# counting points not on elliptic curve [closed]

Given an curve with equation $y^2=x^3+ax+b$, I want to find the number of pairs $(a,b)\in \mathbb{F}_p \times \mathbb{F_p}$ NOT on the curve. How do I do it?

I have an intuition that it is $p$, but don't know how to rigorously show it.

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## closed as unclear what you're asking by CodesInChaos♦, e-sushi♦, otus, poncho, fgrieuOct 22 '14 at 22:15

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Your notation is confusing since you use $a,b$ as curve parameters and as the coordinates of a point. – CodesInChaos Dec 5 '13 at 17:01
Actually, it is still confusing since the notation $E({\mathbb F}_p)$ is typically used for denoting the set of points $(r,s)\in {\mathbb F}_p^2$ satisfying the curve equation $E$ (plus the point at infinity). – DrLecter Dec 5 '13 at 17:07
Do you mean: Given an elliptic curve with equation $y^2=x^3+ax+b$, and i want to find the number of points $(r,s)\in \mathbb{F}_p^2\cup\{\infty\}$ NOT on the curve? – figlesquidge Dec 5 '13 at 17:12
i believe you get the idea of the problem(Basically to prove there are $p$ points which belonging to the field not on the curve)? Could you tell me the technique to solve it? – user12345 Dec 5 '13 at 17:12
Right, the newest interpretation of the question doesn't make sense again. Are you asking how many pairs $(a,b)$ there are such that $E$ is an elliptic curve? – figlesquidge Dec 5 '13 at 18:19

At this point in time the question asked how many points were there that were not on the curve

Whats wrong with doing $p^2 +1 - |E(\mathbb{F}_p)|$?

Taking an example from wikipedia (who reference #1):

The curve defined by $$y^2 = x^3 - x$$ over $\mathbb{F}_{71}$ has $72$ points ($71$ affine points including $(0,0)$ and one point at infinity)

So there are $70*71$ points not on the curve - far more than just $71$.

Ref #1: Koblitz, Neal (1994). "Chapter 6". A Course in Number Theory and Cryptography. Graduate Texts in Mathematics 114 (2nd ed.). Springer-Verlag. ISBN 0-387-94293-9)

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i think the question actually meant (a,b) and not (r,s) if it was (r,s) i think it is trivial as u say! – user12345 Dec 5 '13 at 17:48
Right :/ Well I've Commented on the question, and will leave this answer in case the question changes back again! – figlesquidge Dec 5 '13 at 18:20