Number of real solution of the equation $x^2+y^2+2xy-2014 x-2014y-2015=0$

Question

Number of real solution of the equation $x^2+y^2+2xy-2014 x-2014y-2015=0$

solution i try: $2x^2+2y^2+4xy-4028 x-4028y-4030=0$

$(x-2014)^2+(y-2014)^2+(x+y)^2=(2014)^2+(2014)^2+4030$

how i solve it, please help

Answer 1

Hint $$x^2+y^2+2xy-2014 x-2014y-2015=0 \iff (x+y)^2-2014(x+y)-2015=0 \iff ...$$ note $z=x+y$ and $$... \iff z^2-2014z-2015=0 \tag{1}$$ Because $\Delta=2014^2+4\cdot2015>0$ there will be 2 solutions $z_1,z_2 \in \mathbb{R}$ of $(1)$ and 2 lines to look at $y=z_1-x$ and $y=z_2-x$.

Answer 2

By inspection we find the two solutions $(-1,0)$ and $(0,-1)$.

As a conic cannot consist of two isolated points, we can conclude an infinity of solutions.

Answer 3

Hint: your equation is equivalent to $$(x+y-2015) (x+y+1)$$