IRL_ABooklet_2023

By definition, $n=x-\{x\}$ is an integer. Then $x=n+\{x\}$ and $x^2=n^2+2n\{x\}+\{x{\}}^2$. Hence $\{x^2\}=\{2n\{x\}+\{x{\}}^2\}$. If $\{x{\}}^2=\{x^2\}$ then $2n\{x\}$ is an integer.

As $x<2023$, the largest possible $n$ is $n=2022$. We then look for the largest $\{x\}$ for which $4044\{x\}$ is an integer. As $\{x\}<1$, $4044\{x\}<4044$. The largest possible value for $\{x\}$ therefore is $\{x\}=\frac{4043}{4044}$. Hence the largest solution is $x=2022+\frac{4043}{4044}$.