# Finding $n\in\mathbb{Z}$ such that $\frac{(n^2+3)(n^2-5)}{16n}\in\mathbb{Z}$

I'm trying to follow a step in a proof, which involves finding $n\in\mathbb{Z}$ such that $\frac{(n^2+3)(n^2-5)}{16n}\in\mathbb{Z}$.

The proof then states that

1. $\text{hcf}(n,n^2+3)$ divides 3, and

2. $\text{hcf}(n,n^2-5)$ divides 5.

3. Hence n divides 15.

I can see 1. and 2. hold, as $\text{hcf}(b,a+mb)=\text{hcf}(a,b)$ But I'm not exactly sure what argument to use to deduce 3.

Any hints would be appreciated. Thanks :)

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What is hcf? The greatest common divisor is standardly denoted gcd. –  Andrea Mori May 11 '12 at 9:42
It might be a good idea to give a link to the proof you mention. –  Rankeya May 11 '12 at 9:42
@AndreaMori: hcf = highest common factor. I have come across sources that use this notation for gcd. –  Rankeya May 11 '12 at 9:43
For that quotient to be an integer, what must the highest common factor of $(n^2 + 3)(n^2 - 5)$ and $(16 n)$ be? –  Hurkyl May 11 '12 at 9:44
@Andrea: I think hcf is British. –  Hurkyl May 11 '12 at 9:44

The conclusion follows only with the additional assumption that $\frac{(n^2+3)(n^2-5)}{16n}\in\mathbb{Z}$.

If $\frac{(n^2+3)(n^2-5)}{16n}\in\mathbb{Z}$, it must be the case that $n$ divides $(n^2+3)(n^2-5)=n^4-2n^2-15$, and therefore $n$ divides $15$. You don't really need (1) and (2) at all.

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Ah, thanks. I guess I was misled by those unnecessary steps. –  maliky0_o May 11 '12 at 10:13
well how about thinking this @Brian well $\frac{(n^2+3)}{5}$ since$hcf(n,n^2+3) divides 3$ $\frac{(n^2-5)}{3}$ since $hcf(n,n^2-5) divides 5$ implies$\frac{n^4-2n^2-15}{5*3}=\frac{n^4-2n^2-15}{15}$