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.
 A: Actually, the statement is true even if $b=0$: just take $(x,y)=\left(\pm\sqrt a,0\right)$ if $a\geqslant0$ and  $(x,y)=\left(0,\pm\sqrt{-a}\right)$.
Otherwise, your approach is fine, but not your computations. You should have obtained, when $b\neq0$,$$\pm\left(\sqrt{\frac{a+\sqrt{a^2+b^2}}2},\frac b{|b|}\sqrt{\frac{-a+\sqrt{a^2+b^2}}2}\right),$$where, of course,$$\frac b{\lvert b\rvert}=\begin{cases}1&\text{ if }b>0\\-1&\text{ otherwise.}\end{cases}$$
A: If $b=0$, then either $x$ or $y$ is zero.
Suppose $a\ge0$; then $x^2-y^2=a$ has the solution $x=\sqrt{a}$, $y=0$.
If $a<0$, then $x^2-y^2=a$ has the solution $x=0$, $y=\sqrt{-a}$.
In both cases, the constraint $xy=0$ is satisfied.
For the case $b\ne0$, you can substitute $y=b/(2x)$ and get the equation
$$
4x^4-4ax^2-b^2=0
$$
which is a biquadratic; the associated equation $4z^2-4az-b^2=0$ has a positive root, because $-b^2<0$, so also the biquadratic has a solution (actually two).
Your computations are wrong. The positive root of $4z^2-4az-b^2=0$ is
$$
\frac{4a+\sqrt{16a^2+16b^2}}{8}=\frac{a+\sqrt{a^2+b^2}}{2}
$$
Thus
$$
x=\pm\sqrt{\frac{\sqrt{a^2+b^2}+a}{2}}
$$
From $y=b/(2x)$ we get
\begin{align}
y^2=\frac{b^2}{4x^2}
&=\frac{b^2}{2(\sqrt{a^2+b^2}+a)} \\[6px]
&=\frac{b^2}{2(\sqrt{a^2+b^2}+a)}\frac{\sqrt{a^2+b^2}-a}{\sqrt{a^2+b^2}-a} \\[6px]
&=\frac{b^2(\sqrt{a^2+b^2}-a)}{2(a^2+b^2-a^2)}\\[6px]
&=\frac{\sqrt{a^2+b^2}-a}{2}
\end{align}
Therefore
$$
y=\pm\sqrt{\frac{\sqrt{a^2+b^2}-a}{2}}
$$
The signs are not arbitrarily chosen: if you take the $+$ for $x$, then you will need to take the $+$ or the $-$ for $y$, according whether $b>0$ or $b<0$.
A: 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.
