Note the factorisation of 34

The quadratic Gauss sum is defined as $\displaystyle g(a;p) = \sum_{n=0}^{p-1} e^{2i\pi an^2/p}$ , where $a$ is an integer and $p$ an odd prime number. Certain fascinating properties of the sum are known, but crucial to this problem are two: we have

1. $g(a;p) = \left( \dfrac{a}{p} \right) g(1;p)$ for all $a$ , where $\left( \dfrac{a}{p} \right)$ is the Legendre symbol; and

2. $g(1;p) = \begin{cases} \sqrt{p} & \text{if } p \equiv 1 \pmod{4} \\ i\sqrt{p} & \text{if } p \equiv 3 \pmod{4} \end{cases}$ .

Equipped with these, we are set to evaluate our given sum.

First, note that

$\sum_{n=0}^{18} \sin\left(\frac{34}{19} \cdot 2\pi n^2\right) = \Im \ \sum_{n=0}^{18} e^{2i\pi \cdot 34n^2/19} = \Im [g(34;19)]$

Now, using the properties of the quadratic Gauss sum, we have

$g(34;19) = \left( \dfrac{34}{19} \right) g(1;19) = \left( \dfrac{-1}{19} \right) g(1;19) = -g(1;19)$

(The exact computation is left as an exercise to the reader. One is suggested to note, as the problem title suggests, the factorisation of 34; other methods are, however, possible.)

And so, we also have $g(1;19) = i\sqrt{19}$ since $19 \equiv 3 \pmod{4}$ . Hence,

$\Im [g(34;19)] = \Im [-i\sqrt{19}] = -\sqrt{19}$

Therefore, our sum evaluates to

$\sum_{n=0}^{18} \sin\left(\frac{34}{19} \cdot 2\pi n^2\right) = \boxed{-\sqrt{19}}$