Let $n \ge 2$ be an integer. Show that there exist $n+1$ numbers $x_1, x_2, \ldots, x_{n+1} \in \mathbb{Q} \setminus \mathbb{Z}$, so that $\{ x_1^3 \} + \{ x_2^3 \} + \cdots + \{ x_n^3 \}=\{ x_{n+1}^3 \}$, where $\{ x \}$ is the fractionary part of $x$.
Problem
Source: Romania,2014,First TST,Problem 2
Tags: induction, modular arithmetic, number theory, algebra
andreiromania
18.07.2014 18:56
Solution 1 (the official solution).
Notice that, if $y_1<y_2< \cdots <y_{n+2}$ are positive integers such that $ y_1^3+y_2^3+\cdots+y_{n+1}^3=y_{n+2}^3 \, (*)$, then the numbers
$x_k=\frac{y_k}{y_{n+2}}$ for $1 \leq k \leq n$ and $x_{n+1}=-\frac{y_{n+1}}{y_{n+2}}$ meet the required conditions. We now show by induction (of step $2$) on $n\geq 2$ that numbers satisfying $(*)$ exist for every $n \ge 2$.
The equalities $3^3+4^3+5^3=6^3$ and $3^3+15^3+21^3+36^3=39^3$ settle the cases $n=2$ and $n=3$ respectively.
For the induction step $n \mapsto n+2$, notice that if $3<y_2<\cdots <y_{n+1}<y_{n+2} $ are $n+2$ integers satisfying $(*)$, then $3<4<5<2y_2<\cdots <2y_{n+1}<2y_{n+2} $ are $n+4$ integers satisfying $(*)$ as well.
tahanguyen98
29.01.2015 07:52
My solution
We call a number $l$ is a cubic residue $\pmod{d}$ if there exists $x \in Z$ such that $x^3 \equiv l \pmod{d}$
Lemma: For all $k$ which is a positive integer, then $1,3,5,...,2^k-1$ are cubic residues $\pmod{2^k}$
We prove it by induction
$k=1,2$ it's obviously easy
Suppose that we have proven for $k=m$ we will prove it for $k=m+1$
Indeed, by induction, $1,3,5,...,2^k-1$ are cubic residues modulo $2^k$, so there exists $x_i \equiv i \pmod{2^k}$ for all $i=1,3,5,...,2^k-1$ and $0\le x_i\le 2^k-1$
Let us consider the number $x_i^3$ and $(x_i+2^k)^3$
If $2^{k+1}|x_i^3-i$ then $x_i^3 \equiv i \pmod{2^{k+1}}$ and $(x_i+2^k)^3 \equiv i+2^k \pmod{2^{k+1}}$
If $2^{k}||x_i^3-i$ then $x_i^3\equiv i+2^k \pmod{2^{k+1}}$ and $(x_i+2^k)^3 \equiv i \pmod{2^{k+1}}$
Thus $1,3,...,2^k-1, 1+2^k,3+2^k,...,2^k-1+2^k$ are cubic residues modulo $2^{k+1}$ <QED>
Come back to our problem and notice that $\{\dfrac{p}{q}\}=\dfrac{r}{q}$ where $r$ is the remainder of the division of $q$ from $p$ $(1)$
If $n$ is odd then we choose the number $s^3$ such that $s$ is odd, thus we can write $s^3=l_1+l_2+...+l_n$ with $l_i$ is odd for all $i=1,n$ and choose the number $2^{3k}$ with $k$ is large enough such that $2^{3k}>l_1,l_2,...,l_n,s^3$ then applying the lemma, we obtain that there exists $q_1,q_2,...,q_n$ such that $q_i^3 \equiv l_i \pmod{2^{3k}}$ then we choose $x_i=\dfrac{q_i}{2^k}$ for all $i=1,n$ and $x_{n+1}=\dfrac{s}{2^k}$ and using $(1)$ we get the desired properties...
If $n$ is even, simply we choose $s$ is even and $s^3$ is not divisible by $2^{3k}$
<QED>
I think this topic must be posted in Number theory olympiad section, not in Algebra
nikolapavlovic
25.03.2017 19:52
Pick a very large $p\equiv 2\pmod 3$.Will be looking for numbers in the form $x_i=\tfrac{a_{i}}{p}$,note that $\{\tfrac{a_i^3}{p}\}=\tfrac{(a_i)_{p^3}}{p^3}$ ,where $x_{p^3}$ is the residue class modulo of $x$ modulo $p^3$.Let $j$ be the generator of $\mathbb{Z}_{p^3,\times}$ and let $y=j^k=\left( j^{(k\cdot 3^{-1})}\right)^3$ where the equality is regarded in $\mathbb{Z}_{p^3}$ and $3^{-1}\equiv \frac{1}{3} \pmod{p^2(p-1)}$ (one surely exists as $(p^2(p-1),2)=1$).Notice that by the previous every proper residue in $\mathbb{Z}_{p^3}$ is a cubic one.Now $a_i^3\equiv i\pmod {p^3}$ and note that $LHS=\tfrac{\binom{n+2}{2}}{p^3}$ and select $a_{n+1}\equiv \binom{n+2}{2}_{p^3}$.Now set $p\gg \binom{n+2}{2}$ and hence we reach the equality.$\blacksquare$ If I am not mistaken the above should be able to generalize to the existence of $x_i$ so that : $$\sum_{i=1}^n \{x_i^m\}=\{x_{n+1}^m\}$$Where $m>1$ is an arbitrary odd number.