## Franklin's Notes

A quadratic Guass sum is a finite sum taking the form

where $\zeta_m=e^{2\pi i/m}$ is a primitive $m$th root of unity. Using facts about quadratic residues in the modular rings $\mathbb Z_m$, we can compute the exact values of these sums for certain special values of $m$, in particular primes. For instance, we have the following formula:

Proposition 1. If $p\equiv 1\bmod 4$ is prime, then

Proof. Let $S$ represent the quadratic Gauss sum in question. Then we have and Now, since $p\equiv 1\bmod 4$, we have that $-1$ is a quadratic residue modulo $p$. If $x$ is a square root of $-1$ modulo $p$, we may substitute $s\mapsto xs$ in the sum, which will permute the values $s=1,\cdots,p-1$, giving the sum Now let us include the value $j=0$ in the sum again, and subtract off the same value: Then let us make the substitution $s\mapsto s-r$ in the inner sum, which will permute the values $s=0,\cdots,p-1$ for each fixed value of $r$ in the outer sum, yielding or Now, if we again extract the sum when $s=0$, we obtain and if we make the substitution $r\mapsto r/2s$, which permutes the values of $r=1,\cdots,p-1$, we have This double sum is now equal to a sum of roots of unity, equal to $-1$, times a Gauss sum which is missing the $r=0$ term. That is, we have If we simplify this equation, we obtain $S^2=p$, implying $S=\pm\sqrt{p}$ as claimed. $\blacksquare$

This suggests the question: when exactly is the sign of the above quadratic Gauss sum equal to $+\sqrt{p}$, and when is it $-\sqrt{p}$?

Knowing the value of Gauss sums of the above form modulo a prime, we can calculate the values of a more generalized kind of Gauss sum taking the form where $\alpha$ is an integer. Using the fact that exactly half of the units of $\mathbb Z_p$ are quadratic residues for $p>2$, we can deduce that if $\alpha$ is a QR mod $p$, then whereas if $\alpha$ is not a QR mod $p$, then

Using these sums and some handy symmetries of the trigonometric functions, we can deduce cool trigonometric identities such as the following: