x.(y-1)-y=2
x.(y-1)-y+1=2+1
x.(y-1)-(y-1)=3
(x-1)(y-1)=3
=> x-1;y-1 E Ư(3)={-3;-1;1;3}
Có bảng
| x-1 | -3 | -1 | 1 | 3 |
| x | -2 | 0 | 2 | 4 |
| y-1 | -1 | -3 | 3 | 1 |
| y | 0 | -2 | 4 | 2 |
Vậy (x;y) E {(-2;0);(0;-2);(2;4);(4;2)}
Co tat ca 125 so tu nhien ma ca 3 chu so deu le nhe