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 $\#E(F\_{p\_n})=p\_{n+1}$?



Elliptic curves and prime numbers
(Sorry, I misread the question at first.) The following result reduces your question to a problem of analytic number theory:


**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 
$|p\_{n+1} - (p\_n+1)| \leq 2 \sqrt{p\_n}$?


According to [this esteemed source](https://en.wikipedia.org/wiki/Prime_gap#Upper_bounds), having this kind of upper bound on the prime gap (always) is conjectured to be true but far from being proven. In fact, if I have it right this precisely [Andrica's Conjecture](https://en.wikipedia.org/wiki/Andrica's_conjecture). As Dror Speiser points out, it is not even known conditionally on the Riemann Hypothesis. 


