On the function $f_m(p)=\left|\left\{1\leqslant k<\frac p2:\ \left\{\frac{k^m}p\right\}>\frac12\right\}\right|$ Let $m>1$ be an integer and let $p$ be an odd prime. Can we say something nontrivial about 
$$f_m(p):=\left|\left\{1\leqslant k<\frac p2:\ \left\{\frac{k^m}p\right\}>\frac12\right\}\right|$$
(where $\{x\}$ denotes the fractional part of a real number $x$)?
QUESTION 1. Is it true that $f_m(p)\sim p/4$ (as $p\to\infty$) for each $m=2,3,4,\ldots$?
QUESTION 2. Is $f_5(p)$ always even for every prime $p\not\equiv1\pmod5$?
QUESTION 3. Is it true that $f_3(p)\in\{(p+1)/6+2n:\ n=0,1,2,\ldots\}$ for each prime $p\equiv 5\pmod 6$?
QUESTION 4. Is it true that $f_6(p)\in\{(p+7)/12+2n:\ n=0,1,2,\ldots\}$ for each prime $p\equiv 5\pmod {12}$?
My computation suggests that all the four questions should have positive answers.
 A: At least QUESTION 2,3.4 have positive answers. Let $A=\underset{0<k<\frac{p}{2}}\Pi k$, then $A^2\equiv(-1)^{\frac{p-1}{2}}\times(p-1)!\equiv(-1)^{\frac{p+1}{2}}$.
In QUESTION 2 and 3 , since gcd(m,p-1)=1 and m is odd, $\{k^m, 0<k<\frac{p}{2}\}\cup\{-k^m, 0<k<\frac{p}{2}\}$ forms a complete residue system modulo p. For each $0<k<\frac{p}{2}$, if $\{\frac{k^m}{p}\}>\frac{1}{2}$, then $0<\{\frac{-k^m}{p}\}<\frac{1}{2}$, there are exact $f_m(p)$ such k. When $k$ runs through $1,2,...,\frac{p-1}{2}$, the corresponding $k^m$ or $-k^m$ (mod p) also runs through $1,2,...,\frac{p-1}{2}$ (mod p). Taking the product, we have  $\underset{0<k<\frac{p}{2}}\Pi k^{m}\equiv(-1)^{f_m(p)}\times \underset{0<k<\frac{p}{2}}\Pi k$, so $(-1)^{f_m(p)}\equiv(A^2)^{\frac{m-1}{2}}\equiv(-1)^{\frac{p+1}{2}\times\frac{m-1}{2}}$, hence QUESTION 2 and 3 are positive.
In QUESTION 4, the problem is easier. Since gcd(m,p-1)=2 and $p\equiv 1$ (mod 4), $\{k^m, 0<k<\frac{p}{2}\}$ runs over all the quadratic residue classes modulo p (compare the number of class on both sides and ......), but when $p\equiv 1$ (mod 4), x (mod p) is a quadratic residue classes modulo p if and only if -x is, so exact half of them $\equiv$ some integer  k, where $0<k<\frac{p}{2}$. So $f_m(p)=\frac{p-1}{4}$ and then it is not hard to see that QUESTION 4 is positive.
