Special quadratic reciprocity? $(-3/p)_2$ and $(5/p)_2$ in addition to $(-1/p)_2$ and $(2/p)_2$?

Question

Context: since $\mathbb Z/p^\times$ is cyclic for $p$ prime, $-1$ is a square mod $p$ if and only if $(-1)^{{p-1\over 2}}=1$. A significantly subtler, but still classical, case is determining the quadratic symbol $(2/p)_2$: the almost-universal argument evaluates $(1+i)^p$ in two different ways... The rest of the simplest proof I know for quadratic reciprocity (over $\mathbb Q$, anyway) uses Gauss sums. Ok.

But: just a few days ago, on MathStackExchange, someone (and I'd give credit, but I've lost the link...) remarked that there is also a "special" method for $(-3/p)_2$... but without further comment. Indeed, since $\omega+\omega^{-1}=\sqrt{-3}$ (with $\omega$ a primitive cube root of unity), without understanding the general version of the Gauss sum argument, on one hand $(\omega+\omega^{-1})^p=\omega^p+\omega^{-p}$ in $\mathbb Z[\omega]\mod p$. On the other hand, it's $$ (\sqrt{-3})^{p-1}\cdot \sqrt{-3}\;=\;3^{{p-1\over 2}}\cdot \sqrt{-3}\;=\; (-3/p)_2\cdot \sqrt{-3} $$ So, indeed, $(-3/p)_2={\omega^p-\omega^{-p}\over \sqrt{-3}}$.

Of course, the idea being exploited is that we have "geometric" ways to understand that certain square roots lie in cyclotomic fields. And, yes, an analogous thing works for $(5/p)_2$... It is charming! And, yes, the $(2/p)_2$ argument can be rewritten in a slightly less obscure analogous fashion, using $\omega+\omega^{-1}=\sqrt{2}$, for $\omega$ an $8$th root of unity.

My question is: while I do know that people knew about $(-1/p)_2$ before Gauss, ... what about these others? The most plausible is $(2/p)_2$, but my superficial looking around (and superficial memory) gives nothing on this. Did people know $(-3/p)_2$? $(5/p)_2$?

Yes, these can be viewed as simple precursors of the Gauss sums argument... one main point of which is proof that $\sqrt{\pm p}$ lies in a suitable cyclotomic field... But, without that hindsight, hadn't Lagrange or even Euler observed some of these special cases?

Of course, if Franz Lemmermeyer happens to be looking here, I'd imagine that he knows very well, but I think he does not visit this site, unfortunately. :)

Answer 1

Franz Lemmermeyer was kind enough to send me some information by email, which I quote in part:

"Euler as well as Lagrange could indeed determine the quadratic characters of $2$, $-2$, $-3$ and $3$, mainly using cyclotomic techniques in disquise. In particular they could write the cyclotomic polynomial Phi_p(x) in the form $Y^2 -p^* Z^2$ for $p^* = -3, 5, -7$ and $-11$. I covered Euler's contributions in the edition of his correspondence with Goldbach here: https://edoc.unibas.ch/58842/ Everything else on the quadratic character of small primes is post-Gauss; I'll attach my copy of Cooper's thesis as a pdf file, where Chapter 2 covers these things."

As Franz explains in the preface to that edition of Euler's correspondence with Goldbach, Cooper was a student or protege of L. Dickson at Chicago, and the absence of (history of) quadratic reciprocity from Dickson's Number Theory volumes is probably due to a failed expectation that Cooper would do a good write-up of it. That part is treated in chapter 2 of Cooper's thesis, but apparently it did not meet Dickson's writing standards. :)

(Thanks again to Franz L.)

Answer 2

You do not need Gauss sums or arguments with $p$-th powers to determine when $-3 \bmod p$ is a square for $p > 3$: it happens if and only if there is a nontrivial cube root of unity mod $p$, since in a field $F$ of characteristic not $2$ or $3$, $a^2 = -3$ in $F$ if and only if $(-1+a)/2$ is a nontrivial cube root of unity. Thus $(-3|p) = 1$ if and only if $3 \mid (p-1)$, or equivalently $p\equiv 1 \bmod 3$.

Knowing the quadratic character of $-1\bmod p$ depends on $p\bmod 4$ and of $-3\bmod p$ depends on $p\bmod 3$ tells us how the quadratic character of $3\bmod p$ is determined by $p \bmod 12$.

I am not saying this approach was known to Euler or Lagrange since they did not have a proof that $(\mathbf Z/(p))^\times$ is cyclic, but we have that proof (first established by Gauss) and I suspect this is the approach intended at the MSE link you lost.

The quadratic character of $-1 \bmod p$ for $p > 2$ can be determined without using cyclicity of $(\mathbf Z/(p))^\times$. If $-1 \equiv b^2 \bmod p$ then $(-1)^{(p-1)/2} \equiv b^{p-1} \equiv 1 \bmod p$, so $(p-1)/2$ is even and thus $p \equiv 1 \bmod 4$. Conversely, Wilson’s theorem $(p-1)! \equiv -1 \bmod p$ can be rewritten as $(-1)^{(p-1)/2}((p-1)/2)!)^2 \equiv -1 \bmod p$ by collecting terms at $k$ and $p-k$ together on the left side of Wilson’s theorem. Therefore when $p\equiv 1 \bmod 4$, we get an explicit realization of $-1$ as a quadratic residue mod $p$: it is the square of $((p-1)/2)! \bmod p$. This argument could have been known to Lagrange (but I don’t know if it was) since he proved Wilson’s theorem.

For an account by Weil of what Euler had done, see the first few pages of Section IX of Chapter III of his Number Theory: An Approach Through History.