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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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    Weird, I never encountered the notation hcf. If it is a British usage, must be in Hardy and Wright. I'll check it. Anyway, $3\mid hcf(n,n^2+3)$ implies that $3\mid n$ and in the same lines $5\mid n$. Since $3$ and $5$ are primes, then $3\cdot5=15\mid n$.2012-05-11

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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.2012-05-11