是否存在4个全不同的正整数，其中两个的平方倒数和等于另两个的？
如 \ \[\frac{1}{10^2} + \frac{1}{55^2} = \frac{1}{11^2} + \frac{1}{22^2}\]\[\frac{1}{525749^4} + \frac{1}{1407938^4} = \frac{1}{619913^4} + \frac{1}{624574^4}\] 这个是有现成资料的：
https://oeis.org/A355812 \[\frac{1}{a_n^2}\ +\frac{1}{(a_n+1)^2}=\frac{1}{b_n^2}\ +\frac{1}{\bigl(a_n*(a_n+1)\bigr)^2}\]
a(n)=1, 8, 49, 288, 1681, 9800, 57121, 332928, 1940449,11309768, 65918161, 384199200, 2239277041,13051463048, 76069501249, 443365544448, 2584123765441, 15061377048200,
b(n)=1, 6, 35, 204, 1189, 6930, 40391, 235416, 1372105, 7997214, 46611179, 271669860, 1583407981, 9228778026, 53789260175, 313506783024, 1827251437969, 10650001844790, 是否存在4个全不同的正整数，其中两个的平方倒数和等于另两个的？
当4个全不同的正整数中，出现 1, 2, 3, 4, 23, 43, 47, 53, 69, 73, 83, 89, 107, 109, 131, 139, 149, 151, 157, 163, 167, 173, 179, 191, 211, 214, ...则无解。
各位网友！我们能把这串数找出来吗？ \(\D\frac{1}{a_{n}^2\ }+\frac{1}{\big(a_{n}+1\big)^2\ \ \ }=\frac{1}{b_{n}^2\ }+\frac{1}{\big(a_{n}*(a_{n}+1)\big)^2\ \ \ \ \ }\)
a(n)=1, 8, 49, 288, 1681, 9800, 57121, 332928, 1940449,11309768, 65918161, 384199200, 2239277041,13051463048, 76069501249, 443365544448, 2584123765441, 15061377048200,
b(n)=1, 6, 35, 204, 1189, 6930, 40391, 235416, 1372105,7997214, 46611179, 271669860, 1583407981,9228778026, 53789260175, 313506783024, 1827251437969, 10650001844790,
\(\D a_{n}=\frac{\cosh\big(2n*arccsch(1)\big)-1\ \ \ \ \ \ \ \ \ \ \ }{2}\)
\(\D b_{n}=\frac{\sinh\big(2n*arccsch(1)\big)\ \ \ \ \ \ \ \ \ }{\sqrt{\ 8\ }}\)
\(\D \frac{a_{n}}{b_{n}}=\sqrt{\ 2\ }\)
谢谢 xiaoshuchong！
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