希尔伯特认为是难题之答案

费尔马1

解函数不定方程A^（32n+2）+B^（32n+6）=C^（32n+10）
求出此不定方程的正整数通解式？
其中一个答案是：
A=2^(32n^2+14n+1)*u*v*[u^(32n+2)-v^(32n+2)]^[(256n^2+128n+15)k+64n^2+36n+5]*[u^(32n+2)+v^(32n+2)]^[(256n^2+128n+15)k+160n^2+78n+9]
B=2^(32n^2+10n)*[u^(32n+2)-v^(32n+2)]^[(256n^2+96n+5)k+64n^2+28n+2]*[u^(32n+2)+v^(32n+2)]^[(256n^2+96n+5)k+160n^2+58n+3]
C=2^(32n^2+6n)*[u^(32n+2)-v^(32n+2)]^[(256n^2+64n+3)k+64n^2+20n+1]*[u^(32n+2)+v^(32n+2)]^[(256n^2+64n+3)k+160n^2+38n+2]
其中，n、k为0或正整数，u、v为正整数，u>v