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This question is from Dudley's Elementary Number Theory. "Show that for $k\gt 0$, $m \ge 1$, and $x \equiv 1\ (mod\;m^{k})$ implies $x^{m} \equiv 1\ (mod\;m^{k+1})$."
I have come up with an expression for x: $$x\ =\ (jm^{k}+1)$$ ($j \ge 1$). But I don't know how to proceed. I have substituted values for k and m, and it works to a point, but I would have to use the binomial theorem to express $x^m$, and I don't know how to show it applies generally. I believe that the result should be that every term except the last is divisible by $m^{k+1}$, and since the last term is one, $x^m \equiv 1 (mod \; m^{k+1})$.

k endres
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  • See $\mu$LTE in the linked dupe. $\ \ $ – Bill Dubuque Jun 24 '24 at 02:54
  • And if I have questions about these other questions? I have found that when someone links my question to a previous question and closes mine, I have no avenue for asking for clarification on the linked question (which is also closed). Part of this is being a newbie and not understanding how things work (I am learning as I go, every time I ask another question). The closed questions most often will not allow me to ask questions about the answers given. I also tried to search for this question, but my search did not match any other questions. Thanks for the continued help. – k endres Jun 24 '24 at 04:40
  • If you need further elaboration on the linked answers then please ask for such in comments there (that helps to improve the answers). If that does not succeed then you can pose a further more specific question on the point this is not clear. – Bill Dubuque Jun 24 '24 at 06:18
  • Thanks. I don't have enough reputation points to post a comment. When I have asked questions that intended to clarify previous questions, they were quickly linked to the previous question and closed. – k endres Jun 24 '24 at 12:56
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    In that case you can post any such comments here. – Bill Dubuque Jun 24 '24 at 16:04

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