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I'm referring to this proof. The key formula ("Eisenstein's Lemma") is $$\left(\frac{q}{p}\right)=(-1)^{\sum_{u}\lfloor\frac{qu}{p}\rfloor},\text{ where $u=2,4,\ldots,p-1$}$$
The sum in the exponent is easily seen to count the number of lattice points in this rectangle that are below the diagonal and having even $x$-coordinate,

where $p$ and $q$ are our primes in question, and via some clever flipping, quadratic reciprocity pops out.

I seem to recall in my first summer at PROMYS, some counselors tried to work out a version for quartic reciprocity, using (if I remember correctly) a similarly-defined lattice in $\mathbb{Z}[i]\times\mathbb{Z}[i]$. However, I don't remember the details, or if they were successful. I'd be interested to see if a "low-tech" proof using lattice point counting can work for cubic or quartic reciprocity laws (or even more generally, but I suspect that would be overly optimistic).

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That proof is indeed marvelous. –  awllower Feb 9 '11 at 12:12

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up vote 16 down vote accepted

Gauss's (unpublished and largely unknown) proof of the quartic reciprocity law probably used lattice point arguments. The details were supplied by several authors at the end of the 19th century (for references, see e.g. Hill's article below).

A modern approach using geometric ideas similar to those above was provided in several articles by R. Hill, such as this one.

Edit (2015). For reconstructions of Gauss's ideas see the recently published book Gauss's reciprocity laws in number theory (in German).

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Wow, I didn't even know that Gauss had a proof for the quartic reciprocity law! –  awllower Feb 9 '11 at 12:10
He intended to write 3 papers on the quartic reciprocity law, of which only two appeared. He probably cancelled his plans of writing the third one containing his proof after Jacobi had shown how to derive the theorem very easily from properties of quartic Gauss sums. All that is left are a couple of pages in his Nachlass, later published in his Werke by Dedekind. –  Franz Lemmermeyer Feb 9 '11 at 12:15

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