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?

Mathematics Asked by Aeternal on November 18, 2021

2 AnswersSuppose 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

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.

Answered by saulspatz on November 18, 2021

2 Asked on July 28, 2020 by lucas

0 Asked on July 27, 2020 by strictly_increasing

numerical calculus numerical methods ordinary differential equations probability theory stochastic differential equations

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