TITLE: Can a pre-calculus student prove this?
QUESTION [33 upvotes]: 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?

REPLY [6 votes]: My answer is YES, a pre-calculus student can prove this. I came to think not and I was almost convinced, after seeing the answer of Robert Israel --obviously made by a machine, not by a man--, it was impossible to solve the problem by elementary means. However back to the problem and I could find a "pre-calculus solution" unexpectedly easy.
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We have straight away $$a^3+4a^2b=4a^2+b^4\iff a=\left(\frac{b^2+2a}{2b+a}\right)^2$$ so $a$ is a square of a rational (because $a^3+4a^2b=4a^2+b^4=(b^2+2a)^2-4ab^2$ so…….) and it is asked that $\sqrt a-1$ is a square of a rational.  We change notation making $ (\sqrt a, b)=(x,y)$ and reformulate the question the following equivalent way, in which $x,y$ are rational:
$$\text {If}\space x=\frac{y^2+2x^2}{2y+x^2}\text{ then}\space x-1\space\text {is the square of a rational}.$$
In fact 
$$ x=\frac{y^2+2x^2}{2y+x^2}\iff x^3+2yx=y^2+2x^2\iff y^2-2xy+(2x^2-x^3)=0$$
It follows (from pre-calculus student formula) $$y=x\pm\sqrt{x^2+x^3-2x^2}=x\pm x\sqrt{x-1}$$
The proof is evident now: $y$ is rational if and only if $x-1$ is a square. If $x-1$ is not a square, then we deny the hypothesis with rational $x, y$.
Of course, equivalently, we could also complete the square in $y$ and get directly
 $$x-1=\left(\frac{y-x}{x}\right)^2$$
We parametrize the variables a and b in a more comfortable way than the parametrization of Maple: $$\begin{cases}a=(1+t^2)^2\\b=(1+t)(1+t^2)\end{cases}$$