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I stumbled upon this neat formula for sums of cubes with arbitrary $x,y\in\mathbb{Z}$$$(x^2+9xy-y^2)^3+(12x^2-4xy+2y^2)^3=(9x^2-7xy-y^2)^3+(10x^2+2y^2)^3\tag1$$ With $1729=1^3+12^3=9^3+10^3$ as its first instance. And I believe that this formula was used by Ramanujan to find a formula for$$a^3+b^3=c^3\pm1$$ So my question?

Questions:

  1. What would be someone's thinking process when finding other formulas such as $(1)$?
  2. Are there any other formulas similar to $(1)$?

I'm thinking along the lines of starting with $$(x^2+axy+by^2)^3+(cx^2+dxy+ey^2)^3=(fx^2+gxy+hy^2)^3+(ix^2+jxy+ky^2)^3$$ But even Mathematica can't solve the ensuing system that follows. So for the moment, I'm stuck.

Frank
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  • https://en.wikipedia.org/wiki/Euler's_sum_of_powers_conjecture#Background They seem to be giving Binet's version of Euler's solution, pages 552-555 in volume II of Dickson's History – Will Jagy Dec 31 '16 at 20:48
  • @WillJagy Know an online pdf version I can get to see? – Frank Dec 31 '16 at 21:52
  • https://books.google.com/books?id=dO7C02z4LlcC&pg=PA552&lpg=PA552&dq=dickson+history+of+the+theory+of+numbers+he+would+be+content+if+brouncker+would+divide&source=bl&ots=bQ6fXIiRMZ&sig=X-eoxAcMNregRd3APGZpoStLCV4&hl=en&sa=X&ved=0ahUKEwiLz8Det5_RAhWCKGMKHdMECOIQ6AEIHDAA#v=onepage&q=dickson%20history%20of%20the%20theory%20of%20numbers%20he%20would%20be%20content%20if%20brouncker%20would%20divide&f=false – Will Jagy Dec 31 '16 at 22:04
  • @WillJagy I don't think $(1)$ is Binet's solution. According to the book, Binet's solution is$$x^3+y^3=z^3+u^3$$Where $$\begin{align} & x=m\alpha-n^2\ & y=-m\beta+n^2\ & z=n\alpha-m^2\ & u=-n\beta+m^2\end{align}$$And$$\alpha^2+\alpha\beta+\beta^2=3mn$$It seems more like Ramanujan's solution... – Frank Jan 01 '17 at 22:16

2 Answers2

6

The formula that Ramanujan actually recorded, as well as Euler's solution, are discussed by Ono in https://arxiv.org/abs/1510.00735

The original formula was $$ \left( 6 a^2 - 4ab + 4 b^2 \right)^3 = \left( 3 a^2 +5ab - 5 b^2 \right)^3 + \left( 4 a^2 - 4ab + 6 b^2 \right)^3 + \left( 5 a^2 - 5ab -3 b^2 \right)^3 $$

The formula can be made more symmetric still. The two classes of forms of discriminant $85$ are represented by $$ x^2 + 9 xy - y^2, $$ $$ 3 x^2 + 7 xy - 3 y^2. $$ The latter is equivalent to $3 x^2 - 5 xy - 5 y^2, $ off by a single minus sign.

The two classes of (primitive) forms of discriminant $-20$ are represented by $$ x^2 + 5 y^2, $$ $$ 2 x^2 + 2 xy + 3 y^2. $$

That is, one may pass between Ramanujan's version and yours by using Gauss composition in order to multiply by $3,$ which passes between the principal genus and the other genus, both for discriminant $85$ and $-20.$

On page 2 they give Ramanujan's modern version of Euler's complete solution. When $$ \alpha^2 + \alpha \beta + \beta^2 = 3 \lambda \gamma^2, $$ $$ \left( \alpha + \lambda^2 \gamma \right)^3 + \left( \lambda \beta + \gamma \right)^3 = \left(\lambda \alpha + \gamma \right)^3 + \left( \beta + \lambda^2 \gamma \right)^3 $$

http://esciencecommons.blogspot.com/2015/10/mathematicians-find-magic-key-to-drive.html

Will Jagy
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3

Actually, Mathematica can solve, $$\small (x^2+axy+by^2)^3+(cx^2+dxy+ey^2)^3+(fx^2+gxy+hy^2)^3+(ix^2+jxy+ky^2)^3 =0\tag2$$ One may be guided by the principle of fait accompli (accomplished fact). Ramanujan and others already found solutions therefore $(2)$, approached the right way, must be solvable.

What you do is expand $(2)$ and collect powers of $x,y$. The Mathematica command is Collect[P(x,y),{x,y}] to get,

$$P_1x^6+P_2x^5y+P_3x^4y^2+P_4x^3y^3+P_5x^2y^4+P_6xy^5+P_7y^6 = 0$$

where the $P_i$ are polynomials in the other variables. The hard part is then solving the system,

$$P_1 = P_2 = \dots =P_7 = 0$$

After much algebraic manipulation (which I don't have the strength to type all down), one ends up with the simple identity (which I gave to Mathworld back in 2005),

$$(ax^2-v_1xy+bwy^2)^3 + (bx^2+v_1xy+awy^2)^3 + (cx^2+v_2xy+dwy^2)^3 + (dx^2-v_2xy+cwy^2)^3 = \color{blue}{(a^3+b^3+c^3+d^3)}(x^2+wy^2)^3\tag3$$ where, $$v_1= c^2-d^2\\ v_2= a^2-b^2\\ w= (a+b)(c+d)$$

Thus, if the $RHS$ is zero, or you find a single instance of $\color{blue}{a^3+b^3+c^3+d^3 = 0}$, then the $LHS$ yields a quadratic parameterization that guarantees an infinite more. So to answer your question, $(3)$ can be used to generate infinitely many Ramanujan-type formulas like $(1)$.

Example: The two smallest taxicab numbers are,

$$1^3+12^3=9^3+10^3\\ \color{blue}{2^3+16^3=9^3+15^3}$$

Using the second one and formula $(3)$, and after scaling the variable $y' \to y/12$ to reduce coefficient size, one gets, $$(\color{blue}2 x^2 + 12 x y - 48 y^2)^3 + (\color{blue}{16} x^2 - 12 x y - 6 y^2)^3 + (\color{blue}{-9} x^2 - 21 x y + 45 y^2)^3 + (\color{blue}{-15} x^2 + 21 x y + 27 y^2)^3 = 0$$ and you can see its "parents" in blue.

Tito Piezas III
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