<s>[INST] 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?
 [/INST] Given any $(a,b)$ such that $a^3 + 4a^2b = 4a^2 + b^4$.
(1)  If $a = 0$, then $b$ is clearly 0, $\sqrt{a} - 1 = -1$ is not a rational square.
(2) If $b = 0$, then $a^3 = 4a^2 \implies a = 0, 4$. 


*

*In the first sub case $a = 0 \implies \sqrt{a}-1 = -1$ is not a square. 

*In the second sub case, $a = 4 \implies \sqrt{a} - 1 = 1$ is a square.


Assuming $a, b \ne 0$, we have
$$a(a+2b)^2 = ( a^3 + 4a^2b ) + 4ab^2 = (4a^2 + b^4) + 4a^2b = (2a+b^2)^2\tag{*1}$$
(3) Assuming $a + 2b \ne 0$, this leads to 
$$a = \left(\frac{2a + b^2}{a+2b}\right)^2
\implies \sqrt{a} = \left|\frac{2a + b^2}{a+2b}\right| \quad\text{ is rational }$$
Divide both sides of $(*1)$ by $a^2$ and taking square root, we get
$$\sqrt{a} + \frac{2b}{\sqrt{a}} = \pm \left( \frac{b^2}{a} + 2 \right)$$
The sign on RHS cannot be -ve. otherwise, we will have
$$\sqrt{a} + \frac{2b}{\sqrt{a}} = - \left( \frac{b^2}{a} + 2 \right)
\implies \sqrt{a} + \left(\frac{b}{\sqrt{a}}+1\right)^2 = -1
\quad\text{ which is absurb! }$$
So the sign on RHS is +ve and we have
$$\sqrt{a} - 1 = \frac{b^2}{a} - \frac{2b}{\sqrt{a}} + 1
= \left(\frac{b}{\sqrt{a}} - 1\right)^2
= \left(b\left|\frac{a+2b}{2a + b^2}\right| - 1\right)^2$$
which is the square of a rational number.  
(4) This leaves us with the case $a + 2b = 0$. In this case,
$$( 2a + b^2 )^2 = 0 \implies 2a + b^2 = 0 \implies b(b-4) = 0$$
Since we have assume $a, b \ne 0$, we get $b = 4$ and hence $(a,b) = (-8,4)$.
where $\sqrt{a} - 1$ is not defined.
Summary
Aside from the special case $a = 0$ and $(a,b) = (-8,4)$, $\sqrt{a}-1$ is a the square of a rational number.
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