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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).

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).

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).

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).

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,
http://upload.wikimedia.org/wikipedia/en/7/72/Eisenstein-quadratic-reciprocity-1.svg
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).

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).