Rational solutions of $x^3+y^3=9$

Question

An old alchemist had two sphercial flasks, one with a circunference of $12$ inches and the other one with a circumference of $24$ inches. He desired to transfer their contents into two other spherical flasks whose sizes are different from the first two. What were the circumferences of the two new flasks in rational numbers?

After some easy algebraic operations I have arrived at the problem:

Find solutions of $ x^3+y^3=9$ with $x$ and $y$ in $\mathbb{Q}$.

Are there any other solutions distinct from the $(2,1)$ and $(1,2)$? I have tried very much but I haven't made progress in doing this. How can I proceed with that? What's the name of the theme? What can I do?

Answer 1

This is from Albert Beiler, Recreations In The Theory Of Numbers: The Queen Of Mathematics Entertains. Chapter 25 is called Tilts and Tourneys. This is problem 83, on page 303. The answer is on pages 335-336, and confirms that the desired quantity is rational and positive $x,y$ such that $x^3 + y^3 = 9,$ smallest common denominator, but $(x,y) \neq (2,1)$ and $(x,y) \neq (1,2).$ He gives the specific answer, the common denominator is quite large.

The doubling construction through the point $(2,1)$ gives the new rational point $$ \left( \frac{20}{7} , \; \;\frac{-17}{7} \right) $$

It is not necessary to have a cubic curve in any special format to perform doubling, maybe it is called duplication. It is just calculus. We begin with $x^3 + y^3 = 9.$ We implicitly find $3 x^2 + 3 y^2 y' = 0,$ or $$ y' = - \left( \frac{x}{y} \right)^2 $$ For example, at the rational point $(2,1)$ we find $y' = -4.$ Next we parametrize the tangent line through $(2,1)$ by variable $t$ as $$ x = 2 + t, \; \; y = 1 - 4t $$ We set $x^3 + y^3 = 9$ and arrive at $$ -63t^3 + 54 t^2 = 0.$$ That is, $$ 9 t^2 (6 - 7 t) = 0. $$ Therefore, for nonzero $t,$ we have $t = 6/7.$ then $$ x = 2 + (6/7) = 20/7, $$ $$ y = 1 - (24/7) = -17/7 $$

Joining to $(1,2)$ gives $$ \left( \frac{919}{438} , \; \;\frac{-271}{438} \right) $$

ADDENDUM: I worked out duplication for a rational point $(a,b)$ on a curve where we make $x^3 + y^3 = \mbox{constant}.$ The value of the constant does not enter; we get $$ (a,b) \mapsto \left( \frac{a(a^3 + 2b^3)}{a^3 - b^3}, \; \; \frac{-b(2 a^3 + b^3)}{a^3 - b^3} \right). $$ In particular, $$ \left( \frac{20}{7} , \; \;\frac{-17}{7} \right) \mapsto \left( \frac{-36520}{90391} , \; \;\frac{188479}{90391} \right) $$ is not in the first quadrant. $$ \left( \frac{919}{438} , \; \;\frac{-271}{438} \right) \mapsto \left( \frac{676702467503}{348671682660} , \; \;\frac{415280564497}{348671682660} \right). $$ This is the answer Beiler supplies.

I drew a picture. It seems quite believable that Beiler's answer really is the first rational point that returns to the first quadrant, after leaving it.

I rotated the diagram by $45^\circ$ with $u = x-y, \; v = x+y.$ This was for my convenience in drawing the picture using points from a calculator, as $v^3 + 3 u^2 v - 36 = 0$ and Cardano gives $$v = \sqrt[3]{ \sqrt{324 + u^6} + 18} - \sqrt[3]{\sqrt{324 + u^6} - 18}$$ Then I made another picture with more points, including Beiler's solution. Most of the lines were light green, I made the line that duplicates and produces that in light blue.

Answer 2

Suppose $a/b$ and $c/d$ are slutions to $x^3+y^3=9$, where the fractions are assumed to be irreducible. I suppose for your problem $a,b,c,d$ are also all positive?

In any case, we can then rearrange the equation to$${(ad)}^3+{(bc)}^3=9{(bd)}^3$$ where $a,b,c,d$ are positive integers. Now, look into the prime factorization of $b$ and of $d$. Consider a prime $p$ and let $k_b,k_d$ be the order of $p$ in the factorization of $b,d$, respectively.

Suppose $k_b>k_d\geq 0$. Then $p^{\displaystyle 3k_b+3k_d}$ is a factor of $9{(bq)}^3$. We may divide both sides by $p^{\displaystyle 3k_b}$; the RHS will be an integer, so the LHS must too. ${(bc)}^3$ clearly is divisible by $p^{\displaystyle 3k_b}$, so ${(ad)}^3$ too must be. For this to happen, we'd need $p$ to be a factor of $a$ with exponent at least $k_b-k_d>0$, but this contradicts the irreducibility of $a/b$. It follows that one may not have $k_b>k_d \geq0$.

Similarly, we may not have $k_d > k_b \geq 0$, so it must be that $k_b=k_d$. Since this reasoning applies to every prime $p$, we conclude that $b$ and $d$ have the same prime factorization, and thus are the same.

So, while this is not a full answer, any solution to the problem over the rationals is of the form $x=a/q$ and $y= b/q$, where the fractions are irreducible. We may consider thus the analagous problem of finding solutions to

$$a^3+b^3=9q^3$$

over the integers. Notice that, by a reasoning similar to the one above, any solution must be so that $a$ and $b$ do not share prime factors.