求 A^2+(A+23)^2=C^2 的本原勾股数通项公式
设 Pn= -2,1, 0, 1, 2, 5, 12, 29, 70, 169, ...
设 Qn= -1,1, 1, 3, 7, 17, 41, 99, 239, 577, ...
设 Un=P(n+5)+Q(n+1)=6,13,32,77,186, 449, 1084, ...
设 Vn=P(n+4)+Q(n+0)=1, 6, 13,32, 77, 186, 449, ...
设 Mn=P(n+2)+Q(n+4)=7,18,43,104, 251, 606, 1463, ...
设 Nn=P(n+1)+Q(n+3)=4, 7, 18, 43, 104, 251, 606, ...
得到 A^2+(A+23)^2=C^2 的本原勾股数通项公式
其一:(U^2 - V^2)^2+(2UV)^2 = (U^2+V^2)^2
其二:(M^2 - N^2)^2+(2MN)^2= (M^2+N^2)^2
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