Number of points on elliptic curves over finite fields
Hi all, I noticed that for elliptic curves of the form
[I]y[/I][SUP]2[/SUP] ≡ [I]x[/I][SUP]3[/SUP] + [I]a[/I] (mod [I]p[/I]) sometimes the number of points is always [I]p[/I]+1 for any choice of [I]a[/I]. This seems to be the case for all [I]p[/I] ≡ 5 (mod 6). Moreover, when this does not happen, i.e., for [I]p[/I] ≡ 1 (mod 6), it looks like there are exactly zero curves of such form where the number of points is [I]p[/I]+1. Can someone point me towards the right direction as to why this happens? 
[QUOTE=RedGolpe;570326]Can someone point me towards the right direction as to why this happens?[/QUOTE]
[URL]https://en.wikipedia.org/wiki/Hasse%27s_theorem_on_elliptic_curves[/URL] 
I am familiar with Hasse's theorem, but it says nothing about the specific number of points: it only gives a bound.

[URL="https://math.stackexchange.com/questions/875983/solutionstothemordellequationmodulop"]This[/URL] looks like a good answer.

If p == 5 (mod 6) then gcd(3, p1) = 1, so x > x^3 (mod p) is invertible. In fact, x > x^((2p1)/3) (mod p) is the inverse map.
Thus if p == 5 (mod 6), x^3 is "any residue mod p" and x^3 + a is "any residue mod p." If x^3 + a is one of the (p1)/2 quadratic nonresidues (mod p) there are no points (x, y) on the curve y^2 = x^3 + a. If x^3 + a is one of the (p1)/2 nonzero squares (mod p) there are two points (x,y) and (x, y) on the curve. If x^3 + a = 0 (mod p) there is one point (x, 0) on the curve. That makes p points in all. I seem to be missing one point. 
You are just missing the identity point, or the point at infinity.

The key term here is supersingular elliptic curves. For p = 2 mod 3, any curve of the form y^2 = x^3 + B is supersingular, p > 3.
Another common case is p = 3 mod 4, in which case y^2 = x^3 + x is also known to be supersingular. 
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