Show $\,a^n \mid b^n\Rightarrow a \mid b,\,$ i.e. $\,r\in \Bbb Q,\,r^n\in\Bbb Z\Rightarrow r\in\Bbb Z$

Question

I would like to show that $a^n \mid b^n$ implies $a \mid b$

I thought I could convert it to congruences and work backwards, but as far as I remember, $a \equiv b \pmod{m}$ implies $a^n \equiv b^n \pmod{m}$, not the opposite, unless $m$ is prime. Is that right?

So I am not sure how to approach this one. Any ideas?

Thanks!

Answer 1

Hint $\, $ For $\rm\ c =\Large\frac{b}a\, $ it is $\rm\ c^{\large n} = d\in \mathbb Z\ \Rightarrow\ c\in \mathbb Z.\:$ Now apply the Rational Root Test to $\rm\ x^{\large n}-d$

Notice again, just as in your prior few questions, converting it from divisibility relational form into equational form helps us to solve it efficiently. Here it allows us to notice that the problem is simply the special monic case of the Rational Root Test. Recall that this says that the leading coefficient of a polynomial $\rm\ f(x)\in \mathbb Z[x]\ $ serves as a denominator for all its rational roots. In particular, when the leading coefficient is one this implies that every rational root can be written with denominator $= 1$, i.e. is an integer. This is a fundamental property in number theory. Indeed it is the property that is used to generalize the notion of integer from rationals to algebraic number fields.

For some generalizations see my post here. Here's a pertinent excerpt:

Theorem $\:$ TFAE for $\rm\: a,b\: $ in domain $\rm\:Z\:,\ \: r \in Q \:=\:$ fraction field of $\rm\: Z\:,\ n\in \mathbb N$

$(1)\ \ \rm\ \ r = \sqrt[n]a \ \Rightarrow\ r \in Z$

$(2)\ \ \rm\ \ r^n \in \:Z \ \ \Rightarrow\ r \in Z$

$(3)\ \ \rm\ \ \ a^n\:|\:b^n \ \ \Rightarrow\:\ a\:|\:b$

$(4)\ \ \rm\ \ (a^n) = (b^n,\: c^n) \ \Rightarrow\ (a) = (b,c)\ $ as ideals in $\rm\:Z$

Answer 2

Hint:

Write $b^n = q_1^{n \alpha_1} \dots q_k^{n \alpha_k}$, where $\alpha_i > 0$. If $a^n | b^n$, then $a^n = q_1^{n \beta_1} \dots q_k^{n \beta_k}$, where $0 \leq \beta_k \leq \alpha_k$.

Answer 3

Any prime $p$ that divides $a$ $k$ times, is such that $p^n$ divides $a^n$ $nk$ times and thus $b^n$ as well. But if a prime divides $b^n$ $nk$ times, it divides $b$ $k$ times as well. So every prime divisor of $a$ is one of $b$, with at least the same multiplicity.

Answer 4

Here is a different, though inelegant way to prove the proposition.

Suppose that $a \nmid b.$

Let $a = kx$, $b = ky$, with $k = \gcd(a,b)$. Then, it is easy to show that $x\nmid y.$

$$\implies x^n \nmid y^n $$

since if $\gcd(x^n, y^n) \neq 1$, then for some prime $p$, $\:p \mid x^n, y^n \implies p\:| \:x, y$, which is in contradiction with the coprimality of $x$ and $y$.

$$\implies k^nx^n \nmid k^ny^n $$

$$\implies a^n \nmid b^n $$

Thus $a^n \mid b^n \implies a \mid b$.