把 a=a1/b1a=a_1/b_1a=a1/b1 与 b=a2/b2b=a_2/b_2b=a2/b2 写成既约分数。由 abc=1abc=1abc=1 得 c=b1b2/(a1a2)c=b_1b_2/(a_1a_2)c=b1b2/(a1a2)。通分后,ax+by+cza^x+b^y+c^zax+by+cz 为整数这一假设变为整除关系 a1za2zb1xb2y∣a1x+za2zb2y+a1za2y+zb1x+b1x+zb2y+za_1^za_2^zb_1^xb_2^y\mid a_1^{x+z}a_2^zb_2^y+a_1^za_2^{y+z}b_1^x+b_1^{x+z}b_2^{y+z}a1za2zb1xb2y∣a1x+za2zb2y+a1za2y+zb1x+b1x+zb2y+z。