Let $p>1$ be a natural number. Prove that $n=\sqrt[p-1]{p}$ can't be a natural number if $p>2$.

Question

Let $p>1$ be a natural number. Prove that $n=\sqrt[p-1]{p}$ can't be a natural number if $p>2$. $n=2$ for $p=2$ of course !

Equivalently show that $p=n^{p-1}$ admits for $p>1$ $(n,p)=(2,2)$ as the only pair of non zero natural numbers as solution.

The question is related to exercise n°5 of Bourbaki's first treatise on Algebra [chapter I-III]. At this stage the reader is supposed to have read the treatise covering Set theory. He/She isn't supposed to know what a square root nor even know what a real number is :-)

Thanks

Answer 1

Claim: For all $n\geq 3$, $2^{n-1}\gt n$.

This holds for $n=3$: $2^{3-1} = 4\gt 3=n$.

Assume that $2^{n-1}\gt n$, $n\geq 3$. Then $2^{n}\gt 2n$, and $2n = n+n\gt n+1$. Therefore, $2^{(n+1)-1} \gt n+1$.

This establishes the claim by induction.

Now, this means that for all $p\gt 2$, $n\geq 2$, $n^{p-1}\geq 2^{p-1}\gt p\gt 1^{p-1}$. Thus, $p$ cannot be a $(p-1)$st power of a natural number.

Answer 2

By contradiction, suppose $2<p\in \Bbb N$ and $n^{p-1}=p$ with $n\in \Bbb N.$ Then $n>1,$ so let $n=1+x$ with $x>0.$ Since $2\le p-1\in \Bbb N,$ by the Binomial Theorem we have $$p=n^{p-1}=(1+x)^{p-1}=\sum_{j=0}^{p-1}\binom {p-1}{j}x^j\ge$$ $$\ge 1+\binom {p-1}{1}x+\binom {p-1}{2}x^2>$$ $$>1+\binom {p-1}{1}x=1+(p-1)x$$...because $x>0$ so each term in the Binomial Expansion of $(1+x)^{p-1}$ is positive.

Therefore $p>1+(p-1)x,$ that is, $p-1>(p-1)x,$ and since $p-1>0,$ this implies $1>x.$ But $x>0,$ so $n=1+x$ is a natural number with $1<n<2,$ which is absurd.

Remark: We can, instead, employ induction on $p\ge 3$ to show that if $x>0$ and $3\le p\in \Bbb N$ then $(1+x)^{p-1}>1+(p-1)x.$