Question. ¿Does there exist an integer $n>1$ such that there exist positive integers $a,b$ such that $\{\sqrt[n]{a}\}=\{\sqrt[n]{b}\},a\neq b$ and $a$ and $b$ aren't perfect n-th powers? ( $\{x\}$ is the fractional part of $x$)
Here is what i've done so far. First we start with a tricky lemma:
Lemma. If $z$ is a root of the polynomial $ax^2+bx+c=0$ with $b\neq0$($a$ can be $0$) and for some positive integer $n$, $z^n\in \mathbb{Z}$ then $z\in\mathbb{Z}$.
Proof. If $a=0\implies bz=-c\implies b^nz^n=(-c)^n\implies z^n$ is a n-th power from which $z^n=r^n$ for some $r\in\mathbb{Z}$, therefore $z=r\in\mathbb{Z}$.
If $a\neq 0$ then we set $z=\frac{p\pm\sqrt{q}}{r}$ for some integers $p,q,r$ with $p,q\neq 0$. Then $zr=p\pm \sqrt{q}\implies z^nr^n=(p\pm\sqrt{q})^n=i+j\sqrt{q}$ for some integers $i,j$. If $q$ is a perfect square then $z$ is rational and since $z^n$ is integer, $z$ is also integer. If $q$ is not a perfect square then $j=0$ from which $(p\pm\sqrt{q})^n=i=(p\mp\sqrt{q})^n\implies |p\pm\sqrt{q}|=|p\mp\sqrt{q}|$ then $p=0$, which is impossible since $p=-b\neq0$, or $q=0$ but $q$ isn't a perfect square. In any case the lemma is proven.
Back to the problem. Suposse that for some integer $n>1$, there exist positive integers $a,b,,m$ such that $a$ and $b$ aren't n-th powers and $\sqrt[n]{a}-\sqrt[n]{b}=m\quad{(*)}$ Then $\begin{align*} \sqrt[n]{a}^2-\sqrt[n]{ab}&=m\sqrt[n]{a}\\ \sqrt[n]{a}^2-m\sqrt[n]{a}-\sqrt[n]{ab}&=0 \end{align*}$ So if we prove that $\sqrt[n]{ab}$ is integer, then applying the lemma we conclude that the answer to the initial question is negative.
I'll prove that the answer is negative for $n=5$. Suppose that $(*)$ is true, then setting $x:=\sqrt[5]{a}$ and $y:=\sqrt[5]{a}$: $\begin{align*} x-y&=m\\ x^5-5x^4y+10x^3y^2-10x^2y^3+5xy^4-y^5&=m^5\\ -5xy(x^3-2x^2y+2xy^2-y^3)&=m^5-x^5+y^5\\ -5xy((x-y)^3+x^2y-xy^2)&=m^5-x^5+y^5\\ -5xy(m^3+mxy)&=m^5-x^5+y^5\\ 5mx^2y^2+5m^3xy+m^5-x^5+y^5&=0 \end{align*}$ Now we apply the lemma to $xy$ and we get that $xy$ is integer, so we are done.
Applying a similar argument to $n=2,3,4$ we get what we want. But when $n>5$, we cannot use this trick anymore since we get a polynomial in $xy$ that has a degree greater than $2$.
I'd like to find an elementary answer to this question, but if you have some more advanced method don't hesitate in posting it.