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Let $E$ be the elliptic curve $y^2=x^3+1$ and $p \equiv 1 \pmod{3}$ a prime. Computing the number of points mod $p$ of $E$ using the naive method gives: $$ \#E(\mathbb F_p) = 1+ \sum_{x=0}^{p-1} \left( 1 + \left(\frac{x^3+1}{p}\right)\right) \equiv p+1 + \sum_{x=0}^{p-1}(x^3+1)^{\frac{p-1}2}\pmod{p}$$ developing the right hand side and simplifying there is only one term of the sum surviving and we get: $$ \#E(\mathbb F_p) \equiv p+1-\binom{\frac{p-1}2}{\frac{p-1}3}\pmod{p}$$ Hasse theorem now implies that if $p$ is large enough (say $p>16$) then the minimal residue of $$ \binom{\frac{p-1}2}{\frac{p-1}3}\pmod{p}$$ is smaller than $2\sqrt{p}$. As this last result seems completely unrelated to elliptic curves my question is if there is a more direct proof of this fact not using Hasse's theorem.

Using instead the elliptic curve $y^2 = x^3+x$ and a prime $p \equiv 1 \pmod{4}$ one finds the similar result that the least minimal residue of $$ \binom{\frac{p-1}2}{\frac{p-1}4}\pmod{p} $$ is also bounded in absolute value by $2\sqrt{p}$.

My motivation is simple curiosity, I fell on it playing with some examples and now I can't take it out of my head.

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    $\begingroup$ You don't need the full strength of Hasse's theorem for this, because the elliptic curves you consider have complex multiplication by the cyclotomic rings $\mathbb{Z}[\omega]$ and $\mathbb{Z}[i]$, respectively. Because of that, the sums you mention are essentially the real parts of Jacobi sums: e.g. in the first case, the Jacobi sum is $\sum_{a \mod p}χ(a)ψ(1-a)$, where χ is the quadratic residue character, and ψ is a nontrivial cubic character. Jacobi sums have absolute value equal to √p, as can be shown by explicit calculation. Ireland & Rosen is a good reference. $\endgroup$ – Alison Miller Apr 20 '14 at 4:31
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    $\begingroup$ Also: if p=A^2+3B^2 (which can be done uniquely for any p that is 1 mod 3), the remainder you are getting is equal to 2A. This can be proved using the Jacobi sum mentioned above, which is equal to A + B √-3. $\endgroup$ – Alison Miller Apr 20 '14 at 4:45
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    $\begingroup$ The second result you mention goes back to Gauss, and is related to writing $p$ as a sum of two squares. Jacobi seems to have discussed identities related to the first congruence, which as Alison Miller notes above is related to $x^2+3y^2$. This paper by Hudson and Williams discusses these and other such congruences using Jacobi sums: ams.org/journals/tran/1984-281-02/S0002-9947-1984-0722761-X/… $\endgroup$ – Lucia Apr 20 '14 at 5:11
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This is really just my comment above, but maybe it's a reasonable answer. Congruences between certain binomial coefficients and the representations of primes by some quadratic forms go back to Gauss and Jacobi. The two examples given in the question and a number of other examples are discussed in the paper Binomial coefficients and Jacobi sums by Hudson and Williams (which also gives references to the original works of Gauss, Jacobi and others). One main idea is to work out congruences for binomial coefficients in terms of Jacobi sums (Theorem 5.1 of the paper). Here are a couple more interesting facts from the paper:

If $p\equiv 1 \pmod{7}$ then $$ \binom{3(p-1)/7}{(p-1)/7} \equiv -2x \pmod{p}, $$ where $p=x^2+7y^2$. This is due to Jacobi. Also if $p\equiv 1 \pmod{12}$, then $\binom{(p-1)/2}{(p-1)/12}$ is $\pm \binom{(p-1)/2}{(p-1)/4} \pmod{p}$ and the sign can be determined by looking at the representation of $p$ as the sum of two squares. (Corollary 4.2.2 of the paper.)

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