Proof that if $\gcd(a,b) = 1$ then $\gcd(a^n, b^m) = 1$

Question

Proof via induction for $n,m \geq 0$ that if $\gcd(a,b) = 1$ then $\gcd(a^n, b^m) = 1$ for any $n,m$ that are positive integers.

Answer 1

In the proof, I will use the following theorem:

If a prime p divides a*b then p either divides a or b or both

Now the proof

Say by way of contradiction gcd(aⁿ,b^m) = k > 1

Recall for all integers k greater than 1, there is a prime p such that p divides k. let p be such a prime.

k is the gcd of aⁿ and b^m thus k divides aⁿ and b^m
p divides k thus p divides aⁿ and b^m
Recall that if a prime divides the product of two integers it must divide one of them, as a corollary we have that if p divides aⁿ, p must divide a. similarly p divides b.

Thus p divides a and b, so p must divide gcd(a,b) however gcd(a,b) = 1. Thus by way of contradiction, gcd(aⁿ,b^m) = 1.

As your question suggests, there is induction in this proof. I have hidden the induction step in the corollary.

Try proving the corollary of the theorem I stated via induction.