Let $p$ be a prime with $p \equiv 1 \mod 4$. Evaluate $\sum_{k = 1}^{\frac{p - 1}{2}} \big\{ \frac{k^2}{p}\big\}$. Checking values of $p$ indicates that the answer should be $\frac{p - 1}{4}$. If tried to instead evaluate $\sum_{k = 1}^{\frac{p - 1}{2}} \lfloor \frac{k^2}{p} \rfloor$ since if I can show that this is $\frac{(p - 1)(p - 5)}{24}$ then the result will hold.
I've tried lattice counting arguments like in the proof of quadratic reciprocity but to no avail. I'm also trying to understand the signifigance of $p$ having a residue $1$. Maybe since there are an even amount of terms some sort of manipulation could be used. Also maybe writing $p$ as a sum of squares yields something.
I would preferably like a hint, but would not mind a full solution.