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$ \frac{x^\alpha+1}{x+1} = K$ where $K \in \Bbb N$ ,or is it true $\forall \alpha \in \Bbb N$ or their are specific values of $\alpha$ it is true for? and all the assumes values of $x \in\Bbb N$.

Américo Tavares
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Mukil
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  • Do you look for values of $\alpha$ such that $ \frac{x^{\alpha}+1}{1+x}$ is an integer for all integer values of $x$? – Oussama Boussif May 13 '17 at 11:11
  • yes exactly, i was reading a book where they said for $ a^n +1$ to be prime, a is even and $ n has to be of the form n=2^n$ – Mukil May 13 '17 at 11:15
  • The problem you mention in your comment is a classical one, but the one you posed in the question, isn't very clear, so, is it like I said in the comment or something else? – Oussama Boussif May 13 '17 at 11:16

2 Answers2

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Let $y=x+1$. Then, by the binomial theorem, $x^\alpha+1=(y-1)^\alpha+1=yb+(-1)^\alpha+1$, which leaves remainder $(-1)^\alpha+1$ when divided by $y=x+1$. The remainder is $0$ when $\alpha$ is odd and $2$ when $\alpha$ is even.

Therefore, $\frac{x^\alpha+1}{x+1}$ is an integer iff $\alpha$ is odd or $x+1$ divides $2$, that is, when $x=0$ or $x=1$.

lhf
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  • But, shouldn't we prove that if $\alpha$ is not an integer, then the fraction wouldn't be too? – Oussama Boussif May 13 '17 at 11:13
  • is it necessary that $\alpha$ has to odd?, what will happen is $\alpha$ is even or if $\alpha$ is even and a odd number divides $\alpha$ – Mukil May 13 '17 at 11:17
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We can use this property of the polynomials: $x-a$ is always a factor of the polynomial $x^{\alpha}-a^{\alpha}$, $\alpha\in \mathbb{N},x\in \mathbb{R}.$

So $x+a=x-(-a)$ is always a factor of the polynomial $x^{\alpha}-(-a)^{\alpha}$, $\alpha\in \mathbb{N}$. For $a=1$ we conclude that $x+1=x-(-1)$ is always a factor of the polynomial $x^{\alpha}-(-1)^{\alpha}$, $\alpha\in \mathbb{N}$.

  • If $\alpha$ is odd, then $(-1)^\alpha=-1 $ and $x+1$ is a factor of $x^{\alpha}-(-1)^{n}=x^{\alpha}+1$.
  • If $\alpha$ is even, then $(-1)^\alpha=1$. So $x+1$ is a factor of $x^{\alpha}-(-1)^{\alpha}=x^{\alpha}-1$, but it is not a factor of $x^{\alpha}+1$.

If $x\in \mathbb{N}$ and $\alpha$ is odd, then $ \dfrac{x^\alpha+1}{x+1} = K\in \mathbb{N} .$

Américo Tavares
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