Let the quadratic Gauss sum be defined by

$\displaystyle G(m;n)=\underset{r=1}{\overset{n}{\sum}}\omega^{mr ^{2}}$, where $\displaystyle \omega=e^{2\pi i/n}$.

I have managed to show that $\displaystyle \omega^{mr^{2}}=\omega^{ms^{2}}$, whenever $\displaystyle r\equiv s\mod\: n)$.

My problem is that I don't know how to deduce that $\displaystyle G(m;n)=\{\sum}\omega^{mr^{2}}$, where the summation extends over ANY complete set of residues.

Could anybody help me with this deduction please?

2. Originally Posted by Cairo
Let the quadratic Gauss sum be defined by

$\displaystyle G(m;n)=\underset{r=1}{\overset{n}{\sum}}\omega^{mr ^{2}}$, where $\displaystyle \omega=e^{2\pi i/n}$.

I have managed to show that $\displaystyle \omega^{mr^{2}}=\omega^{ms^{2}}$, whenever $\displaystyle r\equiv s\mod\: n)$.

My problem is that I don't know how to deduce that $\displaystyle G(m;n)=\{\sum}\omega^{mr^{2}}$, where the summation extends over ANY complete set of residues.

Could anybody help me with this deduction please?

If $\displaystyle S:=\{s_1,...s_n\}$ is a complete set of residues modulo n, then $\displaystyle \displaystyle{\forall i=0,1,...,n\,\,\exists !s_{j(i)}\in S}$ s.t.

$\displaystyle \displaystyle{s_{j(i)}=i\!\!\pmod n\Longrightarrow \omega^{mi^2}=\omega^{ms_{j(i)}^2}}$ and thus the corresponding sums are equal...

Tonio