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

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.

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I did once have the opportunity to ask Serre about this. In the brief exchange, he seemed to think there would be no elementary proof of this sort of thing. –  ndkrempel Dec 2 '10 at 4:49
It's also probably worth remarking that you can do a similar thing with $\binom{\frac{p-1}{2}}{\frac{p-1}{3}}$ and another elliptic curve. –  ndkrempel Dec 2 '10 at 4:53
I realised I wasn't completely clear: the fact that I'm asking for a proof of is that this binomial coefficient lies in a small range mod $p$, not the fact that it relates to the number of points on an elliptic curve (which you can prove by an elementary summation trick as Charles Matthews has mentioned.) ... –  ndkrempel Dec 2 '10 at 5:31
... Gerry Myerson has pointed out that the desired fact follows from its relation to the $s$ in $p = s^2 + 4t^2$. I have a vague memory that the usual proof of this relates $s$ to the number of points on the curve, rather than with the binomial coefficient, so this may still not answer the original question. –  ndkrempel Dec 2 '10 at 5:32

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.

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There is an elementary kind of proof based on summations in which this drops out as the non-vanishing term. What that proof does (I learned it from Cassels) is probably to compute the Hasse invariant of the curve. Then by general theory the Hasse invariant is congruent mod p to the number of points. Stringing everything together, I wouldn't say this was a "deep" fact. There is a p-adic theory of Gross-Koblitz analogous to Chowla-Selberg. –  Charles Matthews Dec 2 '10 at 5:12
Yes, thank you. That result was floating in the back of my mind. It may even go back to Legendre or Gauss? In any case, now we have three forms for the same quantity (a binomial coefficient, the number of points on a curve, and the $s$ in $p = s^2 + 4t^2$, and I'm not sure offhand which equalities here are easy and which are harder. –  ndkrempel Dec 2 '10 at 5:24
Thanks for the reference. I probably can't access (or understand) the original German references of Gauss, but I would be interested to see the statements and/or proofs if anyone has them to hand. –  ndkrempel Dec 2 '10 at 5:38
Gauss' collected works are available, in German, at gdz.sub.uni-goettingen.de/dms/load/toc/?PPN=PPN235957348 –  Gerry Myerson Dec 2 '10 at 5:55
Another reference is the text by Ireland and Rosen. –  Robin Chapman Dec 2 '10 at 7:53

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 $y^2=x^3+1$.

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I believe this is related to the quadratic partition $4p=a^2+27b^2$. The sources already cited for the other binomial coefficient don't have anything about this one? –  Gerry Myerson Dec 8 '10 at 11:36
Lemmermeyer indeed cites some of Gauss's works for many other congruences that are very similar to the first one discussed here. But at the moment, I cannot access his collected works on the server of the University of Göttingen. I'll have to try again later. –  Harald Kümmerle Dec 9 '10 at 0:02
Lemmermeyer uses Jacobi sums. Looking for proofs of the other binomial coefficient statement that was given by me and failing to do so, I checked the general techniques involved. Chapter 8 in Ireland-Rosen's book develops a theory that suffices to prove the estimates for the first binomial coefficient (Exercise 8.26), the other one should be tackled similarly, I think. But that is not necessary, as one can directly prove Hasse's Theorem for all $y^2 = x^3 + Dx$, $y^2 = x^3 + D$ for all $F_p$ with $p>3$ and almost arbitary nonzero integers D, using Jacobi sums only! See Chapter 18, §3 and \$4. –  Harald Kümmerle Dec 16 '10 at 2:53