Raising positive integer to $c\in\mathbb{R}-\mathbb{N}$ rarely gives an integer!

Question

Problem: Let $c>0$ be a real number, and suppose that for every positive integer $n$, at least one percent of the numbers $1^c,2^c,3^c,\dotsc,n^c$ are integers. Prove that $c$ is an integer.

My progress: At first we will deal with the case when $c$ is a rational number. Suppose $c=\frac{a}{b}$. It indeed suffices to prove the statement for rationals of the form $\frac{1}{a}$. Observe that there are $\lfloor{M^{\frac{1}{a}}}\rfloor$ integers of the form $n^{\frac{1}{a}}$ between $1$ and $M$. So the percentage of integers of the form $n^{\frac{1}{a}}$ among the first $M$ integers is $$\frac{\lfloor{M^{\frac{1}{a}}}\rfloor}{M}\times 100\le \frac{M^{\frac{1}{a}}}{M}\times 100=\frac{100}{M^{1-\frac{1}{a}}}$$ which will be less than 1 for sufficiently large $M$.

But I am unable to prove the problem for any real $c$. I tried approximating reals with a sequence of rational numbers, but it didn't work well.

I was recently working on an open problem of similar kind, and I stumbled upon this sub-problem. How to solve this one (preferably not requiring too much heavy tool)?

Answer 1

We may modify the finite differences argument. Let $0=t_0<t_1<t_2<\ldots<t_d$ be fixed rational numbers. Our local goal is to find such rational coefficients $\lambda_0,\lambda_1,\ldots,\lambda_d$ that the mean value theorem $$\exists \tilde{x}\in [x,x+t_d]\colon\quad f^{(d)}(\tilde{x})=\sum \lambda_i f(x+t_i)$$ holds for any real $x$ and any $d$ times continuously differentiable on $[x,x+t_d]$ function $f$. This may be done as follows: consider the interpolating polynomial $$h(y)=\sum_{i=0}^d f(x+t_i)\prod_{j\ne i}\frac{y-x-t_j}{t_i-t_j}, \quad \deg h\leqslant d, \quad h(x+t_i)=f(x+t_i).$$ The function $g(y):=f(y)-h(y)$ has zeroes at $y=x+t_i,i=0,\ldots,d$. Therefore by Rolle's theorem there exists $\tilde{x}\in [x,x+t_d]$ such that $g^{(d)}(\tilde{x})=0$. This rewrites as $$ f^{(d)}(\tilde{x})=\sum \lambda_i f(x+t_i),\quad \lambda_i=\frac{d!}{\prod_{j\ne i} (t_i-t_j)}. $$

Now back to the problem. Choose a positive integer $d$ such that $c-d<0$ and integer $M>1000 d$. Call an integer $k$ nice if the set $[k\cdot M+1,k\cdot M+2,\ldots,k\cdot M+M-1]$ contains at least $d+1$ integers $y$ for which $y^c$ is integer. By density condition there exist arbitrarily large nice integers. Denoting the corresponding $d+1$ integers by $x+t_0,x+t_1,\ldots,x+t_d$, where $0=t_0<t_1<\ldots<t_d<M$, and applying our formula we get $$c(c-1)\ldots (c-d+1)\tilde{x}^{c-d}=\sum \lambda_i (x+t_i)^{c}\in (M!)^{-1}\mathbb{Z}.$$ But LHS becomes arbitrarily small while non-zero for large $x$.

Answer 2

To expand on a comment of Lucia, when $c$ is irrational, we can show that there are at most $O((\log N)^2)$ values of $n\leq N$ such that $N^c$ is rational, let alone an integer.

Let $\mathcal{A}$ be the set of $n\leq N$ such that $n^c$ is rational and $n > 1$. Let $n_1$ be the smallest element of $\mathcal{A}$ and let $n_2$ be the smallest element of $\mathcal{A}$ that is not a rational power of $n_1$. The number of elements of $\mathcal{A}$ when no such $n_2$ exists is $O((\log N)^{1+\varepsilon})$ (exercise to the reader).

Since $n_2$ is not a rational power of $n_1$, given a prime $p\mid n_1n_2$, there exists another prime $q\mid n_1n_2$ such that $$ \tag{1}\label{1} \nu_p(n_1)\nu_q(n_2) - \nu_p(n_2)\nu_q(n_1) \neq 0, $$ where $\nu_p(m)$ is the exponent of $p$ in the prime factorization of $m$.

Let $n_3$ be a third element of $\mathcal{A}$ that is not a rational power of $n_1$ or $n_2$ (we're done if this element doesn't exist). Then since $1$ and $c$ are linearly independent over $\mathbb{Q}$, and since $$ \begin{gathered} \exp\left(\log n_1\right),\ \exp\left(\log n_2\right),\ \exp\left(\log n_3\right),\\ \exp\left(c\log n_1\right),\ \exp\left(c\log n_2\right),\ \exp\left(c\log n_3\right) \end{gathered} $$ are all rational, the six exponentials theorem implies that the three real numbers $\log(n_1),\log(n_2),\log(n_3)$ are linearly dependent over $\mathbb{Q}$. That is, there exist rational numbers $A,B,C$ not all zero such that $$ A\log(n_1) + B\log(n_2) + C\log(n_3) = 0. $$ Actually, none of $A,B,C$ are zero, since this would imply that two of $n_1,n_2,n_3$ are rational powers of one another. Dividing by $C$ and renaming the coefficients, and exponentiating, we have $$ \tag{2}\label{2} n_3 = n_1^A n_2^B $$ for some nonzero rational $A,B$. Since $n_3 > 1$, we must have $(n_3,n_1n_2) > 1$. Let $p\mid (n_3,n_1n_2)$ and let $q$ be given by \eqref{1}. Then $$ \begin{aligned} \nu_p(n_3) &= A\nu_p(n_1) + B\nu_p(n_2), \\ \nu_q(n_3) &= A\nu_q(n_1) + B\nu_q(n_2). \end{aligned} $$ There are $O((\log N)^2)$ possible values for the pair $(\nu_p(n_3),\nu_q(n_3))$ with $\nu_p(n_3)\neq (0,0)$. For each pair, \eqref{1} implies that this system has a unique solution in $A$ and $B$. Since any element of $\mathcal{A}$ corresponds to a choice of $A$ and $B$ in \eqref{2}, and since any solution to \eqref{2} corresponds to at most one solution to the system above, the set $\mathcal{A}$ contains at most $O((\log N)^2)$ elements.