The number of solutions to $y^2\equiv x^3+nx\pmod p$ is exactly $p$ 
Question
Suppose $p\equiv3\pmod4$ and $n\in\mathbb Z$, is the number of solutions to $y^2\equiv x^3+nx\pmod p$ exactly $p$? I have numerically confirmed it when $|n|\le10$ and $p\le229$.

Thoughts
Fix an $x$, the number of solutions to $y^2\equiv x^3+nx\pmod p$ is $$1+\left(\frac{x^3+nx}p\right),$$
Summing this formula from $0$ to $p-1$, the question boils down to proving $$p+\sum_{x=0}^{p-1}\left(\frac{x^3+nx}p\right)=p$$ or $$\sum_{x=0}^{p-1}\left(\frac{x^3+nx}p\right)=0.$$
Then I have difficulty going further.
Here $\displaystyle\left(\frac\cdot\cdot\right)$ denotes the Legendre symbol.

Note that $$\sum_{x = 0}^{p -1} \left(\frac{x^3 + nx}{p}\right) = \sum_{x = 1}^{(p -1)/2} \left(\frac{x^3 + nx}{p}\right) + \left(\frac{(-x)^3 -nx}{p}\right)$$ and $$\left(\frac{(-x)^3 -nx}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{x^3 +nx}{p}\right) = - \left(\frac{x^3 + nx}{p}\right),$$ since $p \equiv 3 \pmod 4$.
