Find the period of $[x]^2+\sqrt{\{x\}^2}$, where $[.]$ represents greatest integer function of $x$ and $\{x\}$ represents fractional part of $x$.
$\sqrt{\{x\}^2}$ can be written as $\{x\}$ because it's always positive, and its period is $1$.
Let the required period be $T$. So, $$[x+T]^2+\sqrt{\{x+T\}^2}=[x]^2+\sqrt{\{x\}^2}$$ $$\implies([x]+T)^2+\sqrt{\{x\}^2}=[x]^2+\sqrt{\{x\}^2}$$ $$\implies[x]^2+T^2+2T[x]+\{x\}=[x]^2+\{x\}$$ $$\implies T^2+2T[x]=0$$ $$\implies T+2[x]=0$$ $$\implies T=-2[x]$$
Since $T\gt0\implies[x]\lt0$. For minimum $T$, $[x]$ should be maximum i.e. $-1$. So, $T=2$
But I am not able to confirm it. $[x+2]^2+\{x+2\}=[x]^2+4+4[x]+\{x\}$. Not getting $[x]^2+\{x\}$
It's clear that $\lim_{x\to\infty} \lfloor x\rfloor^2+\sqrt{\{x\}^2}=\infty$. Therefore it cannot be periodic.