已知正整数 $p,q$ 满足 $|2p^2-q^2|=1$，且数列 $\left\{a_{n}\right\}$ 中，$a_{1}=1$，$a_{n+1}=\dfrac{1}{2} a_{n}+\dfrac{p^2}{q^2a_{n}}$，求证：当 $n\geqslant 2$ 时，$\dfrac{2p}{\sqrt{q^2a_n^2-2p^2}}$ 为正整数．
解析 记 $t=\dfrac pq$，当 $n\geqslant 2$ 时，设 $b_n=\dfrac{2t}{\sqrt{a_n^2-2t^2}}$，则\[a_n^2=\dfrac{4t^2}{b_n^2}+2t^2,\]而\[b_{n+1}=\dfrac{2t}{\sqrt{a_{n+1}^2-2t^2}}=\dfrac{2t}{\sqrt{\left(\dfrac 12a_n+\dfrac{t^2}{a_n}\right)^2-2t^2}}=\dfrac{2t}{\sqrt{\left(\dfrac{t^2}{b_n^2}+\dfrac{t^2}2\right)+\dfrac{t^2}{\dfrac{4}{b_n^2}+2}-t^2}}=b_n\sqrt{2b_n^2+4},\]因此\[b_{n+2}=b_{n+1}\sqrt{2b_{n+1}^2+4}=b_{n+1}\sqrt{2b_n^2(2b_n^2+4)+4}=2b_{n+1}(b_n^2+1).\]而\[a_2=\dfrac 12+t^2,\quad b_2=\dfrac{4t}{|2t^2-1|},\quad b_3=b_2\cdot \dfrac{4(2t^2+1)}{|2t^2-1|},\]注意到\[\dfrac{1}{|2t^2-1|}=\dfrac{q^2}{|2p^2-q^2|}=q^2\in\mathbb N^{\ast},\]于是 $b_2,b_3$ 均为正整数，进而可得当 $n\geqslant 2$ 时，数列 $\{b_n\}$ 中的各项均为正整数，命题得证．