Croatian Mathematical Society Competitions

Note that $$\frac{x^2+1}{x+2}+\frac{x-1}{2}=\frac{x(3x+1)}{2(x+2)}.$$ Since $\lfloor t\rfloor \le t$ holds for all real numbers $t$, and the equality is attained if and only if $t$ is an integer, it follows that both $$\frac{x^2+1}{x+2}\text{and}\frac{x-1}{2}$$ must be integers. The latter is an integer if and only if $x$ is an odd integer, hence considering $$\frac{x^2+1}{x+2}=x-2+\frac{5}{x+2}$$ we get $x+2\in \{1,-1,5,-5\}$, i.e. $x\in \{-1,-3,3,-7\}$.