求所有的正整数数列 $\left\{a_n\right\}$ 满足：对任意的 $n \geqslant 3$，都有\[ \frac{1}{a_1 a_3}+\frac{1}{a_2 a_4}+\cdots+\frac{1}{a_{n-2} a_n}+\frac{1}{a_1^2+a_2^2+\cdots+a_{n-1}^2}=1.\]
解析 当 $n=3$ 时，有\[\dfrac1{a_1a_3}+\dfrac{1}{a_1^2+a_2^2}=1\iff (a_1^2+a_2^2-1)(a_1a_3-1)=1,\]于是\[a_1^2+a_2^2-1=a_1a_3-1=1\implies (a_1,a_2,a_3)=(1,1,2).\] 设\[\begin{cases} S_n=\sum\limits_{k=3}^n\dfrac{1}{a_{k-2}a_k},\quad n\geqslant 3,\\ T_n=\sum\limits_{k=2}^na_{k-1}^2,\quad n\geqslant 2,\end{cases}\]则当 $n\geqslant 4$ 时，有\[S_{n-1}+\dfrac1{a_{n-2}a_n}+\dfrac{1}{T_{n-1}}=1\implies a_n=\dfrac{1}{\left(1-S_{n-1}-\dfrac1{T_{n-1}}\right)a_{n-1}},\]因此满足条件的正整数数列如果存在，则是唯一的．下面证明初值 $a_1=1$，$a_2=1$，递推关系为 $a_n=a_{n-1}+a_{n-2}$（$n\geqslant 2$）的数列 $F_n$（即斐波那契数列）符合要求．
求 $T_n$ 注意到\[a_{k-1}^2=a_{k-1}(a_k-a_{k-2})=a_{k-1}a_k-a_{k-2}a_{k-1},\]于是\[T_n=a_{n-1}a_n.\]
求 $S_n$ 注意到\[\dfrac{1}{a_{k-2}a_k}=\dfrac{a_k-a_{k-2}}{a_{k-2}a_k\cdot a_{k-1}}=\dfrac{1}{a_{k-2}a_{k-1}}-\dfrac{1}{a_{k-1}a_k},\]于是\[S_n=\dfrac1{a_1a_2}-\dfrac{1}{a_{n-1}a_{n}}=1-\dfrac{1}{a_{n-1}a_{n}}=1-\dfrac1{T_n}.\] 这样就证明了 $F_n$ 是满足题意的唯一数列．