20210128, 11:34  #1 
Aug 2006
Monza, Italy
73 Posts 
Number of points on elliptic curves over finite fields
Hi all, I noticed that for elliptic curves of the form
y^{2} ≡ x^{3} + a (mod p) sometimes the number of points is always p+1 for any choice of a. This seems to be the case for all p ≡ 5 (mod 6). Moreover, when this does not happen, i.e., for p ≡ 1 (mod 6), it looks like there are exactly zero curves of such form where the number of points is p+1. Can someone point me towards the right direction as to why this happens? Last fiddled with by RedGolpe on 20210128 at 11:46 Reason: more info 
20210128, 11:58  #2  
Dec 2012
The Netherlands
711_{16} Posts 
Quote:


20210128, 12:06  #3 
Aug 2006
Monza, Italy
1001001_{2} Posts 
I am familiar with Hasse's theorem, but it says nothing about the specific number of points: it only gives a bound.

20210128, 13:33  #5 
Feb 2017
Nowhere
2^{2}×3^{2}×173 Posts 
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. Last fiddled with by Dr Sardonicus on 20210128 at 13:44 Reason: xifnig posty 
20210128, 14:03  #6 
Aug 2006
Monza, Italy
73 Posts 
You are just missing the identity point, or the point at infinity.

20210129, 22:29  #7 
Oct 2007
2×53 Posts 
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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