$4-$ [ GCD and LCM. Mutual simplicity ]

Find integer solutions of the equation $x^{2} y=10000 x+y$.

#
$10000 x=y\left(x^{2}-1\right)$. Since $x$ and $x^{2}-1$ are coprime, 10000 is divisible by $x^{2}-1$.

If $x$ is odd, then $x^{2}-1=(x-1)(x+1)$ is divisible by 8 as the product of two consecutive even numbers. If $x$ is even, then $x^{2}-1$ is odd, and therefore must equal a power of five with some sign. $x-1$ and $x+1$ cannot both be divisible by 5, so $x-1= \pm 1$ or $x+1= \pm 1$. Checking these options, we find the solution $x=0$. Then $y=0$.

It remains to consider all divisors of 10000 that are divisible by 8, and check which of them can be represented as $x^{2}-1$. There are 10 such divisors: $2^{3}, 2^{4}, 5 \cdot 2^{3}, 5 \cdot 2^{4}, 5^{2} \cdot 2^{2}, 5^{2} \cdot 2^{4}, 5^{3} \cdot 2^{3}, 5^{3} \cdot 2^{4}, 5^{4} \cdot 2^{3}, 5^{4} \cdot 2^{4}$. Among them, only 8 and 80 fit. This gives us four more solutions: ( $\pm 3, \pm 3750$ ) and ( $\pm 9, \pm 1125$ ).

## Answer

$(-9,-1125),(-3,-3750),(0,0),(3,3750),(9,1125)$.