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The proof as given in " David M. Burton " is as follows:

Suppose that $a^n \equiv a \pmod n$ for every integer a, but $k^2\mid n$ for some $k > 1.$ If we let $a = k,$ then $k^{n} \equiv k \pmod n.$ Because $k^2\mid n$, this last congruence holds modulo $k^2$; that is $ k \equiv k^{n} \equiv 0 \pmod {k^2}$, whence $k^2\mid k$, which is impossible. Thus, $n$ must be square-free.

But I do not understand this statement :

"this last congruence holds modulo $k^2$; that is $ k \equiv k^{n} \equiv 0\pmod k^2$"

Could anyone explain it for me please? why this last congruence holds modulo $k^2$ ? and why this leads to that $ k^{n} \equiv 0$?

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    $\begingroup$ This is just the definition of congruence. Saying $b\equiv c \pmod m$ means $m\,|\,(b-c)$. Of course, if this is true then, if $d\,|\,m$ we must also have $d\,|\,(b-c)$ or $b\equiv c \pmod d$. $\endgroup$ – lulu May 22 at 23:32
  • $\begingroup$ @lulu and why the equivalence to $0$? $\endgroup$ – Secretly May 22 at 23:52
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    $\begingroup$ Just think about it. If $k^2\,|\,k^n-k$ for $n≥2$ deduce that $k^2\,|\,k$. $\endgroup$ – lulu May 22 at 23:53
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    $\begingroup$ It has nothing to do with that. I think you need to review the basic properties of congruences. $\endgroup$ – lulu May 23 at 0:02
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    $\begingroup$ Please review the basic properties of congruences. Getting other people to do homework (or homework level) problems for you is a terrible way to learn a subject. $\endgroup$ – lulu May 23 at 0:06
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Here's my steps.

  • $a^n\equiv a\bmod n\implies nx+a=a^n$
  • $k^2\mid n \implies n=k^2c$
  • $a=k\implies k^2cx+k=k^n\implies k^n-k^2cx=k^2(k^{n-2}+cx)=k\implies k^2\mid k$

Hopefully you can get it in polynomial form. All I used was: conversion of modular congruence to a linear polynomial, divisor pairing, substitution, and factoring out ( reverse of distribution). The portion you quote breaks down to: $$k^n\equiv 0\bmod k^2$$ and, $$k\equiv k^n\bmod k^2$$ The latter of which follows from $k^2$ being a divisor of n, the former from $n>1$

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  • $\begingroup$ how is the former from $n > 1$, what if $n=3$? or it can not be because we are assuming that $k^2 | n$? $\endgroup$ – Secretly May 23 at 2:19
  • $\begingroup$ if $n<2$ then $k^{n-2}=k^m, m<0$ $\endgroup$ – Roddy MacPhee May 23 at 2:23
  • $\begingroup$ No this is not what I am asking, I am saying if n is as odd number greater than 2, like 3 for example $\endgroup$ – Secretly May 23 at 2:30
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    $\begingroup$ if n is greater than 2, then $ n-2 $ is greater than 0 leading to $k^n$ having $k^2$ as a factor. $\endgroup$ – Roddy MacPhee May 23 at 2:33
  • $\begingroup$ got it thanks :) $\endgroup$ – Secretly May 23 at 2:39
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It's a special case of: congruences persist mod factors of the modulus, i.e.

$$ \bbox[5px,border:1px solid red]{a\equiv \bar a\!\!\pmod{\!bm}\ \Rightarrow\ a\equiv \bar a\!\!\pmod{\! m}}\qquad\!$$

via defining divisibility persists: $\, m\mid bm\mid a-\bar a\,\Rightarrow\, m\mid a-\bar a\,$ by transitivity of "divides".

So in the OP the congruence $\,k\equiv k^{\large n}\pmod{\!n}\,$ remains true $\!\bmod k^2\,$ by $\,k^2\mid n$.

Thus $\bmod k^2:\,\ k\equiv k^{\large n}\equiv 0\,$ by $\,k^{\large 2}\mid k^{\large n}\,$ by $\,n\ge 2$

Remark $ $ You can find a full proof here of this criterion for Carmichael numbers, where I present this part concisely as follows:

If $\rm\,n\,$ isn't squarefree then $\rm\,1\neq \color{#0a0}{a^{\large 2}}\!\mid n\mid \color{#0a0}{a^{\large e}}\!-\!a\, \Rightarrow\: a^{\large 2}\mid a\:\Rightarrow\!\Leftarrow$ $\rm\: (note\ \ e>1\: \Rightarrow\: \color{#0a0}{a^2\mid a^{\large e}})$

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