Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 :)

share|improve this question
    
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
show 2 more comments

2 Answers

up vote 4 down vote accepted

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.

share|improve this answer
    
Ah, thanks. I guess I was misled by those unnecessary steps. –  maliky0_o May 11 '12 at 10:13
add comment

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$\frac{n^4-2n^2-15}{5*3}=\frac{n^4-2n^2-15}{15}$

hence n divides 15

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.