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. :)