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 , x , y ∈ Z {\displaystyle 4p(3a^{4}+p^{2})(3b^{12}+4p^{2}(3a^{4}+p^{2})^{2})={x}^{3}+{y}^{3},x,y\in Z}
{ 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}}}
if：p,q是素数,(p,q)=1,则： q p − 1 + p q − 1 ≡ 1 ( mod p q ) {\displaystyle q^{p-1}+p^{q-1}\equiv 1{\pmod {pq}}}
http://mathworld.wolfram.com/Fermats4nPlus1Theorem.html
http://mathworld.wolfram.com/GenusTheorem.html
<<School:李煌数学研究院
