Example 2 Let the perfect square $y^{2}$ be the sum of the squares of 11 consecutive integers. Then the minimum value of $|y|$ is $\qquad$ .
Solution: Let the middle number of 11 consecutive integers be $x$. Then
$$
\begin{aligned}
y^{2}= & (x-5)^{2}+(x-4)^{2}+\cdots+x^{2}+\cdots+ \\
& (x+4)^{2}+(x+5)^{2} \\
= & x^{2}+2\left(x^{2}+1^{2}\right)+2\left(x^{2}+2^{2}\right)+\cdots+2\left(x^{2}+5^{2}\right) \\
= & 11 x^{2}+2\left(1^{2}+2^{2}+3^{2}+4^{2}+5^{2}\right) \\
= & 11\left(x^{2}+10\right) .
\end{aligned}
$$

Since $11\left(x^{2}+10\right)$ is a perfect square, and 11 is a prime number, it follows that 11 divides $\left(x^{2}+10\right)$.
Thus, $x^{2}+10 \geqslant 11$, which means $x^{2} \geqslant 1$.
Therefore, $y^{2}=11\left(x^{2}+10\right) \geqslant 11^{2}$, which implies $|y| \geqslant 11$.
The equality holds when $x^{2}=1$.
Hence, the minimum value of $|y|$ is 11.
Note: This problem is adapted from the 1998 National Junior High School Mathematics Competition. The original problem asked for the minimum value of $y$, but using it as a junior high school mathematics competition problem is questionable. Although it has been shown that $|y| \geqslant 11$, i.e., $y \leqslant-11$ or $y \geqslant 11$, it is not sufficiently justified to conclude that the minimum value of $y$ is -11 (the original answer).

In fact, since $y^{2}=11\left(x^{2}+10\right)$ is a perfect square, it follows that $x^{2}+10=11 k^{2}\left(k \in \mathbf{N}_{+}\right)$, at which point, $y= \pm 11 k$. If the equation $x^{2}+10=11 k^{2}$ has only a finite number of integer solutions, let the solution that makes $k$ the largest be $\left(x_{0}, k_{0}\right)$, then the minimum value of $y$ is $-11 k_{0}$; if the equation $x^{2}+10=11 k^{2}$ has an infinite number of integer solutions, then the minimum value of $y$ does not exist. Clearly, discussing all integer solutions to the equation $x^{2}+10=11 k^{2}$ goes beyond the scope of junior high school mathematics competition knowledge.

Moreover, taking the equation $x^{2}+10=11 k^{2}$ modulo 11, we get $111\left(x^{2}+10\right)$, thus, $111\left(x^{2}-1\right)$.
Therefore, 11 divides $(x-1)$ or 11 divides $(x+1)$.
When 11 divides $(x-1)$, taking $x=23$ satisfies the condition, at which point, $k=7$, and the corresponding $y= \pm 77$.

When 11 divides $(x+1)$, taking $x=43$ satisfies the condition, at which point, $k=13$, and the corresponding $y= \pm 143$.
It is evident that $y_{\text {min }}=-11$ is clearly incorrect.