Number of points on $Y^2 = X^3 + A$ over $\mathbb{F}_p$

Question

Let $p\equiv 2\pmod{3}$ be prime and let $A\in\mathbb{F}^{∗}_p$ . Show that the number of points (including the point at infinity) on the curve $Y^2 = X^ 3 + A$ over $\mathbb{F}_ p$ is exactly $p + 1$

I'm having trouble bring in the fact that $p \equiv 2 \pmod 3$

Any help greatly appreciated!

@tc1729 How do you know it's $p \equiv 1$ rather than $p \equiv 2$? Let's see if the OP will clarify. — Erick Wong, Oct 23 '13 at 16:00

Jyrki Lahtonen · Accepted Answer · 2013-10-23T18:49:47.507

2

The multiplicative group of $\Bbb{F}_p$ is cyclic of order $p-1$. As $p\equiv2\pmod3$, this is not a multiple of three. Therefore cubing is a bijection from $\Bbb{F}_p$ to itself. Hence for each $Y$ ($p$ choices) there is a unique solution $X\in\Bbb{F}_p$, namely the cube root of $Y^2-A$. Including the point at infinity gives a total of $p+1$ solutions.

edited Oct 23 '13 at 18:49

answered Oct 23 '13 at 16:28

Jyrki Lahtonen

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Number of points on $Y^2 = X^3 + A$ over $\mathbb{F}_p$

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