Solving a cubic diophantine equation $ax^3+bx^2=cy^3$

Question

I want to know how to solve the equation $ax^3+bx^2=cy^3$ in positive integers $x$ and $y$, assuming $a, b, c$ are positive integers and $gcd(a, b, c) = 1$. If there is not a general algorithm, I would like to solve more specific ones, such as when $a$ and/or $b$ and/or $c$ equal $1$. I have tried searching online, but could not find an algorithm to solve this.

Edit: How would the answer change if a and c are both 1, while b is a positive integer?

Answer 1

This equation gives you a (singular) genus 0 curve for any choice of $a,b,c$. Now, there is a famous theorem by Siegel which asserts that there are a finite number of integral solutions if the number of points at infinity (over $\mathbb{C}$) is 3 or larger. In you case the points at infinity are the points (in projective coordinates $[x:y:z]$) equal to $[1: \xi: 0]$, with $\xi^3=(c/a)$, which are clearly 3.

So for any given $a$, $b$ and $c$ there is only a finite number of integer solutions.

I expect there is no general formula for this solutions.

Except may be some special values of $a$, $b$ and $c$, I guess that the equation will only have between 0 and 1 solution in coprime strictly positive integers ($x>0$ and $y>0$).

The usual "method" to find all the solutions is to do a search on "small solutions", and then try to show there are no others: either because you have a theorem telling you that all the solutions must be small, or by congruence considerations, or with arguments working over some cubic field.

Edit

For the special case $a=c=1$, observe that, if $b=r^3-s^3$ for some integers $r>s>1$, there is a "trivial solution" $x=s^3$ and $y=rs^2$. This can be extended to the case $b=t(r^3-s^3)$ for some integers $r>s>0$ and $t>0$, and we get the solution $(x,y)=(s^3t,rs^2t)$.

Guess: Any solution $(x,y)$ with strictly positive integers of $y^3=x^3+bx^2$, for $b>0$ integer, verifies that, if $t=gcd(x,y,b)$, $gcd(x,y)=ts^2$ for some $s>0$, and $x=s^3t$, $y=rs^2t$ for some $r>0$. In particular, if $gcd(x,y)=1$, then $x=1$.

I tested this guess for $b$ up to 800 and "small" solutions and it seems it holds.

Answer 2

This is a special case of the generalized Fermat equation $$ ax^p+by^q=cz^r $$ for $(p,q,r)=(2,3,3)$. This case is spherical, because $$ \frac{1}{p}+\frac{1}{q}+\frac{1}{r}>1. $$ So there is either no non-trivial solution, or infinitely many solutions, which can be constructed by a finite set of polynomial parametrisations of the equation, see here:

F.Beukers, The diophantine equation $Ax^p + By^q = Cz^r$, Duke Math.J. 91(1998), 61-88.

Then you can specify the solutions satisfying $y=x$. For example, when $a=b=c=1$, then $x^2+x^3=y^3$ has only the trivial solution $x=y=0$. This fact also follows from this question:

Integer solutions of $x^3+y^3=z^2$

Answer 3

Equation, $ax^3+bx^2=cy^3$

"SI" above has given parametric solution (and is shown below):

$a (s n k^3)^3 + (ne h^3 - an s k^3) (s n k^3)^2 = (s^2e) (n k^2 h)^3$

For $(a,s,n,k,e,h)=(3,1,5,1,3,2)$ the solution is:

$(a,b,c)=(1,35,1)$

$(x,y)=(5,10)$

Another numerical solution is:

$2(6)^3+4(6)^2=9(4)^3$