I came across the result that $(x^a+1)|(x^b+1)$ if and only if $\frac{b}{a}$ is odd. Any intuitive reason why, though? What about $\frac{b}{a}$ being odd makes this true?
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https://math.stackexchange.com/questions/641443/proof-of-anbn-divisible-by-ab-when-n-is-odd – lab bhattacharjee Jul 03 '20 at 07:06
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Hint : Substitute $y=x^a$ , then it should be easier to see it. – Peter Jul 03 '20 at 07:06
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This follows from the fact that, if $k$ is odd,$$x^k+1=(x+1)\left(x^{k-1}-x^{k-2}+\cdots-x+1\right).$$
José Carlos Santos
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What about cases like $x^{12}+1$, though? That factors into $(x^4+1)(x^8-x^4+1)$ even though neither 4 nor 12 are odd. – user940293884 Jul 03 '20 at 07:11
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@user940293884 what matters is the fraction $b/a$. In your case $12/4=3$ that is odd. You can use the formula of José with $x \mapsto x^4$, $k=3$, and obtain yours – Exodd Jul 03 '20 at 07:14
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