**Problem:** Let $c>0$ be a real number, and suppose that for every positive integer $n$, at least one percent of the numbers $1^c,2^c,3^c,\ldots,n^c$ 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)? Thanks.



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


