a question based on polynomials $ \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 $x \in\Bbb N$.
 A: 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 $x=1$.
A: 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 $x^{\alpha}-(-1)^{\alpha}$, $\alpha\in \mathbb{N}$. 


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*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} .$
