Solving $a^2-b^2-a=0$ ∧ $-2ab+b={1\over 2}$ I want to solve
 $$a^2-b^2-a=0 ∧ -2ab+b={1\over 2}$$
I'm a bit stuck in the first part. I thought of the solutions, $a={1±\sqrt {1+4b^2} \over 2}$. But can't figure out solutions to the second part. Am I approaching it the right way?

I would solve it as $b=\pm \sqrt{a^2-a}$ and $b=\frac{1}{2(1-2a)}$.
So $$a^2-a = \frac{1}{4(1-2a)^2}$$
Or:
$$\left(2a-1\right)^2 - 1 = \frac{1}{(1-2a)^2}$$
Letting $u=(2a-1)^2$, you have $u-1=\frac{1}{u}$ or $u^2-u-1=0$ or $u=\frac{1\pm\sqrt{5}}{2}$. Then: 
$$a=\frac{1\pm\sqrt{\frac{1\pm\sqrt{5}}{2}}}{2}$$
Not that $a$ is complex if $u=\frac{1-\sqrt{5}}{2}$.
So for real solutions, we get $a=\frac{1\pm\sqrt{\frac{1+\sqrt{5}}{2}}}{2}$ and $$b=\frac{1}{2(1-2a)}=\frac{1}{\mp 2\sqrt{\frac{1+\sqrt{5}}{2}}}$$
