Average value of the orders of all elliptic curves over the finite field of p-elements

Is true that the average value of the orders of all elliptic curves over $$\mathbb F_p$$ is $$p+1$$?

More precisely, fix a prime $$p$$ and let $$\mathbb F_p$$ be the field of $$p$$ elements. Consider the set $$S=\{(a,b)\in\mathbb F_p\times\mathbb F_p \,:\, 4a^3+27b^2\neq 0\,\, \text{in}\,\,\mathbb F_p\}$$, so that every element $$(a,b)\in S$$ defines an elliptic curve $$E(a,b,p)=\{(x,y)\in\mathbb F_p\times\mathbb F_p \,:\,y^2 =x^3+ax+b\,\, \text{in}\,\, \mathbb F_p\}\cup\{\infty\}$$ Is it true that $$\frac{1}{|S|}\sum_{(a,b)\in S}|E(a, b, p)| = p + 1$$ where $$|S|$$ and $$|E(a, b, p)|$$ denote the orders of these sets.

I have verified it through computation for some small primes and am just wondering if it is true in general.

• Thanks! (Yes, I am working with $p\geq 5$, I should have mentioned that) – jinfosec Jun 3 at 2:39

Yes. If $$E$$ defined by $$y^2=x^3+ax+b$$ has $$p+1-t$$ elements, then the curve $$E_c$$ defined by $$y^2=x^3+ac^2x+bc^3$$ has $$p+1-t$$ points if $$c$$ is a quadratic residue modulo $$p$$ but $$p+1+t$$ points if $$c$$ is a quadratic non-residue. (In the first cases $$E_c$$ is isomorphic to $$E$$, in the second case it's the "quadratic twist" of $$E$$). Averaging out over the $$E_c$$ gives $$p+1$$ points.
Overall the set $$S$$ splits up into various sets of $$E_c$$s each one average with $$p+1$$ points.
(I'm assuming here $$p\ge5$$)
• The quadratic twist is $dy^2=x^3+ax+b$ with $d$ not a square, with $y=dY,x=dX,c=d^{-1}$ we get $Y^2=X^3+ac^2X+bc^3$ – reuns Jun 3 at 4:21