Prove $\Big(\frac{x+1}{x}\Big)^{2/n}$ is irrational for $x,n \in \mathbb{N}, n>2$

Question

For $x,n \in \mathbb{N}, n>2$, prove $$\Big(\frac{x+1}{x}\Big)^\frac{2}{n}$$ is irrational. I'm having trouble with this and I couldn't think of any possible way. Can someone help me or suggest me an idea of how to do this?

Answer 1

If $\left(\dfrac{x+1}{x}\right)^{\dfrac{2}{n}}= \dfrac{p}{q}\implies \dfrac{(x+1)^2}{x^2}= \dfrac{p^n}{q^n}\implies n = 2m\implies \dfrac{x+1}{x} = \dfrac{p^m}{q^m}\implies xq^m+q^m = xp^m\implies x = \dfrac{q^m}{p^m-q^m}$. Observe that we can assume $(p,q) = 1$ to begin with, and if a prime $r$ divides both $q^m$ and $p^m-q^m$,then $r$ divides $p^m$, and thus $r$ divides $p^m, q^m$, hence it divides $p,q$, and divides $(p,q) = 1$, contradiction $r > 1$. So the claim is proven.

Answer 2

Hint: If this is $a/b$ with $a$ and $b$ coprime, you have $a^n x^2 = b^n (x+1)^2$. If $p$ is a prime, how do the highest power of $p$ dividing $a$ and the highest power of $p$ dividing $x+1$ relate? Similarly for $b$ and $x$.