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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.

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Nice question!! – TMM Mar 3 '12 at 21:17
Is there something wrong in the second line? You have $\{\sqrt[n]{a}\}$ twice; and of course, the fractional part of a real number is an integer if and only if the original number is an integer; so you are just saying "and $a$ and $b$ are not perfect $n$th powers", right? – Arturo Magidin Mar 3 '12 at 22:22
@Diego S: There cannot be such $a$ and $b$ for prime $n$. It seems plausible that this extends to $n$ that are not prime, but it does not seem to be as simple as for primes. – André Nicolas Mar 3 '12 at 22:31
@AndréNicolas: Proof please, i also tried that – Zero Mar 4 '12 at 0:06
If $p$ is prime and $a$ is not a perfect $p$-th power, then $x^p-a$ is irreducible over the rationals, so $x^p-a$ is the (unique for monic) minimal polynomial of $\sqrt[p]{a}$. But the hypothesis $\sqrt[p]{b}=m-\sqrt[p]{a}$ gives us another monic polynomial of degree $p$ satisfied by $\sqrt[p]{a}$, not necessarily minimal but that doesn't matter. For more general $n$ there could be a problem, but only when $a$ and $b$ are perfect $k$-th powers where $k$ is a divisor of $n$. I don't think so, but it is Saturday night and there will be no more mathematics for a while. – André Nicolas Mar 4 '12 at 1:24
up vote 6 down vote accepted

More is true: the exponents need not be the same. Let $r$ be rational, and let $a$ and $b$ be positive rationals. Let $m$ and $n$ be positive integers.

Suppose also that at least one of $a^{1/m}$ and $b^{1/n}$ is irrational. We will show that if $r+a^{1/m}=b^{1/n}$, then $r=0$ and therefore $a^{1/m}=b^{1/n}$.

Proof: Without loss of generality we may assume that there is no rational $a'$, and positive integer $m'<m$, such that $(a')^{1/m'}=a^{1/m}$, and that a similar condition holds for $b^{1/n}$. (If there is such a rational $a'$ and integer $m'$, just replace $a$ by $a'$ and $m$ by $m'$. Keep on doing this sort of thing until you can't do it any more.)

The above condition implies that the polynomials $x^m-a$ and $x^n-b$ are irreducible over the rationals, and therefore are respectively minimal polynomials of $a^{1/m}$ and $b^{1/n}$ respectively. (Here we are using some field theory. The remark at the end gives an elementary argument.)

Now suppose first that $m<n$. Then $(a^{1/m})^m=(b^{1/n}-r)^m$, and therefore $b^{1/n}$ is a root of the polynomial $(x-r)^m-a$, which has degree less than $n$, contradicting the fact that the minimal polynomial of $b^{1/n}$ has degree $n$.

Suppose next that $m=n$. Then $b^{1/n}$ is, like above, a root of the polynomial $(x-r)^n -a$, and therefore of the polynomial $[(x-r)^n-a]-[x^n-b]$. If $r\ne 0$, this has degree $n-1$, contradicting the fact that the minimal polynomial of $b^{1/n}$ has degree $n$.

Remark: Suppose that $c$ is rational, and $c^{1/n}$ is irrational, and there does not exist a rational $c'$, and a positive integer $n'<n$, such that $(c')^{1/n'}=c^{1/n}$. We show, without using explicitly machinery from the theory of fields, that there is no non-zero polynomial $P(x)$ with rational coefficients and degree $<n$ such that $P(c^{1/n})=0$.

Suppose to the contrary that there is such a polynomial $P(x)$. Note that $c^{1/n}$ is a root of the polynomial $x^n-c$. Let $D(x)$ be the monic greatest common divisor of $P(x)$ and $x^n-c$, obtained as usual with the Euclidean Algorithm.

Then $D(x)$ has rational coefficients. Let $D(x)$ have degree $n'$. Then $n'<n$. Every root of $D(x)$ is a root of $x^n-c$. The roots of $x^n-c$ all have norm $c^{1/n}$, and therefore so do the roots of $D(x)$. The constant term of $D(x)$ is $\pm$ the product of the roots of $D(x)$, and therefore has norm $c^{n'/n}$. Since $D(x)$ has rational coefficients, we conclude that $c^{n'/n}$ is rational, say $c^{n'/n}=c'$. It follows that $c^{1/n}=(c')^{1/n'}$.

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Great answer, i didn't notice all the details in the remark when working with polynomials. – Zero Mar 5 '12 at 18:11
@DiegoS.: When I first wrote out the answer, I assumed (without even mentioning!) some stuff from field theory which takes a while to build up. So I decided to add at the end enough to make the solution as "elementary" as I now how. – André Nicolas Mar 5 '12 at 20:12
Isn't it, if m = n : $[(x-r)^n-a]-[x^n-b]$. So $r= 0$ and directly $a=b$. – ama Jul 22 '12 at 20:50
@ama: Thank you for the typo-finding. I try, but they keep slipping by. – André Nicolas Jul 22 '12 at 20:57

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