I suppose by contradiction that $x+a$ is a factor of $x^n-a^n$ for all odd $n$. In particular for $n=1$, we have that $x+a$ is a factor of $x-a$, but that is not possible. So that would be a contradiction. If my proof is correct?
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Try to plug in $x=-?.$ – Ryszard Szwarc Feb 13 '22 at 02:42
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Use $,x+a,$ divides $,p(x)\iff p(-a) = 0,,$ by the Factor Theorem in the linked dupes. – Bill Dubuque Feb 13 '22 at 03:15
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This is not a duplicate and should be reopened. – markvs Feb 13 '22 at 10:50
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The statement is not true if $a=0$. It is true for any other $a$. Indeed, $x+a$ is a factor of $x^n-a^n$ iff $-a$ is a root of $x^n-a^n$ or $(-a)^n-a^n=-2a^n=0$ since $n$ is odd. This can happen only if $a=0$.
markvs
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Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Feb 13 '22 at 03:16
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The negation of the given statement is: $x+a$ is a factor of $x^n - a^n$ for some odd $n$.
So, your proof would be correct if the question was: Prove that $x+a$ is not a factor of $x^n - a^n$ for some odd $n$. You proved that this $n$ exists (the case $n=1$), but not that any odd $n$ satisfies the condition.
The correct proof would be:
$P(x)$ being a factor of $Q(x)$ implies that for any $\alpha$ that $P(\alpha)=0$, $Q(\alpha)=0$ also.
We have $P(x)=x+a, \; Q(x)=x^n-a^n$, and $\alpha=-a$.
However, $Q(\alpha)=(-a)^n-a^n=-2a^n \ne 0$ if $n$ is odd. (Assuming $a \ne 0$)
Therefore, if $a \ne 0$, $x+a$ is always not a factor of $x^n-a^n$ for odd $n$.
Saturday
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Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Feb 13 '22 at 03:16