IRL ABooklet

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 $x = 2022 + \frac{4043}{4044}$.