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I'm not sure I've got this right. When proving $a^n \mid b^n \Rightarrow a \mid b$, can we do this indirectly? In short,

"Suppose $a$ does not divide $b$, this implies that $a^n$ does not divide $b^n$. But $a^n \mid b^n$, hence $a$ divides $b$."

How about $n^n \mid m^m \Rightarrow n \mid m$? Can we do this the same way?

Xander Henderson
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Carolus
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5 Answers5

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This is false: $4^4$ divides $10^{10}$ but $4$ does not divide $10$.

Jyrki Lahtonen
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No, your titled implication holds only for $\:\!\rm\color{#c00}{squarefree} \,\ n,\,$ by the Theorem below (which also generalizes all of the counterexamples in the other answers).

Theorem $\rm\quad n\in\mathbb N\:$ is $\rm\color{#c00}{squarefree}$ $\rm\, \iff\, \forall\ m\in \!\mathbb N\!:\, n^n\: |\: m^{\:m} \Rightarrow\ n\:|\:m$

Proof $\ (\Rightarrow)\ \ $ If $\rm\:n\:$ is squarefree then prime $\rm\:p\:|\:n\:|\:n^n\:|\:m^m\ \Rightarrow\ p\:|\:m\:.\:$ So we conclude $\rm n\:|\: m\ $ since $\rm\:m\:$ is divisible by each prime factor of $\rm\:n,\:$ so also by their $\rm\:lcm =\:$ product $\rm = n.$

$(\Leftarrow)\ $ $\rm\ n\,$ not squarefree $\rm\:\Rightarrow n = a\, p^k,\:$ prime $\rm\:p\nmid a,\ k\ge 2\:.\:$ Let $\rm\: j\in\mathbb N,\ p^{k-1}\!\nmid j\:,\:$ e.g. $\rm\ j = 1$

Then $\rm\displaystyle\:\ m\: =\: k\:n+a\:p\;j\ \Rightarrow\ n^n =\: (a\:p^k)^n\ |\ (a\:p)^{k\:n}\ | \ m^m\ \:$ by $\rm\:\ ap\ |\ m \ge k\:n$

but $\rm\ n\nmid m,\ $ else $\rm\displaystyle\ n\:|\:a\:p\:j\ \Rightarrow\: p^{k-1}\:|\ j\:,\: $ contra choice of $\rm\ j$.

Remark $\ $ The counterexamples in the other answers are special cases of this:

E.g. $\rm\ k = 2,\ a = 1\ $ yields $\rm\: m = k\:n + a\:p\:j = 2\:n + p\:j\:,\ p\nmid j\:.\:$

Hence $\rm\:n = 4\:,\ p = 2\ $ yields $\rm\:m = 8 + 2\:j\:,\ 2\nmid j\:.\:$ So $\rm\: j= 1\:$ yields $\rm\:m = 10\ $ (Jyrki);

$\rm\, j = 3\:$ yields $\rm\:m = 14\ $ (link from Chandru = user9413).

$\rm\ n = a\:p^k = 3\cdot 2^2$ yields $\rm\:m = k\:n + a\:p\:j = 24 + 6\:j,\ 2\nmid j\:.\:$ So $\rm\:j = 7\:\Rightarrow\:m = 66\:$ (mixedmath).

Bill Dubuque
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    Great answer, as it shows not only why the argument doesn't work but also where it can work. – Ross Millikan Jul 21 '11 at 05:34
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    this can be the best answer – Dinesh Jul 25 '11 at 18:10
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    It is true for squarefree $n$ and there are counterexamples for all non-squarefree $n$. But which $m$ is it true for? For example $m=1$ or prime or a prime power or I think $m=6$ or $m=15$ or other products of two close primes. Which more are there? – Henry Jan 15 '17 at 13:40
  • Generally if $,n\ge m,$ then $,a^n\mid b^m\Rightarrow a\mid b,,$ by $,a^m\mid a^n\mid b^m\Rightarrow a\mid b,$ by here, but this inference is only vacuously true in the OP where $,a,b = n,m,,$ by $,n>m\Rightarrow n^n > m^m\Rightarrow n^n\nmid m^m\ \ $ – Bill Dubuque Oct 07 '24 at 18:34
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Below are equivalent definitions of $\rm\ q\,$ squarefree. Yours is $(5)$.

Theorem $\ $ Let $\rm\ 0 \ne q\in \mathbb Z\:.\ \ $ The following are equivalent.

$(1)\rm\quad\ \ \ \, n^2\,|\ q\ \ \Rightarrow\ \ n\ |\ 1\qquad\ $ for all $\rm\:\ n\in \mathbb Z $

$(2)\rm\quad\ \ \ \, n^2\, |\, qm^2 \!\Rightarrow n\ |\ m\qquad\! $ for all $\rm\: \ n,m\in \mathbb Z$

$(3)\rm\qquad\ q\ |\ n^2\ \Rightarrow\ q\ |\ n\qquad\ $ for all $\rm\:\ n\in \mathbb Z $

$(4)\rm\qquad\ q\ |\ n^k\ \Rightarrow\ q\ |\ n\qquad\ $ for all $\rm\:\ n\in \mathbb Z,\ k\in \mathbb N $

$(5)\rm\quad\:\ \: q^q\ |\ n^n\ \Rightarrow\ q\ |\ n\qquad\ $ for all $\rm\:\ n\in \mathbb N,\ $ for $\rm\ q > 0 $

Proof $\ \ $ See this answer.

Bill Dubuque
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Not quite. And here's why:

Note that $12 \not | \;66$. Also, $12 = 2^2 \cdot 3$ and $66 = 2 \cdot 3 \cdot 11$. But $12^{12} |\; 66^{66}$, because $66^{66}$ has 66 'different' factors of 2 and 66 'different' factors of 3, and $12^{12}$ has only 24 'different factors of 2 and 12 factors of 3.

So the fact that $a \not | \;\;b$ does not imply that $a^a \not |\; \; b^b$. And I think that was the content of your question, right?

davidlowryduda
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Here is another example taken from this link $$4^{4} \mid 14^{14} \quad \text{but} \ 4 \nmid 14$$