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?
On
We have $$a^3+4a^2b=4a^2+b^4$$ $$a(a^2+4ab)=4a^2+b^4$$ $$a(a+2b)^2-4ab^2=4a^2+b^4$$ $$a(a+2b)^2=b^4+4ab^2+4a^2$$ $$a(a+2b)^2=(b^2+2a)^2$$ $$a=\frac{(2a+b^2)^2}{(a+2b)^2}$$ $$\sqrt{a}-1=\frac{2a+b^2}{a+2b}-1=\frac{a-2b+b^2}{a+2b}$$ I'm stuck there. I also have a PhD in math and I'm an algebraist. Ramanujan could have done it while he was learning precalculus I'm sure!
Edit: as you can see other answers have solutions now.
On
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 $$ \eqalign{\sqrt{a} - 1 &= {\frac { 88445\,{s}^{2}+38961552\,s+21454046784 }{ \left( 146472+133\,s \right) ^{2}}} - 1 \cr &= \left(\dfrac{266 s}{146472 + 133s}\right)^2}$$
On
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.
On
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 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.
On
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 $a^3 = 4a^2 \implies a = 0, 4$.
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.
On
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.
On
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. $$*******$$ 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: $$\begin{cases}a=(1+t^2)^2\\b=(1+t)(1+t^2)\end{cases}$$
There might be a proof that a precalc student could follow, but I'd be rather surprised if there were a proof that a precalc student could invent. I say this having spent 5 minutes on the problem, which isn't much, but I've got a Ph.D. in math, and have had it for over 30 years, so the chances are that stuff I can't do in 5 or 10 minutes, most precalc students can't do at all. It looks as if @Matt Samuel may have a way to provide the former. :)