如果 p = a 2 + b 2 , p ∈ p r i m e {\displaystyle p=a^{2}+b^{2},p\in prime}
则： 4 p ( 3 a 4 + p 2 ) ( 3 b 12 + 4 p 2 ( 3 a 4 + p 2 ) 2 ) = x 3 + y 3 {\displaystyle 4p(3a^{4}+p^{2})(3b^{12}+4p^{2}(3a^{4}+p^{2})^{2})=x^{3}+y^{3}}
{ x = ( a 2 + p ) 3 y = b 6 + 2 p ( 3 a 4 + p 2 ) {\displaystyle {\begin{cases}x=(a^{2}+p)^{3}\\y=b^{6}+2p(3a^{4}+p^{2})\end{cases}}}
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<<School:李煌数学研究院
