Necessary condition for x0 being an integer

Question

I was trying to solve a number theory problem and then I realized that I was needing to verify (prove or disprove) the following ''fact'' about numbers. I would appreciate any help.
Q: Suppose $x >0$ is such that $x^n in mathbb{Z}$ for all $n geq 2$, then $x$ must be an integer? Furthermore, suppose that $ngeq 3$ is an odd integer, does the conclusion holds?

Brian M. Scott · Answer

Suppose that $x^2,x^3inBbb Z$. Then $x^2(x-1)=x^3-x^2inBbb Z$, so $x-1$ is rational. The square of a rational number is an integer if and only if the rational number itself is an integer, so $x-1$ is an integer, and therefore so is $x$.

Answered by Brian M. Scott on November 18, 2021

saulspatz · Answer

If $x^2=kinmathbb{Z}$, then $x=sqrt{k}.$  If $x$ is not an integer, then neither is $x^3=ksqrt{k}$.
A similar argument proves the second statement also.

Necessary condition for x>0 being an integer

2 Answers

Add your own answers!

Ask a Question