<s>[INST] How to get all solutions to equations with square roots I would like to find all solutions to 
$$b-a\sqrt{1+a^2+b^2}=a^2(ab-\sqrt{1+a^2+b^2})$$
$$a-b\sqrt{1+a^2+b^2}=b^2(ab-\sqrt{1+a^2+b^2})$$
I found some solutions. For example, $a = 1, b = \pm i$ and $b = 1 , a = \pm i$. How can I find all solutions? 
 [/INST] The first equation becomes
$$b(1-a^3)=a(1-a)\sqrt{1+a^2+b^2},$$
the second
$$a(1-b^3)=b(1-b)\sqrt{1+a^2+b^2}.$$
If $a=0$, then $b=0$.
Clearly $a=b=1$ is a solution.  Let $a=1$ and $b\neq1$.  Then from the second equation 
$$1+b+b^2=b\sqrt{2+b^2}$$
we derive the solutions $b=\pm i$.
Let $a$ and $b$ not equal 0 or 1.  Then we have
$$b(1+a+a^2)=a\sqrt{1+a^2+b^2},$$
and
$$a(1+b+b^2)=b\sqrt{1+a^2+b^2}.$$
Consider now $1+a+a^2=0\iff a=\frac{-1\pm i\sqrt{3}}{2}$. In this case $1+a^2+b^2$ must be 0, this gives together with $1+a+a^2=0$ the condition $b^2=-a$, from where we get four more solutions.
Now lets divide one equation by the other.  An easy compution yields
$$(b-a)(a+b+ab)=0.$$
Now if you multiply the equations, you'll get
$$ab(1+ab)(a+b+ab)=0\iff (1+ab)(a+b+ab)=0.$$
So either $a+b=-ab$ or $a=b$ and $ab=-1$, that is $a=b=i$.
So we're done. (Remind that if $(a,b)$ is a solution, so is $(b,a)$.
Michael
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