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Let $a,b,n \in \mathbb{N}.$ Prove that if $a^n \mid b^n$, then $a \mid b$.

$\textit{Proof.}$ Consider the prime factorizations for $a$ and $b$ as follows:

$$a=p_1 \cdots p_r, $$ $$b=q_1 \cdots q_s.$$

Where $q_i, p_j$ are all prime numbers. Observe that $p_k$ could be equal to $p_m$ for some $m≠k$, and similarly for $q$. (In other words, for simplicity I decided not to write $a=p_1^{\alpha_1} \cdots p_r^{\alpha_r}$).

Then, we have $$a^n=p_1^n \cdots p_r^n,$$ $$b^n=q_1^n \cdots q_s^n.$$

Note that the prime factors of $b$ are exactly the same prime factors of $b^n$, but each prime factor of $b^n$ is to the $n$-th power.

Since $a^n \mid b^n$, then $b^n=a^n \cdot q$ for some integer $q$, so that

$$b^n = p_1^n \cdots p_r^n \cdot q$$

If $p_i$ for some $i$ doesn't appear in the prime factors of $b$, then $p_i$ couldn't appear in the prime factors of $b^n$. Then, each $p_i$ appearing in $b^n$, will appear in $b$. Hence, $b=p_1 \cdots p_n \cdot p$ for some integer $p$. Then $b=ap$, as desired.

... $\textit{Q.E.D.}$??

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    $\begingroup$ The last step is that $p\mid b^n$ implies $p\mid b$, because $p\mid ab$ always implies $p\mid a$ or $p\mid b$ for $p$ prime. You should write this formally. $\endgroup$ – Dietrich Burde Aug 4 '19 at 16:59
  • $\begingroup$ @DietrichBurde Thanks! $\endgroup$ – rowcol Aug 4 '19 at 17:15
  • $\begingroup$ "If pi for some i doesn't appear in the prime factors of b, then pi couldn't appear in the prime factors of bn." ... yeah... but this doesn't address how many times the $p_i$ occurs. You get $p_i = q_j$ for some $q_j$ but then you have $p_m =p_i$ and so maybe $p_m = q_j$ and you never the other $q_j$. For example $12\not \mid 18$ and yet $12 =3*2*2;p_1=3;p_2=2;p_3=2$ so $12^n = 3^n*2^n*2^n$ and $18=2*3*3;q_1=2;q_2=3;q_3=3$ and we have $p_1=q_2$ and $p_2=q_1$ and $p_3=q_1$. we have $12=q_2q_1q_1$ and $q_2q_1|b$ but we don't have $q_2q_1q_1=q_2q_1$ $\endgroup$ – fleablood Aug 4 '19 at 17:51
  • $\begingroup$ NOT a duplicate as this is a "critique my proof" and not a "how do I prove it" question. $\endgroup$ – fleablood Aug 4 '19 at 18:10
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I think you have a problem with if $p_i=p_j$ and $p_i$ must be one of the $q_r$ and $p_j$ must be one of the $q_s$, you haven't distinguished that if we chose $p_i$ as $q_r$ that we can not also choose $p_j$ as the same instance of $q_r$. You haven't come up with a method of removing $q_r$ from the pool once its been use.

For instance if $a = 12$ and $b =18$ and $p_1 = 2; p_2=2; p_3=3$ and $q_1=2; q_2 =3; q_3=3$ we'd have each $p_j$ equal to some $q_i$: $p_1 = q_1$ and $p_2 = q_1$ and $p_3 = q_2$. We haven't got a method to say if $p_1 = q_1$ we can't use $q_1$ a second time.

I can see why you wanted to avoid powers but... I think you need them.

If $a= \prod p_i^{v_i}$ and $b = \prod q_j^{w_j}$ then $a^n= \prod p_i^{n*v_i}|b^n = \prod q_j^{n*w_j}$ so $\{p_i\}\subset \{q_j\}$.

Relabel the variables and write $b$ as $\prod p_i^{u_i} \prod_{q_j\not \mid a} q_j^{w_j}$. So we have $b^n = \prod p_i^{n*u_i} \prod_{q_j\not \mid a} q_j^{n*w_j}$ and we have for each $n*vi \le n*u_i$. Which means $u_i \le v_i$. So $b = \prod p_i^{u_i\ge v_i} \prod_{q_j\not \mid a} q_j^{w_j}$. And thus $a = \prod p_i^{v_i}| \prod p_i^{u_i\ge v_i} \prod_{q_j\not \mid a} q_j^{w_j} = b$.

......

Note: https://math.stackexchange.com/a/1815338/280126 is a similar proof that doesn't require unique prime factorisation but that if $\frac ba \in \mathbb Q$ and we represent it as $\frac {b'}{a'}$ where $a'$ and $b'$ are relatively prime we get a contradiction if we assume $b' \ne 1$.

I don't know if it is easy, nor do I understand, the frequent complaint "You don't need unique factorization[1]" (Yeah, ... but is there are reason to avoid it?) But it's worth a looksie as basically the same concept, but with much of the tedious mechanics we had to slog through rather simplified.

[1]Of course this requires that rationals can be written in lowest terms and that all integers have a prime factor and those presume unique factorization so it isn't avoided.

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This question have be answered before, here is an answer by Bill Dubuque.

Hint $\,\ \dfrac{b^n}{a^n} = k\in \Bbb Z$ $\ \Rightarrow\ $ $x = \dfrac{b}a\ $ root of $\ x^n\!-k$ $\!\!\!\!\underbrace{\Rightarrow\,x\in\Bbb Z}_{\text{ Rational Root Test}}\!\!\!\!\!$ $\,\Rightarrow\, a\mid b $

Another answer in a different way by awllower,

Prime factorization is not needed: we only need the fact that every integer $\ne\pm1$ has a prime divisor.
Define $r=a/b\in\mathbb Q.$ As $b^n\mid a^n,$ we know $r^n\in\mathbb Z.$
Write $r=a'/b'$ with $\gcd(a',b')=1.$ Let $s=r^n.$ Then $r^n=s$ implies that $(a')^n=(b')^n\cdot s.$ This shows that every prime divisor of $b'$ divides $(a')^n;$ by the definition of a prime, this means that every prime divisor of $b'$ divides $a'.$ This contradicts $\gcd(a',b')=1.$ Therefore $b'$ has no prime divisor, and is equal to $\pm1.$ Thus $r=a/b=a'/b'=\pm a'\in\mathbb Z.$ So $b$ divides $a.$

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    $\begingroup$ The OP asked if his/her proof was valid. He/she didn't ask for an alternative proof. $\endgroup$ – fleablood Aug 4 '19 at 18:11

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