亘古及今第一题

费尔马1

解函数不定方程A^（16n+2）+B^（16n+6）=C^（16n+10）
求出此不定方程的正整数通解式？
其中一个答案是：
A=2^(8n^2+7n+1)*u*v*[u^(16n+2)-v^(16n+2)]^[(64n^2+64n+15)k+16n^2+18n+5]*[u^(16n+2)+v^(16n+2)]^[(64n^2+64n+15)k+40n^2+39n+9]
B=2^(8n^2+5n)*[u^(16n+2)-v^(16n+2)]^[(64n^2+48n+5)k+16n^2+14n+2]*[u^(16n+2)+v^(16n+2)]^[(64n^2+48n+5)k+40n^2+29n+3]
C=2^(8n^2+3n)*[u^(16n+2)-v^(16n+2)]^[(64n^2+32n+3)k+16n^2+10n+1]*[u^(16n+2)+v^(16n+2)]^[(64n^2+32n+3)k+40n^2+19n+2]
其中，n、k为0或正整数，u、v为正整数，u>v。