Proof that if $a^n|b^n$ then $a|b$

Question

I can't get to get a good proof of this, any help?

What I thought was:

$$b^n = a^nk$$ then, by the Fundamental theorem of arithmetic, decompose $b$ such:

$$b=p_1^{q_1}p_2^{q_2}...p_m^{q_m}$$

with $p_1...p_m$ primes and $q_1...q_n$ integers.

then

$$b^n=(p_1^{q_1}p_2^{q_2}...p_m^{q_m})^n= p_1^{q_1n}p_2^{q_2n}...p_m^{q_mn}$$

but here i get stucked, and i can't seem to find a good satisfactory way to associate $a$ and $b$...

Any help will be appreciated

Answer 1

Note: I am making one assumption you did not state. I assume $n \in \{1, 2, ... \}$.

Without loss of generality, let $ a = p_1^{\alpha_1} p_2 ^{\alpha_2} ... p_m^{\alpha_m}$ and $b = p_1^{\beta_1} p_2 ^{\beta_2} ... p_m^{\beta_m}.$ Note this means that some $\alpha_k, \beta_k, k \in \{1, 2, ... m\}$ may be zero. Now assume $a^n | b^n$. Then,

$a^n = p_1^{n \alpha_1} ... p_m^{ n \alpha_m}$ and $b^n = p_1^{n \alpha_1} ... p_m^{n \alpha_m}$ are such that $n \alpha_k \le n \beta_k, \forall k \in \{1, ... , m\}$ (because $a^n$ divides $b^n$). This is true if and only if $\alpha_k \le \beta_k \forall k \in \{1, 2, ... , m\}$ (because $n > 0$). However, this is true if and only if $a | b$.

Answer 2

Start with a single prime $p$.

$p^a|p^b \Leftrightarrow a\leq b$.

So if $p^{na}|p^{nb}$ then $na \leq nb$ which means that $a \leq b$, so $p^a | p^b$. Use prime decomposition for general case.

Answer 3

Show that if $p$ is prime and $p \mid a$, then $p \mid b$.

Then note that $({ a \over p})^n \mid ({b \over p})^n$. This is the induction step.

Answer 4

Assume $p^r\|b$ (i.e. $p^r\mid b$ and $p^{r+1}\not\mid b$) and $p^s\|a$. Then $a^n\mid b^n$ gives aus $ns\le nr$. Provided $n>0$ this implies $s\le r$. Do so for all primes to conclude $a\mid b$

Answer 5

If you know unique factorisation, you can do this by contradiction.

Suppose $a$ is not a factor of $b$. Then there must be a prime $p$ with $p^r$ a factor of $a$, but $p^r$ is not a factor of $b$ (otherwise $a$ is a factor of $b$). Let $p^s$ be the maximum power of $p$ which divides $b$ - and note that $s\lt r$.

Then $a^n=p^{rn}a_1^n$ and $b^n=p^{sn}b_1^n$ with $a_1$ and $b_1$ coprime to $p$. Hence $a^n $ cannot be a factor of $b^n$. Contradiction.

Answer 6

Hint $\ p^{n\alpha}\!\mid p^{n\beta}\! \iff n \alpha \le n\beta \iff \alpha \le \beta \iff p^\alpha\mid p^\beta.\,$ Apply it to prime factorizations of a,b.

Simpler: $\, (b/a)^n = k\in\Bbb Z\Rightarrow b/a\in \Bbb Z\,$ by the Rational Root Test applied to $\,x^n - k.$