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Find all integers such that $\dfrac{n^4 + 1}{n^2 +n + 1}$ is an integer.

I have no idea how to solve things like this and what i tried to do didn't get me anywhere. I'd be grateful for any help!

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4 Answers

Using polynomial division, we can write the expression as $\frac{n+1}{n^2+n+1}+n^2-n$, which is an integer only when $n=0$ or $n=-1$.

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why did you cancel your nomination?! –  user59671 May 9 '13 at 18:32
    
@CutieKrait I understand your disappointment, but my presentation was apparently not serious enough. –  user60725 May 9 '13 at 18:40
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hope you come back at least if there are not enough people! –  user59671 May 9 '13 at 18:56
    
Sorry deleted your user! –  user59671 May 12 '13 at 0:54
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Hint: Divide one polynom by another and consider the remainder.

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Doing the elementary polynomial division we get:

$\frac{n^4 + 1}{n^2 +n + 1}=n^2-n+ \frac{n+1}{n^2+n+1}$

We do not need to consider now $n^2-n$ since it is always an integer(for integers $n$). Then observe $|\frac{n+1}{n^2+n+1}|\leq1$. So we do not have many chances that it will be an integer; it is $-1$, $0$, $1$. Write each equality case explicitly and see easily that the solutions are: $n=0$ and $n=-1$.

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HINT:

$$n^4+n^2+1=(n^2)^2+1^2+n^2=(n^2+1)^2-n^2=(n^2-n+1)(n^2+n+1)$$

$$\implies \frac{n^4+1}{n^2+n+1}=n^2-n+1-\frac{n^2}{n^2+n+1}$$

So, $n^2+n+1$ must divide $n^2$

But $(n^2,n^2+n+1)=1$ (Proof Below)

$\implies n^2+n+1=\pm1$ as $ n^2+n+1$ being denominator $\ne0$

If $ n^2+n+1=-1\implies n^2+n+2=0\implies n=\frac{-1\pm\sqrt{1^2-4\cdot1\cdot2}}2=\frac{-1\pm\sqrt7i}2$ which is not real.

If $ n^2+n+1=1\implies n(n+1)=0\implies n=0,-1$

[

Proof : If prime $p>1$ divides both $n^2, n^2+n+1$

$p$ will divide $(n^2+n+1)-n^2=n+1$

As $p$ is prime, $p$ divides $n^2\implies p$ divides $n$

Then $p$ divides $(n+1)-n=1$ clear contradiction as $p>1$

]

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Please point out the mistake here –  lab bhattacharjee Apr 18 '13 at 3:49
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