TITLE: integer solutions of $z=\frac{x^2}{4}+\frac{x}{y}$
QUESTION [1 upvotes]: 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.

REPLY [0 votes]: Even $x$ case:
Clearly if $x$ is even then it is easy for $x^2$ to divide by 4.
Also $x$ must be at least 88 as $\frac{88^2}{4}+\frac{88}{1}=2024$. Note that $\frac{88^2}{4}+\frac{88}{2}=1980<2012$ so there is only one answer with $x=88$.
For $x=90,92,94,96,98,100$ there is one solution for each divisor of $x$ so that $\frac{x}{y}$ is a whole number. Note however that $\frac{100^2}{4}+\frac{100}{1}=2600$ is too big but $\frac{100^2}{4}+\frac{100}{2}=2550$ is ok.
So count up the divisors of 88-100 and eliminate out of bounds values mentioned above:
$x=88$: one solution, $y=1$
$x=90$: 12 solutions, $y=1,2,3,5,6,9,10,15,18,30,45,90$
$x=92$: 6 solutions, $y=1,2,4,23,46,92$
$x=94$: 4 solutions, $y=1,2,47,94$
$x=96$: 12 solutions, $y=1,2,3,4,6,8,12,16,24,32,48,96$
$x=98$: 6 solutions, $y=1,2,7,14,49,98$
$x=100$, 8 solutions, $y=2,4,5,10,20,25,50,100$
Total: 49 solutions for even $x$ case.
Odd $x$ case:
If $x$ is odd then $x/4$ must have a fractional part of $\frac{3}{4}$.
So for each odd number between 91 and 99 consider values of $y$ which will give this fractional part. So y must be 4 or a multiple of 4 times the multiple of a divisor of $x$.
$x=91$: two solutions, $y=4,52$
$x=93$: two solutions, $y=12,124$
$x=95$: two solutions $y=4,20$
$x=97$: no solutions as its prime
$x=99$: two solutions, $y=4,36$
$x=101$: no solutions as its prime
Total: 8 solutions of $x$ odd.
Grand total: 57 solutions.
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