Can a pre-calculus student prove this? 
a and b are rational numbers satisfying the equation $a^3 + 4a^2b = 4a^2 + b^4$
Prove $\sqrt a - 1$ is a rational square

So I saw this posted online somewhere, and I kind of understand what the question is saying. I'm interesting in doing higher order mathematics but don't quite have the mathematical skills yet to grasp them (hence, precalculus) but is this problem way more complex than it seems or are the tools within the reach of a precalculus student?

We have 
$$a^3+4a^2b=4a^2+b^4$$
$$a(a^2+4ab)=4a^2+b^4$$
$$a(a+2b)^2-4ab^2=4a^2+b^4$$
$$a(a+2b)^2=b^4+4ab^2+4a^2$$
$$a(a+2b)^2=(b^2+2a)^2$$
$$a=\frac{(2a+b^2)^2}{(a+2b)^2}$$
$$\sqrt{a}-1=\frac{2a+b^2}{a+2b}-1=\frac{a-2b+b^2}{a+2b}$$
I'm stuck there. I also have a PhD in math and I'm an algebraist. Ramanujan could have done it while he was learning precalculus I'm sure!
Edit: as you can see other answers have solutions now.
