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Observation

Let $k,m\in\mathbb{Z}_+$

(1)

$$k^m\equiv k^{m+2}\pmod{2!},\ \ m\ge 1$$

(2)

$$k^m\equiv k^{m+2}\pmod{3!} ,\ \ m\ge 1 $$

(3)

$$k^m\equiv k^{m+2}\pmod{4!} ,\ \ m\ge 3 $$

(4)

$$k^m\equiv k^{m+4}\pmod{5!} ,\ \ m\ge 3 $$ (5)

$$k^m\equiv k^{m+12}\pmod{6!} ,\ \ m\ge 4 $$

Example: in observation (2)

Choose any $k\in\mathbb{Z}_+$, now choose any $m\in\mathbb{Z}_{\ge 1}$

Let $k=5,m=3\rightarrow 5^3\equiv 5^5\equiv 5^7\equiv 5^9\equiv\cdots \equiv 5^{d} \pmod{3!}$

Question

For equation $ k^m\equiv k^{m+t}\pmod{q!}$

$(1)$ For every $q$ , does there exist $m$ and $t$ which follows above equation?

$(2)$ Can we generalize relation between $m,t$ and $q$?

Pruthviraj
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    You can use this simple generalization of Fermat & Euler's theorem. – Bill Dubuque Feb 26 '20 at 15:44
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    What do these congruences mean? Definitely for a fixed $r$ and $m\ge 1$, $k^m\equiv k^{m+2}\equiv r\pmod{2!}$ does not hold in general (e.g., when $r=1$, $k^m\equiv k^{m+2}\equiv r\pmod{2!}$ is false when $k=0$). – Batominovski Feb 26 '20 at 16:13
  • @WETutorialSchool note $k\in\mathbb{Z}{\ge 1}$ and $ r\in\mathbb{Z}{\ge 0}$. $r$ is depends on $k^m$ – Pruthviraj Feb 26 '20 at 16:20
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    Then $r$ is very much irrelevant. Why didn't you just write $k^m\equiv k^{m+t}\pmod{q!}$? When you add $r$ into your congruences, it creates a perception that $r$ is arbitrary. – Batominovski Feb 26 '20 at 16:24
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    It can be shown that for every positive integer $n$, the smallest positive integers $m$ and $t$ such that $k^{m}\equiv k^{m+t}\pmod{n}$ for every integer $k$ are the following: $m=\xi(n)$ where $\xi(n) =\max{s_1,s_1,\ldots,s_l}$ if $n=p_1^{s_1}p_2^{s_2}\cdots p_l^{s_l}$ is the prime factorization of $n$, and $t=\lambda(n)$ where $\lambda$ is the Carmichael function. Any other $m$ and $t$ that work must satisfy $m\ge \xi(n)$ and $\lambda(n)\mid t$. – Batominovski Feb 26 '20 at 16:43
  • @WETutorialSchool Thank you. You can write it as an answer. – Pruthviraj Feb 26 '20 at 17:13
  • Oops, I posted the wrong link above. I meant this Theorem. The theorem mentioned by @WETutorialSchool is equivalent to that. If that theorem works for you then we can mark this as a dupe (we already have many prior questions that employ that). But if someone want to work out the specific details in this case for the factorial modulus then it would not be a dupe. – Bill Dubuque Feb 26 '20 at 20:24
  • @BillDubuque It is almost equivalent to my statement. The only difference is $\phi(2^k)\ne \lambda(2^k)$ for $k\ge 3$. – Batominovski Feb 27 '20 at 02:58
  • @WETutorialSchool Correct. There are various versions of these results. – Bill Dubuque Feb 27 '20 at 03:05
  • @BillDubuque Do you think the smallest degree $d$ such that there exists a monic polynomial $f(x)\in\Bbb{Z}[x]$ that satisfies $f(k)\equiv 0\pmod{n}$ for all integers $k$ is $d=\xi(n)+\lambda(n)$? I am not sure how to prove this or whether there is a counterexample. – Batominovski Feb 27 '20 at 03:08
  • @WETutorialSchool I vaguely recall seeing various results about such long ago but, alas, I cannot recall the specifics. It would be a useful addition to the site if you post further results about such. – Bill Dubuque Feb 27 '20 at 03:45
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    @BillDubuque My claim is false. Let $z(n)$ denote the smallest positive integer $z$ such that $n\mid z!$. We have $z(n)\le \lambda(n)+1\le \xi(n)+\lambda(n)$. Then the monic polynomial $f(x)=\prod_{i=1}^{z(n)}(x-i)$ of degree $z(n)$ satisfies $f(k)\equiv 0\pmod{n}$ for every integer $k$. I don't know if $z(n)$ is the smallest possible degree of such a polynomial $f(x)$, though. – Batominovski Feb 27 '20 at 15:52
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    @WETutorialSchool There is much prior work here. Long ago I had a file of many papers on results in this area but - alas - it was among items that were stolen while in storage. You can locate prior work by searching Zbl / AMS math reviews, perusing Dickson's history, etc. Iirc some also occurred in Monthly problems. But it may be quicker to rediscover them in some cases. Here is one paper I just found with a quick search that yields one entry point into the literature. – Bill Dubuque Feb 27 '20 at 15:58
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    @WETutorialSchool And here is another. – Bill Dubuque Feb 27 '20 at 16:00
  • @BillDubuque Thanks, theorem 7 of your second paper is exactly what I was looking for. – Batominovski Feb 27 '20 at 16:04
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    @WETutorialSchool And another Stefan Schwarz, The role of semigroups in the elementary theory of numbers. 1981 Chasing citations on those should locate much of the literature on this and related topics. – Bill Dubuque Feb 27 '20 at 16:04

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