Problem 383

Since $2023=7 imes 17^{2}$, then any perfect square that is a multiple of 2023 must have prime factors of both 7 and 17. Furthermore, the exponents of the prime factors of a perfect square must be all even. Therefore, any perfect square that is a multiple of 2023 must be divisible by $7^{2}$ and by $17^{2}$, and so it is at least $7^{2} imes 17^{2}$ which equals $7 imes 2023$. Therefore, the smallest perfect square that is a multiple of 2023 is $7 imes 2023$. We can check that $2023^{2}$ is larger than $7 imes 2023$ and that none of $4 imes 2023$ and $17 imes 2023$ and $7 imes 17 imes 2023$ is a perfect square.