1-116 The sum of the squares of two consecutive integers can equal the square of another integer (such as $3^{2}+4^{2}=5^{2}$).
(1) Prove: When $m=3,4,5,6$, the sum of the squares of $m$ consecutive integers cannot be the square of another integer.
(2) Find an example where the sum of the squares of 11 consecutive integers is the square of another integer.
[Solution] (1) If $x$ is an integer, then $x^{2} \equiv 0$ or $1(\bmod 3), x^{2} \equiv 0$ or $1(\bmod 4)$.
But
$$
\begin{aligned}
& (x-1)^{2}+x^{2}+(x+1)^{2}=3 x^{2}+2 \equiv 2(\bmod 3), \\
& (x-1)^{2}+x^{2}+(x+1)^{2}+(x+2)^{2} \\
= & 4 x^{2}+4 x+6 \\
= & 2(\bmod 4),
\end{aligned}
$$

Therefore, the sum of the squares of three or four consecutive integers cannot be the square of another integer.
Similarly, $(x-2)^{2}+(x-1)^{2}+x^{2}+(x+1)^{2}+(x+2)^{2}$
$$
\begin{array}{l}
=5 x^{2}+10 \\
=x^{2}+2 \\
\equiv 2 \text { or } 3(\bmod 4) .
\end{array}
$$

Therefore, the sum of the squares of five consecutive integers cannot be the square of another integer. Similarly, the case for $m=6$ can be proven.
(2) The sum of the squares of 11 consecutive integers can be expressed as
$$
\begin{aligned}
& (x-5)^{2}+(x-4)^{2}+\cdots+x^{2}+\cdots+(x+4)^{2}+(x+5)^{2} \\
= & 11\left(x^{2}+10\right) .
\end{aligned}
$$

Clearly, when $x= \pm 1$, this is the square of an integer.