Don't 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, with the double-angle formula in the form $\sin^2 \theta = \dfrac{1-\cos 2\theta}{2}$ , we simplify our given sum:

$\begin{aligned} \sum_{n=0}^{18} \sin^2\left(\frac{34}{19} \cdot 2\pi n^2\right) &= \frac{1}{2} \sum_{n=0}^{18} 1-\cos \left(\frac{68}{19} \cdot 2\pi n^2\right) \\ &= \frac{19}{2} - \frac{1}{2} \Re \sum_{n=0}^{18} e^{2i\pi \cdot 68n^2/19} \\ &= \frac{19}{2} - \Re [g(68;19)] \end{aligned}$

Since $19 \equiv 3 \pmod{4}$ , $g(68;19) = \left( \dfrac{68}{19} \right) g(1;19) \stackrel?= \pm i\sqrt{19}$ is purely imaginary, and as a result we have that the sum is equal to

$\sum_{n=0}^{18} \sin^2\left(\frac{34}{19} \cdot 2\pi n^2\right) = \frac{19}{2} - \Re [g(68;19)] = \boxed{\dfrac{19}{2}}$