$n^a$ integral for all integer $n$ implies $a$ integral

Question

Let $a>0$ be a real number, such that for all integers $n\geq 1$: $n^a \in \mathbb N$
Show that $a$ must be an integer.

It's not difficult to show this when $a$ is a rational number: $2^\frac{p}{q}$ is irrational when the fraction is in lowest terms and $q \neq 1$.

When $a$ is irrational, then for all $n\geq 1$, there exists $m_n\in \mathbb N^*$, such that: $$a = \frac{\log m_n}{\log n}; \quad \text{$m_n$ is not a power of $n$}$$

I think considering $n=2,3$ is enough to show a contradiction, but I can't seem to find it. This is what I get: $$a = \frac{\log p}{\log 2} = \frac{\log q}{\log 3}$$ $$p = 2^{\log q/\log 3} $$

I think the RHS is irrational when $q$ is not a power of $3$, but I can't prove it. The closest thing I have to a solution is this answer on a similar question. But it uses an unproven conjecture, and I was hoping for a more elementary proof.

If $n^c\in\mathbb N$ for every $n\in\mathbb N$, then $c$ is a non-negative integer? — mathlove, Feb 09 '16 at 07:54

score 7 · Accepted Answer · edited Feb 15 '16 at 06:13

7

The idea is decreasing the exponent $a$ below $0$ by taking differences.

For every function $f$, let $(\Delta f)(x)=f(x+1)-f(x)$ be the usual forward difference.

$Lemma$. If $f(x)$ is a $k$ times differentiable function on $[n,n+k]$ then there is some $\xi\in(n,n+k)$ such that $$ f^{(k)}(\xi) = (\Delta^kf)(n) = \sum_{\ell=0}^k (-1)^{k-\ell}\binom{k}{\ell}f(n+\ell). $$

$Proof$. Induction on $k$. For $k=1$ this is exactly Lagrange's mean value theorem.

If the Lemma holds true for $k-1$ then apply it to the function $g(x)=\Delta f$. With some $\zeta\in(n,n+k-1)$, and then $\xi\in(\zeta,\zeta+1)\subset(n,n+k)$, we have $$ (\Delta^k f) (n) = (\Delta^{k-1} g)(n) = g^{(k-1)}(\zeta) = (\Delta f^{(k-1)})(\zeta) = f^{(k)}(\xi), $$ the lemma has been proved.

Now suppose that $a$ is not an integer; let $k$ be a positive integer with $k-1<a<k$. Applying the lemma with an arbitrary positive integer $n$, $$ a(a-1)\dots(a-k+1) \, \xi^{a-k} = \sum_{\ell=0}^k (-1)^{k-\ell}\binom{k}{\ell} (n+\ell)^a. $$ If $n$ is sufficiently large then the LHS is a number between $0$ and $1$. But we have some integer on the RHS, contradiction.

edited Feb 15 '16 at 06:13

DanielWainfleet

57,985

answered Feb 09 '16 at 17:06

G.Kós

14,297

Got it. This looks correct. But could you explain how you got the idea, if you don't mind? – Kitegi Feb 09 '16 at 17:39
1

First I thought to the case $a=0.5$. The idea of taking the difference $\sqrt{n+1}-\sqrt{n}$ was obvious. Then I just needed to take multiple differences. – G.Kós Feb 09 '16 at 18:12
@user141614 how do we know that the LHS is between 0 and 1? – Sandeep Silwal Feb 09 '16 at 19:27
1

@SandeepSilwal $\xi>n$, and $a-k<0$, so the LHS tends to $0$ as $n$ tends to $+\infty$ – Kitegi Feb 09 '16 at 19:29
I edited out the word "and" before "$n$" in the 2nd-last sentence .I think it was just a typo. – DanielWainfleet Feb 15 '16 at 06:15

$n^a$ integral for all integer $n$ implies $a$ integral

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