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.

  • $\begingroup$ Thanks! (Yes, I am working with $p\geq 5$, I should have mentioned that) $\endgroup$ – 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 $p\ge5$)

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  • $\begingroup$ 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 $Y^2=X^3+ac^2X+bc^3$ $\endgroup$ – reuns Jun 3 at 4:21
  • $\begingroup$ Can you recommend a reference? $\endgroup$ – jinfosec Jun 3 at 12:03

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