Problem 1. a) Solve in the set of real numbers the equation: $[x+1]+\left[x+\frac{1}{2}\right]=\frac{2 x+3}{2}$.

a) Equation $\Leftrightarrow[2 x]=\frac{2 x+1}{2}$

$\left\{\begin{array}{l}\frac{2 x+1}{2} \leq 2 x<\frac{2 x+3}{2} \\ \frac{2 x+1}{2}=k \in Z\end{array} \Rightarrow\left\{\begin{array}{l}x \geq \frac{1}{2} \\ x<\frac{3}{2} \\ x=\frac{2 k-1}{2}\end{array} \Rightarrow\left\{\begin{array}{l}\frac{1}{2} \leq x<\frac{3}{2} \\ x=\frac{2 k-1}{2}\end{array} \Rightarrow\left\{\begin{array}{l}\frac{1}{2} \leq \frac{2 k-1}{2}<\frac{3}{2} \\ k \in Z\end{array} \Rightarrow k=1 \Rightarrow x=\frac{1}{2}\right.\right.\right.\right.$

b) Let $n \in \mathbb{N}^{*}$ and $x_{1}, x_{2}, x_{3}, \ldots \ldots . . x_{n} \in\{-2013 ; 2013\}$, determine the set of natural numbers $n$ for which the equality $x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+\ldots \ldots .+x_{n-1} x_{n}+x_{n} x_{1}=0$ is true.
Solution: necessary condition $n=2 p, p \in N$.

$p$ terms from the sum are equal to -2012 and $p$ terms are equal to 2012 implies

$x_{1} x_{2} \cdot x_{2} x_{3} \cdot x_{3} x_{4} \cdot \ldots \ldots . . \cdot x_{n-1} x_{n} \cdot x_{n} x_{1}=(-1)^{p} \cdot 2012^{2 n}$

$\Rightarrow x_{1}^{2} \cdot x_{2}^{2} \cdot x_{3}^{2} \cdot \ldots . . \cdot x_{n}^{2}=(-1)^{p} \cdot 2012^{2 n}>0$. Therefore $p=2 k, k \in N \Rightarrow n=4 k \in N$. It follows that $n \in M_{4}$