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Prove that $x^m + \alpha^m$ is not divisible by $x-\alpha$. I tried it using the principle of mathematical induction but wasn't able to do it.

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Note that $x^m - \alpha^m$ is divisible by $x - \alpha$. If $x^m + \alpha^m$ were too, then $x^m$ would be also. But $x-\alpha$ does not divide $x^m$ unless $\alpha = 0$.

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Assuming your polynomials have coefficients on a field. Divisibility of $f(x)$ by $x-\alpha$ implies $\alpha$ is a root of $f(x)$. But $\alpha^m+\alpha^m\neq 0$ unless $\alpha=0$.

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Hint: For all polynomials $f$, we have $f(x)=(x-\alpha)q(x)+f(\alpha)$. Now take $f(x)=x^m + \alpha^m$.

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$P$ and $Q$ are arbitrary functions, and $a$ is an arbitrary real number. We write $P(x)$ like this: $$P(x) = Q(x - a) * (x - a) + R(a)$$

Here $R(x)$ is just $P(x) mod (x - a)$. So if we calculate $P(a)$, we have:

$P(a) = Q(0) * 0 + R(a) = R(a)$

So $P(a)$ is the remainder. So if we want to calculate the remainder of the division: $\frac{x^m + \alpha ^ m}{x-\alpha}$, we put $x=\alpha$ then we get:

$2 * \alpha^m$

So the remainder is not always zero (since $\alpha$ is not necessarily zero). So it is not divisible.

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