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?

My solution doesn't differ much from the other ones but I want to explain a bit how I did the factorization:
We are mainly interested in $\sqrt{a}-1$ and not so much in $a$, so let's define $c:=\sqrt{a}-1$. Our goal is to show that $\sqrt{c}$ is a rational number, given that $a$ and $b$ are rational, too. We rewrite the equation as
$$(c+1)^6+4\cdot(c+1)^4\cdot b = 4\cdot(c+1)^4+b^4.$$
In a first step, we want to find all the possible $b$ for a given $c$. Since the highest order in $b$ is $4$, we know that there are at most $4$ solutions. In the following, we make some assumptions about the shape of $b$ as an expression of $c$ and hope that we find some solutions:

First, we notice that $(c+1)$ occurs multiple times in our equation. Wouldn't it be cool if $b$ had those $(c+1)$-brackets, too? Well, let's assume that $b$ has the following shape:
$$b=(c+1)\cdot f(c).$$
Then the equation becomes
$$(c+1)^6+4\cdot(c+1)^5\cdot f(c) = 4\cdot(c+1)^4+(c+1)^4\cdot f^4(c)$$
$$(c+1)^2+4\cdot(c+1)\cdot f(c) = 4+ f^4(c).$$
Let's assume that $f(c)$ can be written as
$$f(c)=\alpha\cdot c^\beta+g(c)$$
where $g(c)$ has an order of less than $\beta$. Surely, the highest order on the right side of the equation is $4\cdot\beta$, but what is the highest order on the left side? It is either $2$, coming from the first term which contains $c^2$, or it is $\beta+1$, coming from the second term which contains $4\alpha c \cdot c^\beta$.


*

*If the highest order on the left side was $\beta+1$, then we'd have $2\lt\beta+1$ and $4\beta=\beta+1$ which is impossible.

*If the highest order on the left side is $2$, then we have $\beta+1\lt2$ and $4\beta=2$. This works and we get $\beta=1/2$.


If we compare the coefficients of these two leading terms, we get that $\alpha=\pm1$. Let's plug in everything we have:
$$c^2+2c+1\pm 4\cdot(c+1)\cdot c^{1/2}+4\cdot(c+1)\cdot g(c)$$
$$=4+c^2\pm 4c^{3/2}\cdot g(c)+6c\cdot g^2(c)\pm4c^{1/2}\cdot g^3(c)+g^4(c).$$
If we stare at this a little bit, we see that $g(c)=1$ works. So we have found two solutions for $b$:
$$b_1=(c+1)\cdot(1+c^{1/2})=c^{3/2}+c+c^{1/2}+1$$
$$b_2=(c+1)\cdot(1-c^{1/2})=-c^{3/2}+c-c^{1/2}+1$$

Now we wonder if there are even more real solutions for $b$. Having found two solutions, we can apply polynomial long division on $b^4-4\cdot(c+1)^4\cdot b +4\cdot(c+1)^4-(c+1)^6$ and get:
$$b^4-4\cdot(c+1)^4\cdot b +4\cdot(c+1)^4-(c+1)^6$$
$$=\left(b-(1+c)(1+c^{1/2})\right)\left(b-(1+c)(1-c^{1/2})\right)\left(b^2+2b(c+1)+(c+1)^2(3+c)\right)$$
If we calculate the discriminant of the last term, we'll find that it will be negative, except for $c=-1$, i.e. $a=0$. So, there are no further real expressions, unless $a=0$. (But then we have $c=-1$ which isn't a squared rational number).

In the following we assume that $a\neq0$. Then $b_1$ and $b_2$ are the only solutions to our equation. As Matt Samuel showed, $c=\sqrt a -1$ is a rational number. But we want to show that $\sqrt{c}$ is a rational number, too! Take
$$b=b_1=(c+1)(1+c^{1/2})$$
Then we have
$$c^{1/2}=\frac{b}{c+1}-1$$
which is rational (because $b$ and $c$ are rational). Finally, take
$$b=b_2=(c+1)(1-c^{1/2})$$
Then we have
$$c^{1/2}=-\frac{b}{c+1}+1$$
which is rational, too. We have reached our goal.
