Question 125, the increasing positive sequence $\left\{a_{n}\right\}: a_{1}=a_{2}=3, a_{n+2}{ }^{2}+a_{n+1}{ }^{2}+a_{n}{ }^{2}-a_{n+2} a_{n+1} a_{n}=4 \quad(n \in$ $\mathbb{Z}^{+}$), find the general term formula for $\mathrm{a}_{\mathrm{n}}$.
Question 125, Solution: According to the conditions, we have:
$$
\begin{array}{l}
a_{n+2}^{2}+a_{n+1}^{2}+a_{n}^{2}-a_{n+2} a_{n+1} a_{n}=4 \\
a_{n+3}^{2}+a_{n+2}+a_{n+1}^{2}-a_{n+3} a_{n+2} a_{n+1}=4
\end{array}
$$

Subtracting the two equations, we get:
$$
\begin{array}{l}
a_{n+3}^{2}-a_{n}^{2}=a_{n+3} a_{n+2} a_{n+1}-a_{n+2} a_{n+1} a_{n} \\
\Rightarrow\left(a_{n+3}+a_{n}\right)\left(a_{n+3}-a_{n}\right)=a_{n+2} a_{n+1}\left(a_{n+3}-a_{n}\right)
\end{array}
$$

If \( a_{n+3}-a_{n}=0 \), since the sequence \(\{a_{n}\}\) is increasing, we have \( a_{n+3}=a_{n+2}=a_{n+1}=a_{n} \). Let \( a_{n}=x \), then according to the conditions, \( 3 x^{2}-x^{3}=4 \), solving this gives \( x=2 < a_{1} \), which contradicts the monotonicity of the sequence. Therefore, \( a_{n+3}+a_{n}=a_{n+2} a_{n+1} \), which means
$$
a_{n+3}=a_{n+2} a_{n+1}-a_{n}
$$

According to the conditions, \( a_{3}^{2}+18-9 a_{3}=4 \), solving this gives \( a_{3}=7 \). Hence, we can assume \( a_{0}=2 \), which still satisfies the recursive condition (*). This is: “ \( a_{0}=2, a_{1}=a, a_{2}=a, a_{n+3}=a_{n+2} a_{n+1}-a_{n} \) ” structure.

By trial, we find \( a_{0}=2, a_{1}=a, a_{2}=a, a_{3}=a^{2}-2, a_{4}=a^{3}-3 a, \cdots \). Recalling hyperbolic functions, we assume the general term formula as \( a_{n}=t^{b_{n}}+t^{-b_{n}} \) (where \( t \) is some constant), then
$$
\begin{array}{l}
t^{b_{n+1}}+t^{-b_{n+1}}=a_{n+1}=a_{n} a_{n-1}-a_{n-2} \\
=\left(t^{b_{n}}+t^{-b_{n}}\right)\left(t^{b_{n-1}}+t^{-b_{n-1}}\right)-\left(t^{b_{n-2}}+t^{-b_{n-2}}\right) \\
=\left(t^{b_{n}+b_{n-1}}+t^{-\left(b_{n}+b_{n-1}\right)}\right)+\left(t^{b_{n}-b_{n-1}}+t^{-\left(b_{n}-b_{n-1}\right)}\right)-\left(t^{b_{n-2}}+t^{-b_{n-2}}\right) \\
\text { By comparison, we can take } b_{n+1}=b_{n}+b_{n-1} \text { . At this point, }\left\{\begin{array}{l}
a_{0}=2=t^{b_{0}}+t^{-b_{0}} \\
a_{1}=a=t^{b_{1}}+t^{-b_{1}}, \\
a_{2}=a=t^{b_{2}}+t^{-b_{2}}
\end{array} \text { so we can take } b_{0}=0,\right.
\end{array}
$$
\( b_{1}=1 \), then the sequence \(\{b_{n}\}\) is the Fibonacci sequence \(\{f_{n}\}\). At this point, \( a=t+\frac{1}{t} \), solving this gives \( t=\frac{a+\sqrt{a^{2}-4}}{2} \). Therefore, \( a_{n}=t^{f_{n}}+t^{-f_{n}} \).

Substituting into the original problem, we know \( a=t+\frac{1}{t}=3 \), solving this gives \( t=\frac{3+\sqrt{5}}{2} \). Thus, \( a_{n}=\left(\frac{3+\sqrt{5}}{2}\right)^{f_{n}}+\left(\frac{3+\sqrt{5}}{2}\right)^{-f_{n}} \), where \(\{f_{n}\}\) is the Fibonacci sequence.