Solution

Expanding, we get $2x^2 + y^2 + 7 = 2(x+1)(y+1)$ Writing this as a quadratic in $x$, we get $2x^2 - (2y+2)x + y^2 - 2y + 5$ Since $x$ is real, its discriminant must be $\ge 0$, which means $(2y+2)^2 - (4)(2)(y^2 - 2y+5) \ge 0$ $\implies (y+1)^2 - 2(y^2 - 2y + 5) \ge 0$ $\implies (y-3)^2 \le 0$ which happens only when $y = 3$, substituting into the equation, we get that $x^2 - 4x + 4 = 0 \implies (x-2)^2 = 0 \implies x = 2$ So, the only solution is $(x,y) = (2,3)$

That was a really easy problem even for P1. But I guess it's OK since the rest of the problems were difficult...

Tintarn wrote:

Solution

How do you get these kind of factorizations??

"very easy"

Rukevwe wrote: How do you get these kind of factorizations?? Well, you can think of it as a quadratic equation in one of the variables, with the other one as a "parameter", e.g. as a quadratic in $y$: \[y^2+2(x+1)y+(2x^2-2x+5)=0\]and complete the square to \[(y+x+1)^2=-x^2-4x-4\]and then you are already almost done.

Oh!! I see! Thanks!

It will be a good exercise to do the procedure I described with the role of the variables exchanged. You will get again a sum of two squares expression, but a different one.

very nice problem! Claim:- Only $(x,y)=(2,3)$ satisfies pf:- set $\frac{x}{y}=m$ we get the equation retransform to $y^2(2m^2+1-2m)-y(2m+2)+5=0$ since $y \in \mathbb{R}$ we get $\triangle \ge 0$ so we get $(m+1)^2-5(2m^2+1-2m)\ge 0 \implies -(3m-2)^2\ge 0 \implies m=\frac{2}{3}$ so plugging back we get $y=3$ and hence $x=2$ so $(x,y)=(2,3)$ and claim follows $\blacksquare$

$2x^2+y^2+7=2(x+1)(y+1)$ $2x^2+y^2+7=2(xy+x+y+1)$ $2x^2+y^2+7=2xy+2x+2y+2$ $2x^2+y^2+7-2xy-2x-2y-2=0$ $y^2-2xy-2y+2x^2-2x+7-2=0$ $y^2-2y(x+1)+x^2-2x+x^2+5=0$ $y^2-2y(x+1)+x^2+2x-4x+1+4+x^2=0$ $y^2-2y(x+1)+x^2+2x+1+x^2-4x+4=0$ $y^2-2y(x+1)+(x+1)^2+x^2-4x+4=0$ $(y-(x+1))^2+x^2-2*2x+2^2=0$ $(y-x-1)^2+(x-2)^2=0$ $=>$ $(x-2)^2=0$ , $(y-x-1)^2=0$ $x-2=0$ $x=2$ $(y-x-1)^2=0$ $y-x-1=0$ $y-2-1=0$ $y-3=0$ $y=3$ $(x,y)=(2,3)$