Find all real number $a$ such that sum of the solutions of the equation :$$ \sin\left(\sqrt{ax-x^2}\right)=0$$ is $100$.
Here is my solution :
From the given equation we can say that $\sqrt{ax-x^2}=k\pi$ for some $k\in \mathbb Z$. By simple manipulations we have : $x^2-ax+k^2\pi^2=0$. Using Vieta theorem $x_k+y_k=a , x_ky_k=k^2\pi^2$ (Obviously $x_k$ and $y_k$ are the solutions of the quadratic equation). Now it is easy to notice that $k^2\pi^2\leq\frac{a^2}{4}$ therofore $k\leq \frac{a}{2\pi}$ (We are assuming that $k$ is non-negative number because $-k$ will give us same pair of the solutions). Also for different $k$ we have different set of the solutions (but for every $k$ sum is always $a$). Since number of integers in $[0,\frac{a}{2\pi}]$ is $[\frac{a}{2\pi}]+1$ (where $[x]$ is an integer part of the number) we can say that $a\times([\frac{a}{2\pi}]+1)=100$.From here it is not that hard to show that $a=25$. So we are done.
My solution might be right but I wonder is there easier one? I would like to here every idea. Thank you.