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I'm solving the following equations, $$x+y=zw$$ $$z+w=xy$$ How many solutions $(x,y,z,w)$ exist, if the variables are reals?

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Since your system is underdetermined, there will be infinitely-many solutions. –  Shaun Ault Dec 22 '12 at 14:57
    
Of course, a priori, those infinitely many solutions could all be complex or at infinity (in projective space). –  Hurkyl Dec 22 '12 at 16:29
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The same equations where one wants integer solutions appeared on this forum. math.stackexchange.com/questions/219762/… –  coffeemath Dec 23 '12 at 19:20
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up vote 4 down vote accepted

Solving the 1st equation for $y$ and substituting in the second gives $$z+w=xzw-x^2\iff x^2-xzw+z+w=0$$ This equation has a solution for $x$ when $$\Delta\ge 0\iff z^2w^2-4z-4w\ge 0$$ It remains to check that it always has solutions for $z,w$. Therefore, your system has an infinite number of solutions all of which satisfy: $$z^2w^2-4z-4w\ge 0$$ $$x=\frac{zw-\sqrt{z^2w^2-4z-4w}}2$$ $$y=zw-x$$

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oops... I think you meant $y = zw - x$ in your last line. –  Shaun Ault Dec 22 '12 at 16:20
    
@ShaunAult Now fixed –  Nameless Dec 22 '12 at 16:25
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