Solutions of $ax^3+by^3=1$

Question

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?

Answer 1

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.

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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.