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This question already has an answer here:

Find all positive integers n such that $n + 3$ divides $n^2 + 27$.

I am quite stuck on how to conclusively show all of the solutions. I tried look at the intersections of $y = n^2 + 27$ and $f = x(n + 3)$ where is $x$ is some number. But it still involves setting $x$ values. Any help would be much appreciated.

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marked as duplicate by lab bhattacharjee elementary-number-theory Nov 18 '17 at 14:15

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    $\begingroup$ Hint: $x^2+27=(x-3)(x+3)+36$. $\endgroup$ – lulu Nov 18 '17 at 14:13
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Hint. Note that $$\frac{n^2+27}{n+3}=n-3+\frac{36}{n+3}$$ hence $n+3$ divides $n^2+27$, iff $n+3$ divides $36$, that is $$n+3\in\{1, 2, 3, 4, 6, 9, 12, 18, 36\}.$$

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