gcd between powers of two co-prime numbers

Question

Is it true that $\forall x,y,n\in \mathbb{Z}$, if $\gcd(x,y)=1$ then $\gcd(x^n, y)=1$? If not, is there a counterexample?

Answer 1

Hint: Assume $\gcd(x^n,y)\neq 1$. Then there exists a prime $p$ that divides both terms...

Answer 2

By Bezout there is $u,v$ so that $xu+yv=1$. Using binomial theorem yields: $$(xu+yv)^n=x^n{u^n}+ yv\sum_{k=0}^{n- 1}{{n\choose k} (ux)^k(vy)^{n-k-1}}=1$$ Now putting $U’=u^n$ and $V’= v\sum_{k=0}^{n- 1}{{n\choose k} (ux)^k(vy)^{n-k-1}}$ gives: $$U’x^n+V’y=1.$$Hence $\gcd(x^n,y)=1$