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For how many integers $N$ is the rational function $\frac{N^2-2N-15}{N^2-N-12}$ also an integer?

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closed as off-topic by greenturtle3141, Brandon_J, gabbo1092, Gamow, Glorfindel Jun 6 at 19:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – greenturtle3141, Brandon_J, gabbo1092, Gamow, Glorfindel
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  • $\begingroup$ I don't think this is off-topic. The "obvious" way to approach this would be to write $N^2−2N−15=M(N^2−N−12)$ and then try to reduce possibilities for $M$ and $N$, perhaps by using modular arithmetic or even brute-forcing. The "aha moment" required by the linked policy is to factorise the polynomials, and the "unexpected result" is that only two values of $N$ work and it's easy to find which ones. $\endgroup$ – Rand al'Thor Jun 7 at 13:02
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Since the given expression can be simplified to

$\frac{(N-5)(N+3)}{(N-4)(N+3)}=\frac{N-5}{N-4}=1-\frac{1}{N-4}$,

we simply need to make sure

$N-4$ divides $1$, i.e., $N-4=\pm 1$, which leads to $N=5$ or $N=3$. We can check that both of them work.

So

There are two such $N$.

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  • $\begingroup$ Perfect! I was hoping to lead people on a wild goose chase with $N^2-2N-15=M(N^2-N-12)$ or suchlike, but you're too good. $\endgroup$ – Rand al'Thor Jun 6 at 10:09

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