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I'm solving a math puzzle and arrived at a quadratic:$$ \frac{6000n}{x(x-n)}=c $$ I only just graduated from high school and have very limited knowledge. I'm wondering if it's possible to find all integer solutions to this equation where $n$ and $c$ are variables, not constants. I know that when $n=3$ and $c=100$, $x=15$ and $x=-12$. Also, when $n=3$ and $c=1000$, $x=6$ and $x=-3$. Also again, for $n=3$, as $c \to \infty$ there also seem to be roots at $x=0$ and $x=3$. There is no limit on the domain/ range.

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  • $\begingroup$ Welcome to MSE. Please edit and use MathJax to properly format math expressions. Signing off is redundant (and some consider this as unwanted noise) since there's always the gray block at the bottom right. $\endgroup$ – Lee David Chung Lin Apr 8 '19 at 4:48
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Hint

If your equation is $$\frac{6000\,n} {x(x-n)}=c$$ multiply both sides by $x(x-n)$ and expand to get $$6000\,n=c x^2-c n x\implies cx^2-c nx -6000n=0$$ and use the solution of the quadratic equation.

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  • $\begingroup$ I found the solutions earlier to be: $$x= \frac{cn\pm\sqrt{c^2n^2+24000cn}}{2cn} $$ Is there any way to find all of the cases where $x$ is an integer? $\endgroup$ – Ethan Colgan Apr 8 '19 at 5:16

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