Suppose that -1 is not a square in $\mathbb{Z_p}$. Show that the number of points on the elliptic curve $y^2=x^3+ax$ over $\mathbb{Z_p}$ is $p+1$.
Hint: Use the fact that $x^3+ax$ is an odd function.
I'm not even too sure where to start. From what I understand about a curve like this, is that if a point $(x,y)$ exists on the curve, then a point $(x,-y)$ also exists. So the curve is reflected in the line $y=0$. Since this is in $\mathbb{Z_p}$, we shouldn't run into any such $y$ where $y=-y$, except when $y-0$. So they way I think about it is that there should be an odd number of points, that is an even number (due to the reflection) plus the one point at $(0,0)$. Although I've been staring at this question so long, that I am now mixing up things and trying to make things so basic...
I've also attepted to work with the fact that $\left(\frac{p-1}{p}\right)=-1$, but can't relly see how that fits in properly.