$$\sum_{m=1}^{2n-1} \sin \dfrac{\pi m^{2}}{2n}=\sqrt{n}$$
I figured out this sum on my own by experimenting values on Wolfram Alpha but I am unsure how to prove it.I looked up on several sources about this and found "Gauss Sums" but it looks way above my level for now.If you can give an intuitive approach that be really helpful
I tried converting into euler form and separating even and odd parts but didn't work.
Let $$I = \sum_{m=1}^{2n-1} \sin \frac{\pi m^{2}}{2n}$$ Consider the complex sum $S$. $$S = \sum_{m=0}^{2n-1} e^{i\frac{\pi m^2}{2n}}$$ The imaginary part of $S$ is $$\text{Im}(S) = \sum_{m=0}^{2n-1} \sin \frac{\pi m^2}{2n} = \sin(0) + \sum_{m=1}^{2n-1} \sin \frac{\pi m^2}{2n} = I$$ The terms of $S$ are periodic with period $2n$. $$e^{i\frac{\pi (m+2n)^2}{2n}} = e^{i\frac{\pi (m^2+4nm+4n^2)}{2n}} = e^{i\frac{\pi m^2}{2n}} e^{i(2\pi m+2\pi n)} = e^{i\frac{\pi m^2}{2n}}$$ Consider the quadratic Gauss sum $G$. $$G = \sum_{m=0}^{4n-1} e^{i\frac{\pi m^2}{2n}}$$ Splitting the sum: $$G = \sum_{m=0}^{2n-1} e^{i\frac{\pi m^2}{2n}} + \sum_{m=2n}^{4n-1} e^{i\frac{\pi m^2}{2n}}$$ The first part is $S$. For the second part, let $k=m-2n$. $$\sum_{m=2n}^{4n-1} e^{i\frac{\pi m^2}{2n}} = \sum_{k=0}^{2n-1} e^{i\frac{\pi (k+2n)^2}{2n}} = \sum_{k=0}^{2n-1} e^{i\frac{\pi k^2}{2n}} = S$$ So, $$G = S+S=2S \implies S = \frac{1}{2} G$$ We now evaluate $G$. $$G = \sum_{m=0}^{4n-1} e^{i\frac{\pi m^2}{2n}} = \sum_{m=0}^{4n-1} e^{2\pi i \frac{m^2}{4n}}$$ This is a standard quadratic Gauss sum of the form $\sum_{k=0}^{N-1} e^{2\pi i k^2/N}$ with $N=4n$. For $N \equiv 0 \pmod 4$, the sum evaluates to $(1+i)\sqrt{N}$. $$G = (1+i)\sqrt{4n} = (1+i)2\sqrt{n}$$ Substitute this back into the expression for $S$. $$S = \frac{1}{2} \left( (1+i)2\sqrt{n} \right) = (1+i)\sqrt{n} = \sqrt{n} + i\sqrt{n}$$ The original sum is the imaginary part of $S$. $$I = \text{Im}(S) = \text{Im}(\sqrt{n} + i\sqrt{n}) = \sqrt{n}$$ Thus, $$\boxed{\sum_{m=1}^{2n-1} \sin \dfrac{\pi m^{2}}{2n}=\sqrt{n}}$$