Question

已知a是整数,x,y是方程x的平方—xy—ax+ay+1的整数解,求x—y

Answer 1

x^2 - xy -ax +ay +1

= x(x-y)-a(x-y) +1

=(x-a)(x-y)+1=0

(x-a)(x-y) =-1

既然是整数方程,那么有

x-y=正负1