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The solution of the question A cubic diophantine Equation depends upon being able to find non-zero integer solutions of the equations $ax^3+by^3=1\text{ or }3$, where $a$ and $b$ are given positive integers.

I am aware of Selmer's work on equations of the form $ax^3+by^3+cz^3=0$ but I imagine that there might be much stronger results when we in effect know $z=1$. Does anyone know of such results or see how one might proceed?

  • You can obtain some solutions for x by setting b=0 and assuming a is non-zero. https://www.wolframalpha.com/input/?i=ax%5E3%2Bby%5E3%3D1 – Alessio K Dec 24 '19 at 12:35
  • Thanks - but integer solutions are required with $a,b$ positive. –  Dec 24 '19 at 12:38
  • Equation $aX+bY=1$ has infinite solutions of form $X=b t+X_0$ and $Y=-a t+ Y_0$ , among them $X=x^3$ and $Y=y^3$ may be found. This depend on values of a and b. – sirous Dec 25 '19 at 15:02
  • For example $X=3t+8$ and $Y=-47t-125$ give infinite solutions, one of them is $x=2$ and $y=-5$. There may be more solutions. – sirous Dec 25 '19 at 15:28

1 Answers1

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Equations $F(x,y)= k$ where $F$ is homogenous can be solved via factorizations and solutions to Thue equation.

If factorization is possible, say: $$ G(x,y)H(x,y) = k $$ then the approach is to split into cases and solve for all possible divisors $r$ of $k$, i.e. set $G(x,y) = k/r$ and $H(x,y)=r$.

After factorization degree $\leq 2$ equations are the usual Pell equations and degree $\geq 3$ is solved as Thue equations. In particular Thue equations only have finitely many solutions.

Unfortunately the only easy way I know for solving it is via existing computer programs.

A PARI/GP example: $$5x^3-2y^3=7$$ Since $5x^3-2y^3$ is irreducible so this is already a Thue equation. This PARI/GP command can solve it:

thue(thueinit(5*x^3 - 2,1),7)

Note that the second "1" indicates no assumptions (GRH). You can try it here. This will return

%1 = [[1, -1], [3, 4]]

and indeed $(X,Y)=(1,-1),(3,4)$ are both solutions. Other softwares like MAGMA and Sagemath can solve it too.

Yong Hao Ng
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  • Great - thanks very much for this clear explanation. I've meant to look up Thue's Theorem for a while and not got round to it until now. –  Dec 27 '19 at 13:20
  • I learned it via Theorem 3.35 of Silverman's "Arithmetic of Dynamical Systems", which is pretty short and quite simple, but it uses the really difficult Roth's Theorem which I still don't understand. Maybe there's a more comprehensive version out there. – Yong Hao Ng Dec 27 '19 at 13:45
  • Thanks for this reference. I was thinking of working through the proof given in Dickson, Introduction to the theory of numbers. On a quick skim, the proof there seems elementary and self-contained but very, very long. –  Dec 27 '19 at 14:41