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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?

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    $\begingroup$ where did you find it, exactly? $\endgroup$ – Will Jagy Oct 30 '15 at 22:52
  • $\begingroup$ If $a=b=1$ then the first equality holds, and $\sqrt1-1=0$ which is a rational square. $\endgroup$ – zoli Oct 30 '15 at 23:03
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    $\begingroup$ @zoli of course the point is to show this is true for every such choice of $a$ and $b$. $\endgroup$ – Kevin Carlson Oct 30 '15 at 23:48
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    $\begingroup$ This problem is discussed (and solved) here: artofproblemsolving.com/community/… It's apparently from the Polish 2010 Math Olympiad. $\endgroup$ – Eric Towers Oct 31 '15 at 6:20
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    $\begingroup$ Trying to do things beyond what you think you know is good. Random searching isn't likely to find good problems. Is there a math club or a math competition group at your school? Is there a sympathetic teacher who can guide you? There may be websites that will help - search for "problem of the day" and choose a grade level. $\endgroup$ – Ethan Bolker Oct 31 '15 at 13:09
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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.

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    $\begingroup$ Something that I think is slightly more intuitive than how you set this up: Let $s=\sqrt{a}-1$. We want to show $s$ is a rational square. So in turn, we set $t = \sqrt{s}$. Now, let's substitute for $a$ in terms of $t$. $\endgroup$ – Joel Oct 31 '15 at 15:27
  • $\begingroup$ a comment above, after you answered, gives initial source as Polish 2010 Olympiad: artofproblemsolving.com/community/… $\endgroup$ – Will Jagy Oct 31 '15 at 18:24
  • $\begingroup$ I've edited to include @Joel 's suggestion in the solution. Made it easier for me. $\endgroup$ – Dheeraj Bhaskar Nov 1 '15 at 18:03
  • $\begingroup$ @DheerajBhaskar Thanks, I didn't see your edit so I updated it myself. $\endgroup$ – Henry Cohn Nov 1 '15 at 21:00
  • $\begingroup$ @HenryCohn: The problem is, incredibly, very easy. Look at my answer, please. $\endgroup$ – Piquito Nov 5 '15 at 13:31
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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 $$ \eqalign{\sqrt{a} - 1 &= {\frac { 88445\,{s}^{2}+38961552\,s+21454046784 }{ \left( 146472+133\,s \right) ^{2}}} - 1 \cr &= \left(\dfrac{266 s}{146472 + 133s}\right)^2}$$

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  • $\begingroup$ The problem is, incredibly, very easy. Look at my answer, please. $\endgroup$ – Piquito Nov 5 '15 at 13:32
  • $\begingroup$ @Piquito The problem may be simple, but it is definitely not easy. $\endgroup$ – Nikos Feb 13 '17 at 15:59
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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.

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    $\begingroup$ Wolfram seems to find four solutions for $b$ in terms of $a$, all variations of $b = \sqrt{a}\sqrt{\sqrt{a}-1} + \sqrt{a}$. Together with your proof that $\sqrt{a}$ is rational, this would suggest $\sqrt{a}-1$ is also a rational square. I don't know how it is solving the quartic. $\endgroup$ – Michael Oct 30 '15 at 23:30
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    $\begingroup$ It also says though a solution pair is $(a=-8,b=4)$ but this would mean $\sqrt{a}-1$ is not always rational for rational pairs satisfying the given equation. $\endgroup$ – randomgirl Oct 30 '15 at 23:34
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    $\begingroup$ So much for the computer assisted proof! $\endgroup$ – Matt Samuel Oct 30 '15 at 23:37
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    $\begingroup$ @randomgirl : A simpler counter-example is $a=b=0$. $\endgroup$ – Michael Oct 30 '15 at 23:39
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    $\begingroup$ @Joshua I didn't prove it's a square. $\endgroup$ – Matt Samuel Nov 1 '15 at 21:51
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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. :)

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  • $\begingroup$ artofproblemsolving.com/community/… $\endgroup$ – Will Jagy Oct 31 '15 at 18:26
  • $\begingroup$ @John Hughes: No reason to be surprised. The problem is, incredibly, very easy. Look at my answer, please. $\endgroup$ – Piquito Nov 5 '15 at 13:36
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    $\begingroup$ I like your solution, but in my 45 years of teaching one person or another, from high-school to graduate students, I've come upon no pre-calc student who could have come up with that (although plenty could probably follow it). Of course, maybe you've had better experiences. Mine were mostly with Ivy-league students. :( $\endgroup$ – John Hughes Nov 6 '15 at 1:58
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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 $a^3 = 4a^2 \implies a = 0, 4$.

  • In the first sub case $a = 0 \implies \sqrt{a}-1 = -1$ is not a square.
  • In the second sub case, $a = 4 \implies \sqrt{a} - 1 = 1$ is a square.

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.

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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: $$\begin{cases}a=(1+t^2)^2\\b=(1+t)(1+t^2)\end{cases}$$

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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.

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  • $\begingroup$ It can be shorter. The problem is, incredibly, very easy. Look at my answer, please. $\endgroup$ – Piquito Nov 5 '15 at 13:39
  • $\begingroup$ @Ataulfo: your method is indeed shorter and more elegant. Make sure to handle the case $a=-2b$ though. $\endgroup$ – Will Orrick Nov 5 '15 at 15:52
  • $\begingroup$ Thank you for 2b + a = 0, the solutions (a,b)=(-8,4). I was so convinced that the problem was very difficult that when I realized it was very easy, I neglected possible exceptions and I just wanted to say the problem can be solved by a pre-calculus student. (In truth I was mostly thinking that the curve has an infinity of integer points (which is something that interests me particularly). Regards. $\endgroup$ – Piquito Nov 5 '15 at 22:08
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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.

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