Prove that there are $p+1$ points on the elliptic curve $y^2 = x^3 + 1$ over $\mathbb{F}_p$, where $p > 3$ is a prime such that $p \equiv 2 \pmod 3$.

Let $p > 3$ be a prime such that $p \equiv 2 \pmod 3$. Define the elliptic curve $E$ over $\mathbb{F}_p$ by $y^2 = x^3 + 1$. Prove that $E(\mathbb{F}_p)$ consists of $p+1$ points.

Using Fermat's little theorem you can prove that $x^3 + 1$ is a bijection on $\mathbb{F}_p$. Hence, $$\#E(\mathbb{F}_p) = \#\{(x,y) \in \mathbb{F}_p^2 : y^2 = x^3 + 1\} + \#\{\infty\} = \#\{(x,y) \in \mathbb{F}_p^2 : y^2 = x\} + 1.$$ But then I am stuck trying to prove that $y^2 = x$ has $p$ solutions $(x,y) \in \mathbb{F}_p^2$.

Solution 1:

If any of $x,y$ is zero, then $x=y=0$ is the only solution pair. Otherwise, there are $\displaystyle\frac{p-1}2$ quadratic remainders $x$ and for each one exactly $2$ pieces of $y$'s belong ($\pm y$ once $y^2\equiv x$). It's altogether $1+2\cdot\displaystyle\frac{p-1}2=p$.

Or, an even easier approach: for each $y\in\Bbb F_p$ there is exactly one such $x$.

Solution 2:

As in RSA, the equation $c \equiv x^3 \text{mod} \; n$ has exactly one solution for x, namely $x \equiv c^d \; \text{mod}\; n$ there $d = inv(3, \phi(n))$.

The only condition is $1 = \gcd(3, \phi(n)) = \gcd(3, p-1)$. The condition is true since $p \equiv 2 \; \text{mod}\; 3$, since $p - 1 \equiv 1 \; \text{mod} \; 3$ it follows $\gcd(3, p-1) = 1$.

For $y^2 \neq 0$, we get two points for a specific value of $y$. Since $y^2-1$ has $\frac{p-1} 2$ different values, we get $p - 1$ points plus the point for $y = 0$ the point $(p-1, 0)$ and the point at infinity. So, we have a total of $(p - 1) + 1 + 1 = p + 1$ points.