74th Romanian Mathematical Olympiad

a) $(\lfloor x\rfloor +\{x\}{)}^2+4\lfloor x\rfloor =\{x{\}}^2$ yields $\lfloor x{\rfloor }^2+2\lfloor x\rfloor \{x\}+4\lfloor x\rfloor =0$, (). If $x\ge 0$, then $\lfloor x\rfloor =0$, thus any $x\in [0,1)$ is a solution. If $x<0$, equality () leads to $\{x\}=-\frac{\lfloor x{\rfloor }^2+4\lfloor x\rfloor }{2\lfloor x\rfloor }=-\frac{\lfloor x\rfloor +4}{2}$. Since $\{x\}\in [0,1)$, we conclude that $\lfloor x\rfloor \in \{-5,-4\}$, thus obtaining the solutions $x_1=-\frac{9}{2}$, $x_2=-4$, which verify the equality. Therefore, $x\in \{-\frac{9}{2},-4\}\cup [0,1)$.

b) For $y\in (0,1)$ we have $A(y)=y^2\in (0,1)$, $A(1)=5$, and for $y\in [1,2)$ we have $A(y)=y^2+4\in (5,8)$, thus $A(y)$ is not a perfect square for $y\in (0,2)$. For $y\ge 2$, the condition from the statement leads to $y^2\in \mathbb{N}$, therefore $y=\sqrt{m}$, with $m\in \mathbb{N}$, $m\ge 4$, leading to $m+4\lfloor \sqrt{m}\rfloor =p^2$, with $p\in \mathbb{N}$. If $k=\lfloor \sqrt{m}\rfloor$, then $k\ge 2$ and $k^2\le m<(k+1{)}^2$, hence $m+4k=p^2$. Using the fact that $k^2+4k\le p^2<(k+1{)}^2+4k=k^2+6k+1<(k+3{)}^2$, it follows that (1) $p^2=(k+1{)}^2=1$, for $k=0$ or (2) $p^2=9$ for $k=1$, values that do not verify the equality in the statement, or (3) $p^2=(k+2{)}^2$, for $k\ge 2$.

Therefore, $p=k+2$ and $m=(k+2{)}^2-4k=k^2+4$. Thus, the numbers that verify the equality in the statement are $y=\sqrt{k^2+4}$, $k\in \mathbb{N}$, $k\ge 2$.