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How can I find all ordered pairs $(x,y)$ such that $x + y^2 = 2$ and $y + x^2 = 2$?

A solution/explanation would be greatly appreciated.

If it helps, the solutions given in the textbook are x=-2, y=-2; x=1, y=1; x= (1+root 5)/2, y=(1-root 5)/2; x= (1-root 5)/2, y=(1+root 5)/2

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  • $\begingroup$ subtract the two equations... $\endgroup$ – zwim Jan 6 '18 at 0:23
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    $\begingroup$ I got up to (y-x)(1-x-y)=0 $\endgroup$ – ana556 Jan 6 '18 at 0:33
  • $\begingroup$ @ana556 then deduce $ab=0\implies a=0$ or $b=0$ and continue solving. $\endgroup$ – zwim Jan 6 '18 at 0:34
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If we leave $y$ alone in the second equation, we have $y = 2-x^2$. Then putting it in the first equation, we get $x+(2-x^2) = 2 \implies x^4-4x^2+x+2 = 0$. Now, notice that $x=1$ satisfies this equation, therefore we can say that the polynomial $x^4-4x^2+x+2$ can be factorized as $$x^4-4x^2+x+2 = (x-1)(Ax^3+Bx^2+Cx+D)$$ Here, we can find $A$ and $D$ easily because there is only one term to consider and $A = 1$, $D = -2$. Notice that after that, we can conclude that $B-1 = 0$ and $C-B = -4$ by polynomial equality. Then we have $B = 1$ and $C = -3$, so we have $$x^4-4x^2+x+2 = (x-1)(x^3+x^2-3x-2) = (x-1)(x+2)(x^2-x-1)$$ From here, you can find all $x$ and corresponding $y$'s by using the equations given (Don't forget to consider $x^2-x-1 = 0$ because it has two roots as well).

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HINT

$$y+x^2= x+y^2\iff y-x-y^2+x^2=0\implies y-x-(y-x)(y+x)=0\implies (y-x)(1-x-y)=0$$

thus

$$y-x=0 \implies y=x \implies x^2+x-2=0$$

or

$$x+y=1\implies x^2-x-1=0$$

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Well $x+y^2=y+x^2$ so $x-y=x^2-y^2=(x-y)(x+y) $

If $x-y=0$ then $x=y$ and $x+x^2=2$. We can use the the quadratic equation to solve that.

If $x-y\ne 0$ then $1=(x+y) $ and $y=1-x $

So $x+y^2=x^2+y=x^2-x+1=2$ and we can use the quadratic equation to solve that.

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