<s>[INST] Show that the equation system $I:x^2-y^2=a;II: 2xy= b$ has always a solution $(x,y)\in \mathbb {R}^2$ The statement must be false because if $b=0$ then $x$ or $y$ must be $0$. If $x$ is $0$ we get from $I$ that $-y^2=a\iff y^2 = -a$. Wouldn't that be a contradiction to the statement? Because if $a>0$ then there is no solution for the equation System, vice versa if $y=0$ and $a<0$. 
For the general case I have assumed that $b\neq 0$ and therefore from $II$ we get $y=\frac{b}{2x}$, substituting with $I \Longrightarrow x^2-(\frac{b}{2x})^2=a\iff 4x^4 - b^2 = 4x^2a \iff x^4 -ax^2 - \frac{b^2}{4}=0 (*)$ 
Completing the square 
$(*)\iff x^4 - 2\frac{ax^2}{2}-\frac{b^2}{4} \iff x^4 - 2\frac{ax^2}{2} +\frac{a^2}{4}-\frac{b^2}{4}-\frac{a^2}{4}\iff (x^2 - \frac{a}{2})^2+\frac{-b^2-a^2}{4}=0$
$\Rightarrow x^2-\frac{a}{2}=\sqrt{\frac{b^2+a^2}{4}} \Rightarrow x^2=\sqrt{\frac{b^2+a^2}{4}}+\frac{a}{2} \Rightarrow x = \pm \sqrt{\sqrt{\frac{b^2+a^2}{4}}+\frac{a}{2}}$
$II\Longrightarrow y= \frac{b}{2(\pm \sqrt{\sqrt{\frac{b^2+a^2}{4}}+\frac{a}{2}})}$
Can somebody tell my whether this Solutions are Right or not because when I have tried to verify it I end up with that term:
$\frac{5a^2+a(5\sqrt\frac{b^2+a^2}{4})(4\sqrt{\frac{b^2+a^2}{4}}+\frac{a}{2})}{4}$ which should be equal to $a$
Thank you for your time.
 [/INST] Geometric solution
If $\;a,b\neq 0\;$ the equations define two hyperbolas centered both in $0,$ where 


*

*$I:x^2-y^2=a\;$ has asymptotes $\;y=\pm x,$ and lies in the left and right sectors ($a>0$) or in upper and lower sectors ($a<0$) delimited by the asymptotes;

*$II: 2xy= b\;$ has asymptotes $\;x=0,y=0\;$ and lies in the first and third quadrants ($b>0$) or in the second and fourth if $b<0$ (the quadrants are sectors delimited by asymptotes).
If $a=0$ or $b=0,$ the corresponding equation defines the two asymptotes of $I$ or of $II.$

Thus the two objects $I$ and $II$ necessarilly intersect in $2$ points if at least one of them is a hyperbola, and in one point in the degenerate case $a=0=b.$ The coordinates $(x,y)$ of these points are solutions of the system.
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