## Problem 2

Let $a \in \mathrm{R}$. Solve the equation:

$[\mathbf{x}]^{2}+\left[\mathbf{x}+\frac{1}{2}\right]=\mathbf{a}+[\mathbf{2 x}]$, where $[\mathbf{x}]$ is the integer part of the real number $\mathbf{x}$.

GM nr.11/2013
## Solution:

The Hermite equality for $n=2$ is known:
$[\mathrm{x}]+\left[\mathrm{x}+\frac{1}{2}\right]=[2 \mathrm{x}]$.

The equation in the statement becomes:

$[\mathrm{x}]^{2}+[2 \mathrm{x}]-[\mathrm{x}]=\mathrm{a}+[2 \mathrm{x}][\mathrm{x}]^{2}-[\mathrm{x}]=\mathrm{a}[\mathrm{x}]^{2}-[\mathrm{x}]-\mathrm{a}=0$.

The last equation has real solutions if $1+4 \mathrm{a} \geq 0, \mathrm{a} \in \mathrm{Z}\mathrm{a} \in \mathrm{N}$.

On the other hand: $[\mathrm{x}]^{2}-[\mathrm{x}]=\mathrm{a}[\mathrm{x}]([\mathrm{x}]-1)=\mathrm{a}, \quad \mathrm{a} \in \mathrm{N}$.

Therefore, if $\mathrm{a}=(\mathrm{k}-1) \mathrm{k}, \quad \mathrm{k} \in \mathbf{N}$, then the equation reduces to:

$[\mathrm{x}]=\mathrm{k}, \mathrm{k} \in \mathbf{N}$ and the solution will be $\mathrm{x} \in[\mathrm{k}, \mathrm{k}+1)$.

For other real values of $\mathrm{a}$, the equation has no real solutions.