平方差公式法解二项和函数不定方程

费尔马1

A^(2n+1)+B^(2n+2)=C^(2n+3)
其中一个答案是：
A=2^(4n^2+8n+2)*uv*[(u^(2n+1)-v^(2n+1)]^(4n+6)*[(u^(2n+1)+v^(2n+1)]^(4n^2+8n+4)

B=2^(4n^2+6n)*[(u^(2n+1)-v^(2n+1)]^(4n+4)*[(u^(2n+1)+v^(2n+1)]^(4n^2+6n+2)

C=2^(4n^2+4n)*[(u^(2n+1)-v^(2n+1)]^(4n+2)*[(u^(2n+1)+v^(2n+1)]^(4n^2+4n+2)

其中，u、v为正整数，且u＞v

费尔马1

A^(8n+5)+B^(8n+1)=C^(8n+3)
其中一个答案是：
A=2^(16n^2+6n+1)*uv*[(u^(8n+5)-v^(8n+5)]^(16n^2+6n)*[(u^(8n+5)+v^(8n+5)]^(32n^2+20n+2)

B=2^(16n^2+14n+3)*[(u^(8n+5)-v^(8n+5)]^(16n^2+14n+2)*[(u^(8n+5)+v^(8n+5)]^(32n^2+36n+10)

C=2^(16n^2+10n+1)*[(u^(8n+5)-v^(8n+5)]^(16n^2+10n)*[(u^(8n+5)+v^(8n+5)]^(32n^2+28n+4)

其中，u、v为正整数，且u＞v

费尔马1

请老师们检验，谢谢！