Assume $x>1$ is an even integer, show that. $$\sqrt[5]{x} \notin \mathbb{N}$$
I am not sure if this is actually a true theorem, I am conjecturing based on $2, 4, 6, 8, 10, .... 126$.
I am attempting to try to prove this.
Let $x = 2$, $\sqrt{2} \notin \mathbb{Q}$ hence, $\sqrt{2} \notin \mathbb{N}$
Suppose $\sqrt{x} \notin \mathbb{N}$, the objective is to show: $\sqrt{x+2} \notin \mathbb{N}$ either.
I am not sure of the induction hypothesis since, $x+1$ will always be odd, I chose to go with $x+2$.
As @GitGud and several other Commenters pointed out, the premise of this Question is not true. Any fifth power of an even number is even, so $2^5 = 32$ and many other natural numbers furnish counter-examples.