only values the Legendre symbol takes are $1$ and $-1$.