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Fix an [edit: positive] integer $q$. Let me say that an [edit: positive] integer $m$ is IK over $q$ if $q$ and $m$ are coprime and the (multiplicative) order of $q$ mod $m$ is IK over $q$. Note that that multiplicative order is strictly less than $m$, so the induction terminates.

Is there a non-inductive characterization of which $m$ are IK over $q$?

This question is a follow-up to my question about when a field extension is iterative Kummer, hence the name "IK." In an answer, Will Sawin speculates that the answer to this question is "no," but it seems worth asking a wider audience.

Added in response to comment: The base of the induction is that $1$ is IK over $q$. When you unwrap the induction, it asks the following. Repeatedly replace $m \mapsto $ order of $q$ mod $m$. If you eventually get to $1$, then $m$ is IK over $q$. If you eventually get to a number not coprime to $q$, then $m$ is not IK over $q$.

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    $\begingroup$ I do not understand the definition. What is the base of this induction? $\endgroup$
    – Fedor Petrov
    Commented Jan 4 at 22:37
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    $\begingroup$ It seems that the base is "1 is IK over any q" $\endgroup$
    – Max Alekseyev
    Commented Jan 5 at 13:03
  • $\begingroup$ Please include some examples. It will make it more immediately clear what is going on. $\endgroup$
    – KConrad
    Commented Jan 11 at 1:29

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