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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 :)

maliky0_o
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  • What is hcf? The greatest common divisor is standardly denoted gcd. – Andrea Mori May 11 '12 at 09:42
  • It might be a good idea to give a link to the proof you mention. – Rankeya May 11 '12 at 09:42
  • @AndreaMori: hcf = highest common factor. I have come across sources that use this notation for gcd. – Rankeya May 11 '12 at 09:43
  • For that quotient to be an integer, what must the highest common factor of $(n^2 + 3)(n^2 - 5)$ and $(16 n)$ be? –  May 11 '12 at 09:44
  • @Andrea: I think hcf is British. –  May 11 '12 at 09:44
  • Which explains why we used HCF instead of GCD in India. – Rankeya May 11 '12 at 09:49
  • 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 $3\cdot5=15\mid n$. – Andrea Mori May 11 '12 at 09:50

1 Answers1

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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.

Brian M. Scott
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