# Why does the quadratic Gauss sum $G(a,b,c)$ evaluate to 0 when $gcd(a,c)>1$?

On the Wikipedia page for the quadratic Gauss sum it states that it will evaluate to $$0$$ in the case that $$\text{gcd}(a,c)>1$$, except when $$l|b$$.

The Gauss sum is defined as

$$G(a,b,c)=\sum_{n=0}^{c-1} e^{2\pi i\frac{an^2+bn}{c}}$$

Lets say that $$\text{gcd}(a,c)=l>1$$. Then we could write $$a=a'l$$ and $$c=c'l$$ such that

$$G(a,b,c)=\sum_{n=0}^{c-1} e^{2\pi i\frac{a'ln^2+bn}{c'l}}=\sum_{n=0}^{c-1} e^{2\pi i\frac{a'n^2}{c'}}e^{2\pi i\frac{bn}{c'l}}$$

How can this be utilised to show that $$G(a,b,c)=0$$, except when $$l|b$$?

$$G(a,b,c) = \sum_{n=0}^{c-1} \exp\left( 2\pi i \frac{an^2+bn}{c}\right)$$; Note that the summand $$f(a, b, c; n) = \exp\left( 2\pi i \frac{an^2+bn}{c} \right)$$ satisfies $$f(a,b,c;n) = f(a,b,c;n+c)$$, so we can shift the index to have
$$G(a,b,c) = \sum_{n=0}^{c-1}\exp\left( 2\pi i \frac{a(n+r)^2+b(n+r)}{c}\right)$$ for any integer $$r$$.
Let $$a = \ell A$$ and $$c = \ell C$$; let $$r = C$$ to have $$G(a,b,c) = G(a,b,c) \cdot e^{2\pi i \frac{b}{l}}$$. If $$\ell \not \mid b$$ , $$e^{2\pi i \frac{b}{\ell}} \ne1$$ and so $$G(a,b,c) = 0$$. Here $$\ell$$ is the gcd of $$a$$ and $$c$$.