已知正实数列 $a_1$，$a_2$，$\cdots$，$a_n$，$\cdots$ 满足：
① $a_{n+1}=a_1^2a_2^2\cdots a_n^2-3$（$n \in \mathbb N^+$）；
② $\dfrac 12(a_1+\sqrt {a_2-1})\in \mathbb N^+$． 求证：$\dfrac 12(a_1a_2\cdots a_n +\sqrt {a_{n+1}-1})\in \mathbb N^+$．
解析 对 $n$ 进行归纳，当 $n=1$ 时，命题显然成立．假设当 $n=k$（$k\in\mathbb N^{\ast}$）时命题成立，即 $\dfrac 12(a_1a_2\cdots a_k)+\sqrt{a_{k+1}-1}$ 是正整数．记\[t=\dfrac 12(a_1a_2\cdots a_k+\sqrt{a_{k+1}-1},\]整理得\[a_{k+1}=a_1^2a_2^2\cdots a_k^2-4ta_1a_2\cdots a_k+4t^2+1,\]又\[a_{k+1}=a_1^2a_2^2\cdots a_k^2-3,\]因此\[a_1a_2\cdots a_k=t+\dfrac 1t,\]于是当 $n=k+1$ 时，有\[\begin{split}\dfrac 12\left(a_1a_2\cdots a_{k-1}+\sqrt{a_{k+2}-1}\right)&=\dfrac 12\left((a_1a_2\cdots a_k)a_{k+1}+\sqrt{(a_1a_2\cdots a_{k+1})^2-4}\right)\\ &=\dfrac 12\left(\left(t+\dfrac 1t\right)\left(\left(t+\dfrac 1t\right)^2-3\right)+\sqrt{\left(t+\dfrac 1t\right)^2\left(\left(t+\dfrac 1t\right)^2-3\right)^2-4}\right)\\ &=\dfrac 12\left(t^3+\dfrac1{t^3}+\sqrt{\left(t^3+\dfrac1{t^3}\right)^2-4}\right)\\ &=t^3,\end{split}\]是正整数，因此当 $n=k+1$ 时命题成立． 综上所述，原命题得证．