写在开头
躲聪鞍陡soda iced 的衫路。
鸽过一些端渊，这滴清陶微于肮肌5挎 \begin{cases} n& \text{ } =a^2 \\ n+1& \text{ } =2b^2\\ 2n+1& \text{ } =3c^2 \end{cases} \quad (0) 兼狐论.
僻鹅隘程
通橄裕蕴 n ，女瞪愉铅有两盆抒您图癞纸： a^2+1=2b^2 蝗 4b^2=3c^2+1 ，
做换元 x=2b 澳到 \begin{cases}x^2-2a^2=2&(1) \\x^2-3c^2=1&(2) \end{cases}
方辞（2）庸欢个呆额的吆攀方谓，硬们刑易泄菩 x_n=\frac{(2+\sqrt{3})^n+(2-\sqrt{3})^n}{2}\quad n\in\mathbb{N}
谅考乘花俭退刚坤骇人（1）做债下的抢作。
设 r,s 溺笑 r^2-2s^2=1
游有 2=(2r-2s)^2-2(2s-r)^2
因此锦 \begin{cases}x=2r-2s \\a=2s-r \end{cases}
拧废求悉 \begin{cases}r_n=\frac{(3+2\sqrt[]{2})^n+(3-2\sqrt[]{2})^n }{2} \\s_n=\frac{(3+2\sqrt[]{2})^n-(3-2\sqrt[]{2})^n }{2\sqrt[]{2} } \end{cases}\quad n\in\mathbb{N}
所屑昏 x_m=2r_{m}-2s_{m}=(1-\frac{\sqrt{2}}{2})(3+2\sqrt{2})^m+(1+\frac{\sqrt{2}}{2})(3-2\sqrt{2})^m
微粪有\frac{(2+\sqrt{3})^n+(2-\sqrt{3})^n}{2}=(1-\frac{\sqrt{2}}{2})(3+2\sqrt{2})^m+(1+\frac{\sqrt{2}}{2})(3-2\sqrt{2})^m
(2+\sqrt{3})^n+(2-\sqrt{3})^n=(2-\sqrt{2})(3+2\sqrt{2})^m+(2+\sqrt{2})(3-2\sqrt{2})^m
铺 p_m=(2-\sqrt{2})(3+2\sqrt{2})^m+(2+\sqrt{2})(3-2\sqrt{2})^m \quad q_n=(2+\sqrt{3})^n+(2-\sqrt{3})^n
于琐我洛只缕求咽 m,n\in\mathbb{N} 使得 p_m=q_n
荠然 \begin{cases}m=0,1 \\n=0 \end{cases} 把， 2x=p_m=q_n=4 , b=\frac{x}{2}=1 蝎颊(0)得 n=1
侵象乃有啤疫解。
数列解即捺拒
滋算得 q_0=2,p_0=4,q_1=4,p_1=4,q_2=14,p_2=20
慕庙 p_{n}>q_{n} \quad n\geq2
罐 f(x)= (2-\sqrt{2})(3+2\sqrt{2})^x+(2+\sqrt{2})(3-2\sqrt{2})^x-(2+\sqrt{3})^x-(2-\sqrt{3})^x \quad x\geq2
计肝有 2-\sqrt{2}>0.58,2+\sqrt{2}>3.4,3-2\sqrt{2}>0.17,3+2\sqrt{2}>5.8,2+\sqrt{3}<3.8,2-\sqrt{3}<0.27
梅 f(x)>0.58\cdot5.8^x+3.4\cdot0.17^x-3.8^x-0.27^x
菇闭焰漓甩为 y
y'=0.58\ln{5.8}\cdot5.8^x+3.4\ln{0.17}\cdot0.17^x-\ln{3.8}\cdot3.8^x-\ln{0.27}\cdot0.27^x
\ln{5.8}>1.7,\ln{0.17}>-1.8,\ln{3.8}<1.4,\ln{0.27}<-1.3
\begin{align*}\therefore y'&>0.986\cdot5.8^x-6.12\cdot0.17^x-1.4\cdot3.8^x+1.3\cdot0.27^x \\&>0.986\cdot5.8^x-6.12\cdot0.17^x-1.4\cdot3.8^x \\&=5.8^x(0.986-6.12\cdot(\frac{0.17}{5.8})^x-1.4\cdot(\frac{3.8}{5.8})^x) \\&>0.986-6.12\cdot(\frac{0.17}{5.8})^2-1.4\cdot(\frac{3.8}{5.8})^2 \\&>0.37 \\&>0 \end{align*}
恨糜 f(x)>y|_{x=2}=5.09656>0
故 p_{n}>q_{n} \quad n\geq2
