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Find all real numbers $c$ satisfying the following condition:

For all $n \in \mathbb{N}$, we have $n^c \in \mathbb{N}$.

My attempt: Clearly all $c \in \mathbb{N}$ works while negative integer $c$ does not work.

Suppose $c=\frac{p}{q} \in \mathbb{Q}$ such that $n^c \in \mathbb{N}$. Then denote $\{ x \}$ as fractional part of $x$ while denote $\lfloor x \rfloor$ as integer part of $x$. Then we have $n^{\frac{p}{q}}=n^{\lfloor \frac{p}{q} \rfloor} n^{\{ \frac{p}{q} \}} \in \mathbb{N} \Rightarrow n^{\{ \frac{p}{q} \}} \in \mathbb{Q}$, which is a contradiction. So $c$ is not rational number.

I have trouble proving $c$ is not irrational number. Can anyone give some hint?

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    $\begingroup$ I think you mean, "Suppose $c = \frac{p}{q} \in \Bbb Q - \Bbb N$..." $\endgroup$ – Travis Willse May 22 '15 at 7:24
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    $\begingroup$ $4^{1/2}=2$. Did I miss something? $\endgroup$ – mhp May 22 '15 at 7:24
  • $\begingroup$ @mhp..Yes. For all $n\in \mathbb N$, we should have: $n^c \in \mathbb N$. For example: $3^{\frac12}$ isn't in $\mathbb N$ $\endgroup$ – hamid kamali May 22 '15 at 7:26
  • $\begingroup$ @hamid I see. Thanks. $\endgroup$ – mhp May 22 '15 at 7:28
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    $\begingroup$ There is a mathoverflow question on this topic. $\endgroup$ – TonyK May 22 '15 at 8:02
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If $0<c<1,$ then $1<2^c<2,$ so $c$ doesn't have the property. Suppose now $1<c<2.$ Define $f(x) = (x+1)^c-x^c.$ Because $x^c$ is strictly convex, $f(x)-f(x-1)> 0$ for $x\ge 1.$ That's because it equals

$$\frac{(x+1)^c - x^c}{(x+1)-x}-\frac{x^c - (x-1)^c}{x-(x-1)},$$

i.e., the difference in the slopes of successive chords on the graph of a strictly convex function. On the other hand, the MVT shows $f(x)-f(x-1) = f'(y_x) = c[(y_x+1)^{c-1}-y_x^{c-1}].$ Because $0<c-1<1,$ this latter difference $\to 0$ as $x\to \infty.$ If $c$ had the property in question, then we have $f(n+1)-f(n)$ is always a positive integer and yet $\to 0,$ contradiction.

Well, that's a start ...

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