If $a, b \in \mathbb{N}$ are relatively prime, show that, for any $k \in \mathbb{N}$, $a^k$ and $b$ are relatively prime.

Question

I am given this statement: $a, b \in \mathbb{N}$ are relatively prime if and only if there exist integers $\alpha, \beta$ such that $1 = \alpha \cdot a + \beta \cdot b$.

I know that $\gcd(a^k,b) = 1$ means $a^k$ and $b$ are relatively prime, so if I use this to show $\gcd(a^{k+1}, b) = 1$ then I can prove this by induction.

When $k = 1$, $\gcd(a,b) = 1$ because $a$ and $b$ are relatively prime. So this works for $k = 1$.

Next, I'm assuming the formula holds $\gcd(a^k) = 1$ and now I have to show that $\gcd(a^{k+1}) = 1$.

Because $a^k$ and $b$ are relatively prime, $1 = \alpha \cdot a^k + \beta \cdot b$. I'm pretty sure if I can show that $1 = \alpha \cdot a^{k+1} + \beta \cdot b$, that would complete the proof. This is where I'm stuck.

Answer 1

Hint. You can do it without induction. If $$1=\alpha a+\beta b$$ then by the binomial theorem, $$1=[\alpha^k]a^k+\left[\binom k1\alpha^{k-1}a^{k-1}\beta+\binom k2\alpha^{k-2}a^{k-2}\beta^2b+\cdots+\binom k{k-1}\alpha a\beta^{k-1}b^{k-2}+\beta^kb^{k-1}\right]b\ ,$$ and the terms in square brackets are integers.

See if you can fill in the details.