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:



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> 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?
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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.



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


