How to show that $(2^n-1)^{1/n}$ is irrational for all integer $n\ge 2$?
If $(2^n-1)^{1/n}=q\in\Bbb Q$ then $q^n=2^n-1$ which doesn't seem right, but I don't get how to prove it.
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Bart Michels
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If you know that the $n$-th root of a positive integer is either an integer or irrational … – Daniel Fischer Sep 05 '15 at 12:30
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For contradiction, assume $(2^n-1)^{1/n}$ is rational.
Then $(2^n-1)^{1/n}=p/q$ for some $p,q\in\Bbb Z_{>0}, \gcd(p,q)=1$.
But then $(2^n-1)q^n=p^n\implies q^n\mid p^n\implies q=1$.
$2^n-1=p^n$. We can't have $p=1$, since $n\ge 2$. But then $2^n-1<2^n\le p^n$, contradiction.
user236182
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If the $n$th root of an integer is rational, then it is in fact an integer (any prime occuring in the dneominator of $\frac xy$ occurs also in the denominator of $\frac{x^n}{y^n}$). As $1^n<2^n-1<2^n$, this is not possible.
Hagen von Eitzen
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If $(2^n-1)^{1/n}=a/b$ for $a,b\in\Bbb Z$ and $n\ge 3$, then $$a^n+b^n=(2b)^n,$$ contradicting Fermat's Last Theorem.
Spenser
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Fermat's last theorem isn't strong enough to prove $n=2$, I guess ;) – Caleb Stanford Sep 05 '15 at 13:14