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In his Lost Notebook, Ramanujan exhibits infinitely many integer solutions to $x^3+y^3+z^3=1$. On his webpage (http://www.math.harvard.edu/~elkies/4cubes.html), Elkies determines all rational solutions to this equation. I have two questions: does Elkies' result enable one to determine all integer solutions, and does Elkies' result at least enable one to recover Ramanujan's integer solutions?

Ramanujan's integer solutions are as follows: write $$ \frac{1+53t+9t^2}{1-82t-82t^2+t^3} = x_0 + x_1t + x_2t^2 + ... $$ $$ \frac{2-26t-12t^2}{1-82t-82t^2+t^3} = y_0 + z_1t + y_2t^2 + ... $$ $$ \frac{-2-8t+10t^2}{1-82t-82t^2+t^3} = z_0 + z_1t + z_2t^2 + ... $$ Then, for every $n$, it is true that $$ x_n^3 + y_n^3 + z_n^3 = (-1)^n. $$ (For a proof, see J.H.Han and M.D.Hirschhorn, "Another look at an amazing identity of Ramanujan", Math.Mag. 79 (2006), 302-304, or alternately either of two earlier papers by Hirschhorn cited in that one.)

Elkies showed that all rational solutions to $x^3+y^3+z^3=1$ can be written as $(x,y,z)=(\frac AD,\frac BD,\frac CD)$ where there are integers $r,s,t$ for which $$ A=-(s+r)t^2 + (s^2+2r^2)t - s^3 + rs^2 - 2r^2s - r^3 $$ $$ B = t^3-(s+r)t^2+(s^2+2r^2)t+rs^2-2r^2s+r^3 $$ $$ C = -t^3+(s+r)t^2-(s^2+2r^2)t+2rs^2-r^2s+2r^3 $$ $$ D=(2r-s)t^2+(s^2-r^2)t-s^3+rs^2-2r^2s+2r^3. $$

Does anyone have an idea of how to connect these two results?

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I am not an expert here, so please take this as a remark. But I think, Ramanujan's solution appears to hinge on special circumstances of the solution $$ (x^2 +7xy−9y^2)^3 +(2x^2 −4xy+12y^2)^3 = (2x^2 +10y^2)^3 +(x^2 −9xy−y^2)^3, $$ which is claimed in http://thales.math.uqam.ca/~rowland/papers/Known_families_of_integer_solutions_of_x%5E3+y%5E3+z%5E3=n.pdf. On the site sites.google.com/site/tpiezas/010, the following identity is also attributed to Ramanujan: $$ (3x^2+5xy-5y^2)^3 + (4x^2-4xy+6y^2)^3 + (5x^2-5xy-3y^2)^3 = (6x^2-4xy+4y^2)^3, $$

Edit: However, this cannot give solutions for $x^3+y^3+z^3=1$, see Michael's comment. Piezas also has found a way to generate other Ramanujan-like families of solutions. See also: http://math.stackexchange.com/questions/381111/generalizing-ramanujans-sum-of-cubes-identity

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Thanks for the interesting references! But I don't think the identity you gave is relevant to Ramanujan's integer solutions of $u^3+v^3+w^3=1$. For, suppose one could produce integer solutions by dividing both sides of your equation by the right side. By scaling $x$ and $y$, we may assume they are coprime integers, and we need $6x^2-4xy+4y^2$ to divide each of the three quadratic polynomials on the left side. But the value of $6x^2-4xy+4y^2$ is even, while the value of $3x^2+5xy-5y^2$ is odd, contradiction. So this doesn't yield any integer solutions. –  Michael Zieve Jun 28 '13 at 22:04
    
@Michael: sorry, I confused the two identities. For the other identity, I have not yet checked how the correspondence is ($x=y=1$ gives $1=10^3+9^3-12^3$). –  Dietrich Burde Jun 29 '13 at 12:42
    
thanks! Actually, I see now that this is in the Han-Hirschhorn paper cited in my question: if $h_{n+2}=9h_{n+1}+h_n$ where $h_0=0$ and $h_1=1$, then evaluating the first identity you mentioned at $x=h_{n+1}$ and $y=h_n$ yields $x_n^3+y_n^3=(-z_n)^3+(-1)^n$ where $x_n,y_n,z_n$ are as in my question. In a sense, this explains how Ramanujan might have discovered his result, since he knew the identity you mentioned. Also, given $x,y,z$ satisfying $x^3+y^3+z^3=1$, Elkies shows that these come from his formulas with $r=yz+x$ and $s=x-y+xz+1+z+z^2$ and $t=y^2-xy-y+x^2+x+xz$. <continued...> –  Michael Zieve Jun 29 '13 at 22:27
    
<...cont> But in another sense, this approach says that one could come up with Ramanujan's solutions if one could guess the first identity you mentioned. My question is whether, knowing Elkies' formulas, there's an easy way to come up with Ramanujan's solutions without assuming any other formulas of Ramanujan's. Better yet, is there a characterization of Ramanujan's solutions as all solutions satisfying certain properties? And is it possible to determine all integral solutions to $x^3+y^3+z^3=1$? –  Michael Zieve Jun 29 '13 at 22:46
    
Oops -- I permuted the variables in Elkies' formulas. His result says that, if rational $x,y,z$ satisfy $x^3+y^3+z^3=1$ and $z\ne 1$, then $x,y,z$ are given by his formulas with $r=xy+z$ and $s=(y-x)(x+1)-(xz+1)$ and $t=(y-z)(x+z)+y-y^2$. –  Michael Zieve Jun 30 '13 at 17:54
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I am not saying anything you must not already know, but you can solve the equations in $(r,s,t)$ for the Ramanujan solutions of small $n$.

$(1,2,-2)\rightarrow (0,-1,2)$

$(135,138,-172)\rightarrow (-238,-299,388)$

$(11161,11468,-14258)\rightarrow (-6510,-8269,10742)$

$(926271,951690,-1183258)\rightarrow (-1620976,-2058689,2674342)$

$(76869289,78978818,-98196140)\rightarrow (-44840418,-56948791,73979396)$

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Thanks! Not surprisingly, the $(r,s,t)$ you listed agree, up to multiplying all three of $r,s,t$ by a common nonzero rational number, with the values for $r,s,t$ produced by applying Elkies' formulas to the given $(x,y,z)$. Namely, $r=xy+z$ and $s=(y-x)(x+1)-(xz+1)$ and $t=(y-z)(x+z)+y-y^2$. –  Michael Zieve Jun 30 '13 at 18:05
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