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Let $a$ be an integer, $a\ge2$. If $a^{m}+1\mid a^{n}+1$ then prove that $m\mid n$.

Actually I know a similar proof which is, $a^{m}-1\mid a^{n}-1 \iff m\mid n$, but I can't prove this. I also need some examples of the question. Can't seem to find any correlation between the two proofs.
I seem to not find examples where $a$ is something different from $2$ and taking $m=2$.
Please help. I think 4-5 examples might help me to see the proof.

user26857
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Love Invariants
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  • Specialize this more general result on their gcd (but a direct proof is easier) – Bill Dubuque May 17 '19 at 19:28
  • Note: $3^3+1$ divides $3^9+1, 3^{15}+1, 3^{21}+1,...$ – J. W. Tanner May 17 '19 at 19:31
  • @J.W.Tanner I see a pattern. Don't know if correct but I see it for $2^2+1,3^2+1,3^3+1,3^4+1$. That it is the third multiple,sixth multiple,ninth multiple of m which makes $a^n+1$ divisible by $a^m+1$. I guess it is correct for all primes . Thanks btw for it giving me little hint. Well seeing if I can find the proof or not. – Love Invariants May 17 '19 at 19:38

3 Answers3

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Suppose $\gcd(m,n)=d$. $$s=\gcd(a^{2m}-1,a^{2n}-1)=a^{2d}-1$$ But obviously $a^m+1\mid s$, so $$a^m+1\mid a^{2d}-1,$$ so $$m < 2d\Rightarrow d>\frac m2.$$ But $d\mid m$, so $d=m$ which means $m\mid n$.

user26857
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snowAuoue
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$U_{k} = a^{k} + 1$ are terms of a Lucas sequence; hence, $U_{m} | U_{n}$ iff $m | n$.

DDS
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Theorem $\,\Rightarrow\, c = \overbrace{a^{M}+1\mid a^{(M,N)}+1}^{\small\textstyle \Rightarrow \color{#90f}{M \le (M,N)}}\,\Rightarrow \color{#90f}{(M,N)= M}\Rightarrow M\mid N $

Theorem $\,\ \begin{align}c\mid a^M+1\\ c\mid a^N+1\end{align}\,$ $\Rightarrow\ \begin{align}&c\mid \color{#0a0}{a^{\large d}+1}\\ &\!d = {\small (M,N)}\end{align}\ \ $ Proof $\ $ Let $\,\begin{align} {\small M} &= dm\\ {\small N} &=\, dn\end{align}\,\ $ so $\ (m,n)=1$

$\!\!\bmod c\!:\ a^{\large dm}\equiv -1\equiv a^{\large dn}\!\Rightarrow a^{\large 2d\color{darkorange}m}\equiv 1\equiv a^{\large 2d\color{darkorange}n}\,$ thus $\,{\rm ord}\: a^{\large 2d}$ divides coprimes $\,\color{darkorange}{m,n}\,$ so it is $1,\,$ so $\,\color{#c00}{a^{\large 2d}\equiv 1}.\,$ $\,(m,n)\!=\!1^{\phantom{I^{I^I}}}\!\!\!\!\!\!\Rightarrow m\,$ or $\,n\,$ odd, wlog $\,n = 1\!+\!2j^{\phantom{I^I}\!\!\!\!}\,$ so $\ \color{#0a0}{{-}1}\equiv a^{\large dn}\!\equiv a^{\large d}(\color{#c00}{a^{\large 2d}})^{\large j}\!\equiv \color{#0a0}{a^{\large d}}$

Bill Dubuque
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  • The Theorem is excerpted from this Theorem which computes the gcd $,(a^M+1,a^N+1)\ \ $ More simply we could compute $, a^N+1 \bmod a^M+1,,$ but the Theorem is stronger and more insightful (and not much more work). – Bill Dubuque May 17 '19 at 23:20
  • Sigh, serial political downvoter is still active (they always downvote my answer, then upvote the others - just as done $5$ mins ago here). The above proof is correct (and the easiest direct way). Both of the other answers use unproved results (whose proofs are no shorter than above). – Bill Dubuque Apr 15 '23 at 23:49
  • +1 but I find the logic of your post hard to follow (I am missing linking words; I guess what you mean only because I know it already). And imo the "easiest direct way" would be: let $n=mq+r$ with $0\le r<m,$ then $0\equiv(-1)^qa^r+1\bmod{a^m+1}$ hence $a^r=1.$ – Anne Bauval Jun 13 '23 at 09:53
  • @Anne Thanks. I'm puzzled what logic is missing. The reason for presenting this method of proof is explained in my first comment above. – Bill Dubuque Jun 13 '23 at 12:49