75th Romanian Mathematical Olympiad

Since each fractional part $\{kx\}\in [0,1)$, the right-hand side is a sum of integers, hence nonnegative, so $x\ge 0$.

a) For $n=2$, the equation becomes $\{x\}+\{2x\}=[x]+[2x]+[3x]+[4x]$. The left side is in $[0,2)$ and the right side is an integer number, so the right side is in $\{0,1\}$. If $\{x\}+\{2x\}=0$, then $\{x\}=\{2x\}=0$, so $x\in \mathbb{Z}$. Taking in the original equation $x\in \mathbb{Z}$ yields $10x=0$, so $x=0$, which satisfies the equation.

If $\{x\}+\{2x\}=1$, since $[x]\le [2x]\le [3x]\le [4x]$, we have $[x]=[2x]=[3x]=0$, $[4x]=1$, which implies $3x<1\le 4x$, so $x\in [\frac{1}{4},\frac{1}{3})$. But from $\{x\}+\{2x\}=1$ and $[x]=[2x]=0$, we have $x+2x=1$, which implies that $x=\frac{1}{3}\notin [\frac{1}{4},\frac{1}{3})$, a contradiction. Hence, for $n=2$, the unique solution is $x=0$.

b) Using similar reasoning for general $n$, since $$\{x\}+\cdots +\{nx\}\in [0,n)\cap \mathbb{Z},$$ and the integer parts satisfy $[x]\le [2x]\le \cdots \le [2nx]$, if $[nx]\ge 1$, then the right-hand side sum is at least $n+1$, which contradicts that the left side is less than $n$. So $[x]=[2x]=\cdots =[nx]=0$, and the equation reduces to:

$$x+2x+\cdots +nx=[(n+1)x]+\cdots +[2nx]=k,$$ where $k\in \{0,1,\dots ,n-1\}$. This yields $x=\frac{2k}{n(n+1)}$.

$$n-\sqrt{\frac{n(n-1)}{2}}-\left(n+\frac{1}{2}-\sqrt{\frac{2n^2+2n+1}{4}}\right)<1,$$ there is at most one integer $k$ satisfying this, so there is at most one additional solution besides $x=0$. Therefore, the given equation has at most two real solutions for any $n\ge 2$.