2
$\begingroup$

In article 23 of his first memoir on biquadratic residues, Gauss gave the first example of a binomial-type congruence, which is apparently a quite "deep" result - many authors relate it to the theory of integral points on certain elliptic curves. His result is stated as follows: if $p=4n+1$ is a prime, then $p=a^2+b^2$ for some integers $a,b$ (this is Fermat's two squares theorem). Gauss showed that if $a$ is fixed by $a\equiv 1 \pmod{4} $ then

$$2a\equiv{2n \choose n}\pmod {p}$$ or $2a\equiv{\frac{p-1}{2} \choose \frac{p-1}{4}}\pmod {p}$.

In his treatise on Gauss's number theoretic investigations, Paul Bachmann remarks that other isolated theorems of this kind were also known to Gauss, and refers to p. 39 of volume 10-1 of his collected works. Looking there, I found the following statements of Gauss:

If $p=3n+1=x^2+3y^2$ is a prime number, then $$2x\equiv {\frac{p-1}{2}\choose \frac{p-1}{6}} \pmod {p}$$ Under the same condition, if $4p=l^2+27m^2$ then $$l\equiv - {\frac{2(p-1)}{3}\choose \frac{p-1}{3}}\pmod{p}.$$ If a prime number is $p=8n+1 = a^2+2b^2,$ then $$2a\equiv {\frac{p-1}{2}\choose \frac{p-1}{8}} \pmod {p},$$ and if $p=8n+3=a^2+2b^2$ then $$2a \equiv > {\frac{p-1}{2}\choose\frac{p-3}{8}} \pmod {p}$$

While his published binomial congruence is quite referenced in the literature (it is often refered to as the "classical" example of such congruences), I have not found discussions of the unpublished results from his Nachlass. Are these unpublished congruences not discussed just because they are based on similar principles as his published one? and are their derivations of similar complexity? whatever the answer is, I would like to know on what principles they rest (but only in general terms, as I am not a mathematician and right now I cannot read into the details).

Any useful comments/references will be appreciated!

$\endgroup$
2
  • $\begingroup$ Your condition $a \equiv 1 \bmod p$ doesn't make sense, e.g., $13 = 2^2 + 3^2$ so $a$ can only be $\pm 2$ or $\pm 3$, and none are $1 \bmod 13$. Perhaps you meant to say something about $a \bmod 4$. Whatever you mean, please check it with some actual examples ($p = 5, 13, 17, 29$) to make sure your update is correct. $\endgroup$
    – KCd
    Dec 2, 2023 at 16:55
  • 1
    $\begingroup$ Ohh sorry - I meant $a \equiv 1 \pmod {4}$, but somehow I did not notice and wrote "mod p" instead of "mod 4". $\endgroup$
    – user2554
    Dec 2, 2023 at 17:02

1 Answer 1

4
$\begingroup$

I have seen papers and books on Gauss and Jacobi sums discuss such results, e.g., see https://core.ac.uk/download/pdf/82122545.pdf.

The book "Gauss and Jacobi sums" by Berndt, Evans, and Williams has some results like this too: see page 2 in the introduction and look up "binomial coefficients" in the index. They also related use the Gross-Koblitz formula for $p$-adic Gauss sums to improve some congruences to modulus $p^2$.

Congruences like this appear in some exercises in Ireland & Rosen's A Classical Introduction to Modern Number Theory 2nd ed. (exer. 27 in Chapter 8, exer. 39 & 40 in Chapter 9)

$\endgroup$
3
  • $\begingroup$ [+1] Thanks for your searching effort, but I already saw this article, and immediately noticed that it is too technical for me... I am more interested in explanations that give general perspective - in particular, on what ideas these congruences rest? Do the particular cases of binomial congruences differ in complexity? $\endgroup$
    – user2554
    Dec 2, 2023 at 17:17
  • 2
    $\begingroup$ The idea they rest on is getting a formula for those numbers using Gauss and Jacobi sums and then working out an expression for the real or imaginary part of such a character sum in a second way that brings in binomial coefficients. It is technical. I don't think there is a way to avoid that when seeking some general perspective. When $p\equiv 1 \bmod 4$ or $p \equiv 1 \bmod 3$ or $p \equiv 1 \bmod 8$, you'll be using multiplicative characters on the integers mod $p$ with order $4$ or $3$ or $8$. $\endgroup$
    – KCd
    Dec 2, 2023 at 17:27
  • 2
    $\begingroup$ As an example, say $p\equiv 1 \bmod 4$. Then $4\mid (p-1)$, so there is a multiplicative character $\chi$ of order $4$ on $\mathbf Z/p\mathbf Z$ and the Jacobi sum $J(\chi,\chi)$ is in $\mathbf Z[i]$ with $|J(\chi,\chi)|^2 = p$, so when we write $J(\chi,\chi)=a+bi$ with integer real and imaginary parts, we have $a^2+b^2 = p$. Analyzing the real part $a$ of $J(\chi,\chi)$ could then give an expression for $a \bmod p$. This viewpoint is related to decomposing $p$ in rings like $\mathbf Z[i]$ or $\mathbf Z[\sqrt{-2}]$ and expressing factors of $p$ in those rings as Jacobi sums. It's technical. $\endgroup$
    – KCd
    Dec 2, 2023 at 17:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.