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?

As randomgirl and Michael point out, there are counterexamples, such as $a=b=0$. However, it is true when $a \ge 1$. In principle it can be proved with just pre-calculus mathematics, but only an exceptional pre-calculus student could prove it.

Suppose $a^3 + 4a^2b = 4a^2+b^4$, where $a$ and $b$ are rational numbers with $a \ge 1$. We would like to show that $\sqrt{a}-1$ is the square of a rational number. Let $t = \sqrt{\sqrt{a}-1}$. Then $t$ is a real number because $a \ge 1$, and we would like to show that $t$ is rational. In terms of $t$, we have $a = (1+t^2)^2$, and a first question is whether we can also solve for $b$ in terms of $t$.

Substituting $a = (1+t^2)^2$ into $a^3 + 4a^2b - (4a^2+b^4)$ and factoring, we find that either $b = t^3+t^2+t+1$, $b = -t^3+t^2-t+1$, or $b^2 + 2(1+t^2)b + t^6 + 5t^4 + 7t^2 + 3 = 0$. The latter is a quadratic equation in $b$ with discriminant $-4(t^2+2)(t^2+1)^2$, which is negative, so it has no real root. Thus, $b$ must be either $t^3+t^2+t+1$ or $-t^3+t^2-t+1$, and without loss of generality we can assume $b = t^3+t^2+t+1$ by switching the sign of $t$ if necessary, because replacing $t$ with $-t$ does not affect the equation $a=(1+t^2)^2$.

Thus, the problem reduces to showing that if $a = (1+t^2)^2$ and $b = t^3+t^2+t+1$ with $a$ and $b$ rational, then $t$ must be rational as well. We will prove this by solving for $t$ in terms of $a$ and $b$. To do so, first note that $b = (t+1)(t^2+1)$, from which it follows that $b^2/a = (t+1)^2$. Then $b^2/a+b-2 = t(t^2+2t+3)$, while $b^2/a + 2 = t^2+2t+3$ (which is strictly positive because it equals $(t+1)^2+2$). Putting this together, we find that $t = (b^2/a+b-2)/(b^2/a+2)$, which is rational because $a$ and $b$ are rational.

Hmm. Well, it does seem to be true (though I don't immediately see a way for anyone to prove it by hand, let alone a precalc student).

Maple says the curve has genus $0$ with parametrization

$$ \eqalign{a&={\frac { \left( 88445\,{s}^{2}+38961552\,s+21454046784 \right) ^{2} }{ \left( 146472+133\,s \right) ^{4}}} \cr b&=-{\frac {11763185\,{s }^{3}-7772829624\,{s}^{2}-2853388222272\,s-3142417140546048}{2352637\, {s}^{3}+7772829624\,{s}^{2}+8560164666816\,s+3142417140546048}}} $$ Then $$ \eqalign{\sqrt{a} - 1 &= {\frac { 88445\,{s}^{2}+38961552\,s+21454046784 }{ \left( 146472+133\,s \right) ^{2}}} - 1 \cr &= \left(\dfrac{266 s}{146472 + 133s}\right)^2}$$