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The number of pairs $(x,y)$ that satisfy : $2x^2 + y^2 + 2xy - 2y + 2 = 0$ is

a.) $0$

b.) $1$

c.) $2$

d.) None of the foregoing numbers

My attempt : I am not well versed in number theory , thus I took the most basic approach that I could see , that is I tried to divide the given equation into squares to and see if i could so something from that however I got stuck at $ (x+y)^2 + x^2 - 2y + 2 $

Also I tried putting x = 0 and realised that there exists no real number y which could form the required pair with x = 0 atleast , similarly i could observe the same thing with y = 0.

Please suggest me a solution as well as a more general approach towards solving these kind of problems

My background is a degree in Electrical Engineering , however I have never taken any specific course in number theory.

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3 Answers 3

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$2x^2 + y^2 + 2xy - 2y + 2 = 0$

$2(x+\frac{y}{2})^2+\frac12(y-2)^2=0$

so the answer is unique, and

$y=2,x=-1$

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(in the spirit of the short demonstration of @Takahiro Waki : no need to extract roots...) yet another factorization : $(y + x - 1)^2 + (x + 1)^2$ wich can be null only if the content of the two parentheses are simultaneously equal to zero i.e, for $x=-1$ and $y=2$.

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$$2x^2+y^2+2xy−2y+2=0$$ $$y^2+2(x-1)y+2x^2+2=0$$ $(x,y) - $ pair of rational numbers, then $$y=1-x\pm\sqrt{(x-1)^2-2x^2-2}=1-x \pm \sqrt{-x^2-2x-1}$$ $\sqrt{-(x^2+2x+1)}=\sqrt{-(x+1)^2}$. Then $x=-1$ and $y=2$

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  • $\begingroup$ I am sorry, but the last line of your answer is insufficient: you need to say that you are looking for real (and even rational) solutions, not complex ones ; the only way to have that is to have a null radicand. $\endgroup$
    – Jean Marie
    Commented Apr 11, 2016 at 18:34
  • $\begingroup$ "Number of pairs of rational numbers that satisfy the given relation" $\endgroup$
    – Roman83
    Commented Apr 11, 2016 at 19:46
  • $\begingroup$ I don't want to polemicate. I just wanted to say that $\sqrt{-(x+1)^2}.$ immediately followed by "Then $x=-1$... " without other comments is a little surprising. $\endgroup$
    – Jean Marie
    Commented Apr 11, 2016 at 20:03

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