Giải phương trình nghiệm nguyên x^2 + x + 1 = 2xy + y

x² + x + 1 = 2xy + y

⇔ x² + x + 1 – 2xy – y = 0

⇔ \({x^2} + \frac{1}{2}x + \frac{1}{2}x + \frac{1}{4} - y(2x + 1) + \frac{3}{4} = 0\)

⇔ \(\frac{1}{2}x\left( {2x + 1} \right) + \frac{1}{4}\left( {2x + 1} \right) - y(2x + 1) + \frac{3}{4} = 0\)

⇔ \(\left( {2x + 1} \right)\left( {\frac{1}{2}x + \frac{1}{4} - y} \right) = - \frac{3}{4}\)

⇔ \(\left( {2x + 1} \right)\left( {2x + 1 - 4y} \right) = - 3\)

Ta có bảng:

2x + 1	-3	-1	3	1
2x + 1 – 4y	1	3	-1	-3
x	-2	-1	1	0
y	-1	-1	1	1

Vậy (x;y) = (-2;-1) ; (-1;-1); (1;1); (0;1)}