Let the $10$ consecutive numbers be $n$, $(n+1)$, $\cdots$, $(n+9)$. Then, its sum equals $$S=n^2 + (n+1)^2 +\cdots + (n+9)^2 = 10n^2 + 90 n + 285=5\times(2n^2 + 18n + 57)$$

Therefore, it is sufficient to show that $(2n^2 + 18n+57)$ is not a multiple of $5$. If so, $S$ is a multiple of $5$ but not $5^2$, hence cannot be a square.

• If $n\equiv 0\pmod{5}$, then $2n^2+18n+57\equiv 2\pmod{5}$.
• If $n\equiv 1\pmod{5}$, then $2n^2+18n+57\equiv 2\pmod{5}$.
• If $n\equiv 2\pmod{5}$, then $2n^2+18n+57\equiv 1\pmod{5}$.
• If $n\equiv -1\pmod{5}$, then $2n^2+18n+57\equiv 1\pmod{5}$.
• If $n\equiv -2\pmod{5}$, then $2n^2+18n+57\equiv 4\pmod{5}$.

Therefore, we conclude the claim holds.