<s>[INST] integer solutions of $z=\frac{x^2}{4}+\frac{x}{y}$ Let $x,y,z$ be positive integers and $2012\leq z\leq 2555$.
How to solve the equation $z=\dfrac{x^2}{4}+\dfrac{x}{y}$ ?
By wolfram alpha, I get $x=88,y=1,z=2024$.
Can it be solved by manual?
Thank you.
 [/INST] $$\frac{x^2}4<z<\frac{x^2}4+x+1\\
2012<\left(\frac x2+1\right)^2,\left(\frac x2\right)^2<2555\\
2\sqrt{2012}-2<x<2\sqrt{2555}$$
If $x$ is even, then $y$ can be any factor of $x$.
If $x$ is odd, then $x/y$ must be $n+\frac34$, so $x/(y/4)$ can be any factor of $x$ which is $3\pmod4$
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