0

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$?

DanielWainfleet
  • 57,985
Intuition
  • 3,269
  • 1
    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 $b\equiv c \pmod d$. – lulu May 22 '19 at 23:32
  • @lulu and why the equivalence to $0$? – Intuition May 22 '19 at 23:52
  • 1
    Just think about it. If $k^2,|,k^n-k$ for $n≥2$ deduce that $k^2,|,k$. – lulu May 22 '19 at 23:53
  • I feel it ..... but I do not know how to prove it :( @lulu – Intuition May 22 '19 at 23:55
  • Well, try. These are very basic properties of congruences. Much simpler than something like Carmichael numbers. Just write it out. If $k^2,|,k^n-k$ then there is some integer $c$ with $k^2\times c =k^n-k$. Can you take it from there? – lulu May 23 '19 at 00:00
  • may be because $k^{2}$ & $k-1$ are relatively prime .... which I do not know how to prove. – Intuition May 23 '19 at 00:01
  • 1
    It has nothing to do with that. I think you need to review the basic properties of congruences. – lulu May 23 '19 at 00:02
  • may be I will cancel $k$ from both sides of the congruence .... but then what ?@lulu – Intuition May 23 '19 at 00:05
  • 1
    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. – lulu May 23 '19 at 00:06
  • This is not a homework problem .... it is on pg.91 of David M. Burton "seventh edition" it is in the paragraph before last ..... I am reading this paragraph and did not understand this part @lulu – Intuition May 23 '19 at 00:08
  • since $gcd (k, k^2) = k$ then $k \equiv k^n \pmod{k^2}$ is equivalent to $1 \equiv k^{n - 1} \pmod k$ which means that $ k \mid k^{n - 1} - 1 $ but then what ? @lulu – Intuition May 23 '19 at 00:19
  • I got it ..... I am sorry for my stupidness @lulu – Intuition May 23 '19 at 00:20
  • No I did not got it @lulu – Intuition May 23 '19 at 00:23
  • 1
    My edit was for a typo only. BTW for single keystrokes you can often not need brace-brackets, e.g. k^n instead of k^{n}. – DanielWainfleet Dec 18 '19 at 05:16

2 Answers2

2

It's a special case of: congruences persist mod factors of the modulus, i.e.

$$ \bbox[6px,border:1px solid red]{a\equiv \bar a\!\!\pmod{\!bm}\ \Rightarrow\ a\equiv \bar a\!\!\pmod{\! m}}\qquad\!$$

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 $\rm\,1\neq \color{#0a0}{a^{\large 2}}\!\mid n\mid \color{#0a0}{a^{\large e}}\!-\!a\, \Rightarrow\: a^{\large 2}\mid a\:\Rightarrow\!\Leftarrow$ $\rm\: (note\ \ e>1\: \Rightarrow\: \color{#0a0}{a^2\mid a^{\large e}})$

Bill Dubuque
  • 272,048
1

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 $n>1$