5. The integer part $[x]$ of a number $x$ is defined as the greatest integer $n$ such that $n \leqslant x$, for example, $[10]=10,[9.93]=9,\left[\frac{1}{9}\right]=0,[-1.7]=-2$. Find all solutions to the equation $\left[\frac{x+3}{2}\right]^{2}-x=1$.
# Solution.

From the equation, it follows that $x=\left[\frac{x+3}{2}\right]^{2}-1$ is an integer. Therefore, $x=n \in \mathbb{Z}$, but we need to consider the cases of even and odd $n$ separately.

First, let's move all terms to the left side of the equation.

1) If $x=2 k$.

$$
\left[\frac{2 k+3}{2}\right]^{2}-2 k-1=\left[k+1+\frac{1}{2}\right]^{2}-2 k-1=\quad=(k+1)^{2}-2 k-1=k^{2}=0
$$

The obtained equation has a solution $k=0$, which gives $x=0$.

2) If $x=2 k+1$.

$$
\begin{aligned}
& {\left[\frac{2 k+4}{2}\right]^{2}-(2 k+1)-1=(k+2)^{2}-2 k-2=k^{2}+2 k+2=} \\
& \quad=\left(k^{2}+k\right)+(k+1)+1=k(k+1)+(k+1)+1=(k+1)(k+1)+1 \geqslant 1
\end{aligned}
$$

Since the obtained expression is strictly positive, solutions of the form $x=2 k+1$ are not possible.

Answer. $x=0$.