# Elliptic curves and prime numbers

Let $p_n$ be the $n^{th}$ prime number. Suppose $E(F_{p_n})$ denotes an elliptic curve over the Galois field $GF(p_n)$ which is defined by $y^2=x^3+ax+b$. Is the below claim true?

For each integer number $n>3$, there exist integer numbers $a$ and $b$ such that $\#E(F_{p_n})=p_{n+1}$?

• Why do you believe such a thing is true? – Stanley Yao Xiao Dec 23 '15 at 23:34
• This is stronger than the Riemann Hypothesis – Dror Speiser Dec 23 '15 at 23:34
• I think that with finding such $a,b$, we can generating all prime numbers. – Meysam Ghahramani Dec 23 '15 at 23:38
• @Meysam: I am very confused by your last comment. Note in particular that in your formulation you allow $a$ and $b$ to depend on $n$. If you didn't mean that -- i.e., if you wanted to consider reductions of a fixed elliptic curve defined over $\mathbb{Q}$, the answer is certainly no (it follows from Serre's theorems on the image of the $\ell$-adic Galois representation.) – Pete L. Clark Dec 23 '15 at 23:44
• For each $n$ at first we find such $a,b$, and then we can generate next prime number. – Meysam Ghahramani Dec 23 '15 at 23:57

Theorem (Hasse-Deuring-Waterhouse): For a prime $p$ and $N \geq 1$ the following are equivalent:
(i) There is an elliptic curve $E_{/\mathbb{F}_p}$ such that $\# E(\mathbb{F}_p) = N$.
(ii) We have $|N-(p+1)| \leq 2\sqrt{p}$.
As long as $p > 3$ (i.e., $p = p_n$ for $n \geq 3$) every elliptic curve can be put in "short Weierstrass form" $y^2 = x^3 + ax +b$.
So you are reduced to asking: is it true that for all $n > 3$ we have $|p_{n+1} - (p_n+1)| \leq 2 \sqrt{p_n}$?