Direct proof of special case of Hasse's theorem for elliptic curves

Question

Consider the elliptic curve $y^2 = x^3 + x$ over $\mathbb{F}_p$, where $p \equiv 1 \pmod 4$.

If memory serves correctly, the number of points (excluding the point at infinity) is $p - a$ where $a$ is the residue of the binomial coefficient $\binom{\frac{p-1}{2}}{\frac{p-1}{4}}$ modulo $p$ of smallest absolute value.

Therefore, Hasse's bound implies that $a \leq 2\sqrt{p}$, which for large $p$ is quite a strong statement about a number you might otherwise expect to be anything mod $p$.

My question is:

Is there a direct simple proof of this fact about $\binom{\frac{p-1}{2}}{\frac{p-1}{4}}$ being small mod $p$? Could one try to unwind the proof of Hasse's theorem or more generally a proof of the Riemann hypothesis for curves to get such a proof?

On a related sidenote: of course the $(\frac{p-1}{2})!$ occuring here is a primitive fourth root of unity. The other relevant term is $(\frac{p-1}{4})!$. This leaves me to wonder if there is any useful connection with the $p$-adic $\Gamma$ function evaluated at $\frac{1}{4}$, and hence with some kind of $p$-adic analogue of the Chowla-Selberg formula...? (I don't know anything about this, so I could be clutching at straws here.) If there is anything interesting to say about this part, I should perhaps make it a separate question.

Answer 1

I think it is known, and elementary, that if $p\equiv1\pmod4$, then $p=s^2+4t^2$, where $s\equiv1\pmod4$ and $2s\equiv{(p-1)/2\choose(p-1)/4}\pmod p$. Tom Storer makes use of this result in his book, Cyclotomy and Difference Sets, but it goes back farther than that. I'll try to find a good reference.

EDIT. It goes back to Gauss. Dickson's History, Volume 2, page 234: C F Gauss stated that, if a prime $p=4k+1$ is expressed in the form $e^2+f^2$, $e$ odd, $f$ even, then $\pm e$ and $\pm f$ equal the minimum residues (i.e., between $-p/2$ and $+p/2$) modulo $p$ of $(1/2)r/k!$ and $(1/2)r^2$, respectively, where $$ r=(k+1)(k+2)\cdots(2k). $$ The references given are Gott gelehrte Anz 1, 1825; Comm soc sc Gott recent 6, 1828; Werke II 1863, 168,90-1; Cf Bachmann, Kreisteilung, Ch X.

FURTHER EDIT. Found a source, readily available and in English, with a proof that the binomial coefficient gives you twice the odd part of the quadratic partition of $p$ (and is therefore less than $2\sqrt p$ in absolute value). It's Corollary 6.6 on page 192 of Reciprocity Laws: From Euler to Eisenstein, by our very own Franz Lemmermeyer. Google Books has it, you can probably find it there by searching for Gauss's congruence.

Answer 2

Hasse bound for elliptic curves of the form $y^2=x^3+b$ can be proven without algebraic geometry. It was done by Schrutka in the article Ein Beweis für die Zerlegbarkeit der Primzahlen von der Form $6n +1$ in ein einfaches und ein dreifaches Quadrat. For the curves of the form $y^2=x^3+ax$ the same method was used earlier by Jacobsthal in Über die Darstellung der Primzahlen der Form $4n+1$ als Summe zweier Quadrate. Now corresponding sums are known as Jacobsthal sums and Schrutka sums.

Jacobsthal sums $$J(a)=\sum\limits_{x=0}^{p-1}\left(\dfrac{x^3+ax}{p}\right)$$ have following properties. For $p=4k+3$ they are identically zeroes. For $p=4k+1$ we have

1. $J(0)=0$;

2. $J(a)$ is even for all $a$;

3. $J(ak^2)=\left(\frac{k}{p}\right)J(a)$, in particular, if $\left(\frac{a}{p}\right)=\left(\frac{b}{p}\right)$ then $|J(a)|=|J(b)|$;

4. if $\left(\frac{r}{p}\right)=1$ and $\left(\frac{n}{p}\right)=-1$, then $$\left(\frac{J(r)}{2}\right)^2+\left(\frac{J(n)}{2}\right)^2=p;$$

5. ${J(-1)}/{2 }\equiv (p-3)/ 2\pmod{ 8};$

6. $J(-a)=\left(\frac{2}{p}\right)J(a).$

The forth properiy implies Hasse bound for elliptic curves of the form $y^2=x^3+ax.$ Also it works for the curves $y^2=x^4+a$, because for $a\ne 0$ $$ \sum_{t=0}^{p-1}\left(\frac{t^{4}+a}{p }\right)=J(a)-1. $$

Similar answer is given to the question Congruence for the number of points in the elliptic curve $y^2 = x^3+b \pmod{p}$.

Answer 3

I am looking for elementary proofs of (special cases) of Hasse's theorem. The above discussion was of great help for me in the stated case, but does somebody know where to find a proof for $\binom{\frac{p-1}{2}}{\frac{p-1}{3}}$ being small mod p?

This should correspond with Hasse for the curve $y^2=x^3+1$.