(11) The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{2}=2, a_{3}=3, a_{4}=4, a_{5}=5$. When $n \geqslant 5$, $a_{n+1}=a_{1} a_{2} \cdots a_{n}-1$. How many positive integers $m$ exist such that $a_{1} a_{2} \cdots a_{m}=a_{1}^{2}+a_{2}^{2}+\cdots+a_{m}^{2}$?
11 Let $b_{n}=a_{1} a_{2} \cdots a_{n}-a_{1}^{2}-a_{2}^{2}-\cdots-a_{n}^{2}, n \geqslant 1$. Then
$$
\begin{aligned}
b_{n+1}= & a_{1} a_{2} \cdots a_{n} a_{n+1}-a_{1}^{2}-a_{2}^{2}-\cdots-a_{n}^{2}-a_{n+1}^{2} \\
= & a_{1} a_{2} \cdots a_{n}\left(a_{1} a_{2} \cdots a_{n}-1\right)-a_{1}^{2}-a_{2}^{2}-\cdots \\
& -a_{n}^{2}-\left(a_{1} a_{2} \cdots a_{n}-1\right)^{2} \\
= & a_{1} a_{2} \cdots a_{n}-a_{1}^{2}-a_{2}^{2}-\cdots-a_{n}^{2}-1 \\
= & b_{n}-1 . \quad(n \geqslant 5)
\end{aligned}
$$

This shows that $\left\{b_{n}\right\}$ from the 5th term onwards, forms an arithmetic sequence with the first term $b_{5}=65$ and a common difference of -1. It is easy to calculate that $b_{1}=0, b_{2} \neq 0, b_{3} \neq 0, b_{4} \neq 0$.
When $n \geqslant 5$, $b_{n+1}=b_{n}-1$, only $b_{70}=b_{5}+(70-5) \times(-1)=0$.
Therefore, the positive integer $m$ that satisfies $a_{1} a_{2} \cdots a_{m}=a_{1}^{2}+a_{2}^{2}+\cdots+a_{m}^{2}$ is $m=70$ or $m=1$.