Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $(x,y) \in \mathbb{F}_p^2$.

share|improve this question
    
How this is true :$$\#\{(x,y) \in \mathbb{F}_p^2 : y^2 = x^3 + 1\} + \#\{\infty\} = \#\{(x,y) \in \mathbb{F}_p^2 : y^2 = x\} + 1.$$ –  curious Dec 3 '13 at 11:47
    
@curious Consider the equations $y^2 = x^3 + 1$ and $y^2 = z$ over $\mathbb{F}_p$. The map $(x,y) \mapsto (z = x^3 + 1,y)$ is a bijection between their respective solution sets, because $x \mapsto x^3 + 1$ is a bijection on $\mathbb{F}_p$. In particular, the equations have the same number of solutions. –  Ricardo Buring Dec 3 '13 at 19:08

1 Answer 1

up vote 4 down vote accepted

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 $x$.

share|improve this answer
    
Your easy approach is nice. Thank you. –  Ricardo Buring Nov 11 '12 at 14:33

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.