Expressions $x^a+1$ and $x^b+1$, when $a$, $b$ are odd positive integers

Question

Let $a\geq 3$ and $b\geq 3$ be odd integers. Let $x\geq 2$ be an integer. I would like to know if we can state that $x^a+1|x^b+1 \Longleftrightarrow a|b$. I know how to prove that $a|b$ implies $x^a+1|x^b+1$. Is the other implication also true? I'm especially interested in the case when $x$ is a prime number, but I don't think that this additional assumption matters so much. I've checked the result for particular primes $x$ and it seems that it holds. Still, I would like to know how to prove it or, if it isn't true, which would be a counterexample?

There is a similar result which states: $x^a-1|x^b-1 \Longleftrightarrow a|b$. By following the proof for this similar result, I've tried to apply the same argument (the quotient remainder theorem) to show the implication above, but I did not manage to finish the proof. Any hints are appreciated.

If you have the result that $x^a-1\mid x^b-1\Leftrightarrow a\mid b$, then you can use the facts that for polynomials $f(x),g(x)$ (over a commutative ring), we have $f(x)\mid g(x)\Leftrightarrow -f(x)\mid -g(x)$ and $f(x)\mid g(x)\Leftrightarrow f(-x)\mid g(-x)$. This works since $a,b$ are odd. — Mor A., Mar 19 '21 at 14:06
@MorA. The question is not about $f(x)\mid g(x)$, but about $f(n)\mid g(n)$ with $n\ge2$ — Hagen von Eitzen, Mar 19 '21 at 14:46

score 1 · Accepted Answer · answered Mar 19 '21 at 15:27

Suppose $x^a+1\mid x^b+1$ (where $x$ is either an unknown and we work in polynomials $\Bbb Z[x]$, or $x$ is an integer $\ge 2$ and we work in $\Bbb Z$ itself).

Suppose $x^a+1\mid x^b+1$ where $a,b$ are odd positive integers and $x$ is an integer $\ge 2$. Then necessarily $b\ge a$ and also $$ x^a+1\mid (x^a+1)x^{b-a}-(x^b+1)=x^{b-a}-1.$$ If $b\le 2a$, the right hand side is $<x^a+1$ hence must be $=0$, i.e. $a=b$ and of course $a\mid b$. So far, we have proved $a\mid b$ for $b\le 2a$. Now we can treat the rest by induction on $b$ because we have $$ x^a+1\mid (x^b+1)-(x^a+1)(x^a-1)x^{b-2a}=x^{b-2a}+1$$ where $b-2a$ is an odd positive integer $<a$ so that by induction hypotheseis, $a\mid b-2a$, which then implies $a\mid b$.

Expressions $x^a+1$ and $x^b+1$, when $a$, $b$ are odd positive integers

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