If $\gcd (a,b) = 1$ then $\gcd (a^n,b^k) = 1$

Question

Statement: Suppose $(a,b) = 1$ then $(a^n,b^k) = 1$ for $n,k \geq 1$.

Attempt at Proof: Let $P$ be the set of all primes. Let $P_a$ be the set of primes $p_i$ such that

$$a = \prod_{i=1}^{r_1} p_i^{\alpha_i}.$$ Let $P_b$ be the set of primes $q_i$ such that $$b = \prod_{i=1}^{r_2} p_i^{\beta_i}.$$

Since $(a,b) = 1$, we have $P_a \cap P_b = \emptyset$.

Write $a = \prod_{i=1}^{\infty} p_i^{\alpha_i}$ where $\alpha_i = 0$ if $p_i \notin P_a$. Write $b = \prod_{i=1}^{\infty} q_i^{\beta_i}$, where $\beta_i = 0$ if $q_i \notin P_b$. Thus, $a^n = \prod_{i=1}^{\infty} p_i^{n\alpha_i}$ and $b^k = \prod_{i=1}^{\infty} q_i^{k\beta_i}$. By theorem 1.12 (Apostol Analytic Number Theory), we may write

$$(a^n,b^k) = \prod_{P_a} p_i^{c_i} \prod_{P_b} q_i^{c_i} \prod_{P \setminus (P_a \cup P_b)} s_i^{c_i}.$$

Where $c_i = \min\{n\alpha_i,k\beta_i\}.$ But, $c_i = 0$ for all $P_a,P_b$ by the above discussion and clearly $c_i = 0$ for $P \setminus (P_a \cup P_b).$ Thus, the above product is $1$, so $$(a^n,b^k) = 1$$ as desired.

Are there any issues with this proof? This is my first day of number theory, so I'm a bit unfamiliar with the proof structures.

Answer 1

Much easier: as $p\mid rs, p\text{ prime}\implies p\mid r\text{ or }p\mid s$: $$p\mid a^n\text{ and }p\mid b^k\implies p\mid a\text{ and }p\mid b.$$

Answer 2

Here is a proof without resorting to prime factorisations: $\gcd(a, b)$ means that there are integers $x, y$ such that $$ ax + by = 1 $$ If we raise each side of this equality to the power of $n+k-1$ and use the binomial theorem, we get $$ (a + b)^{n+k-1} = 1^{n+k-1}\\ a^{n+k-1} + \cdots + \binom{n+k-1}{k-1}a^nb^{k-1} + \binom{n+k-1}{k}a^{n-1}b^k + \cdots + b^k = 1\\ a^n\left(a^{k-1} + \cdots + \binom{n+k-1}{k-1}b^{k-1}\right) + b^k\left(\binom{n+k-1}{k}a^{n-1} + \cdots + b^{n-1}\right) = 1 $$ which means that $\gcd(a^n, b^k) = 1$ as well.