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?

Here's another solution that proceeds mostly by the use of divisibility properties.  All steps can be justified if one assumes the fundamental theorem of arithmetic, whose statement (but not proof) will be familiar to many precalculus students.  I personally, however, would have found it tough-going to follow this proof as a precalculus student.
Write $a=\frac{w}{x}$, $b=\frac{y}{z}$, where $w$ and $x$ are relatively prime integers, $y$ and $z$ are relatively prime integers, and $x$ and $z$ are positive.  We then have
$$
a^2(a+4b-4)=b^4\Rightarrow w^2(wz+4xy-4xz)=\frac{y^4x^3}{z^3}.
$$
Since the left side is an integer and $y$ and $z$ are relatively prime, $\frac{x^3}{z^3}=\left(\frac{x}{z}\right)^3$ is an integer and therefore $\frac{x}{z}$ is an integer.  Write $x=nz$, where $n$ is a positive integer.  Since $w$ is relatively prime to $x$, it is also relatively prime to $z$ and $n$.  Substituting $x=nz$ gives
$$
w^2z(w+4ny-4nz)=y^4n^3.
$$
Since $n$ is relatively prime to $w$, it is relatively prime to $w+4ny-4nz$, and therefore $z$ is divisible by $n^3$.  Write $z=n^3m$, where $m$ is a positive integer.  Since $y$ is relatively prime to $z$, it is relatively prime to $n$ and $m$.  Substituting $z=n^3m$ gives
$$
w^2m(w+4ny-4n^4m)=y^4.
$$
Since $m$ is a positive integer, relatively prime to $y$, but $m$ divides $y^4$, the only value $m$ can take is $1$.  Hence $x=n^4$ and $z=n^3$.  This leads to
$$
w^2(w+4ny-4n^4)=y^4.
$$ 
Assuming now that $y\ne0$, $\frac{y^4}{w^2}$ is an integer, and therefore $\frac{y^2}{w}$ is an integer.  Setting $y^2=kw$, with $k$ a nonzero integer, and eliminating $w$ yields
$$
y^2+4nky-4n^4k=k^3.
$$
Solving the quadratic for $y$ gives
$$
y=-2nk\pm\sqrt{4n^2k^2+4n^4k+k^3}=-2nk\pm\sqrt{k(4n^4+4n^2k+k^2)}=-2nk\pm\sqrt{k(2n^2+k)^2}.
$$
So either $2n^2+k=0$, which implies $k=-2n^2$ and $y=4n^3$, or $k$ is a perfect integer square, say $k=r^2$, and
$$
y=-2nr^2+r(2n^2+r^2)=r(2n^2-2nr+r^2)=r(n^2+(n-r)^2).
$$
If the former, then
$$
a=\frac{w}{x}=\frac{y^2/(-2n^2)}{n^4}=\frac{-8n^4}{n^4}=-8,\qquad b=\frac{y}{z}=\frac{4n^3}{n^3}=4.
$$
If the latter, then
$$
a=\frac{w}{x}=\frac{y^2/r^2}{n^4}=\frac{(n^2+(n-r)^2)^2}{n^4},\qquad b=\frac{y}{z}=\frac{r(n^2+(n-r)^2)}{n^3},\qquad n,r\in\mathbf{Z}, n\ge1,
$$
and so $\sqrt{a}-1=\frac{(n-r)^2}{n^2}$.
In conclusion, either $y=0$ (and hence $b=0$ and $a\in\{0,4\}$), or $a=-8$ and $b=4$, or $\sqrt{a}-1$ is a rational square.
