# How to solve $4x^3-3z^2=y^6$ in positive integers?

Solve in positive integers
$$4x^3-3z^2=y^6$$

We are given that $\gcd (x,y) = \gcd (y,z) = \gcd (x,z) = \gcd (x,y,z) = 1$.

I do not have the slightest idea how to even begin this question. Seeing the L.H.S. , all I can think is about the cosine triple angle formula but I don't think that'll help here ☺

Any help will be appreciated.

Thanks.

• Where did you come across this equation?
– quid
Feb 9 '15 at 18:20
• For what it's worth, this stumps Wolfram Alpha. Feb 9 '15 at 21:36

$x = y = z = 1$ is a solution and it is the only positive integer solution.

This can be proved using either Elliptic curve or Fermat's Last Theorem.

Method 1 - Elliptic curve

Let $X= \frac{12x}{y^2}$ and $Y = \frac{36z}{y^3}$. Whenever $y \ne 0$, we have

$$y^6 - 4x^3 + 3z^2 = 0\quad\iff\quad Y^2 = X^3 - 432$$ The RHS is the equation of an elliptic curve. Let me call it $\mathcal{E}$. It is clear for every integer solution of LHS, there is a corresponding rational point $(X,Y)$ on $\mathcal{E}$.

To extract information about $\mathcal{E}$, we feed following commands to the online CAS MAGMA:

Q<x> := PolynomialRing(Rationals());
E00:=EllipticCurve(x^3-432);
E00;
MordellWeilShaInformation(E00);
Generators(E00);


We find $\mathcal{E} : Y^2 = X^3 - 432$ has following properties:

• The rank of $\mathcal{E}$ is $0$ - this means $\mathcal{E}$ has at most finitely many rational points.
• The torsion subgroup of $\mathcal{E}$ is isomorphic to $\mathbb{Z}/3\mathbb{Z}$ - this means $\mathcal{E}$ has two non-trivial rational points.
• $(12,36)$ is a generator of the torsion subgroup - this implies the two non-trivial rational points of $\mathcal{E}$ are $$(X,Y) = (12,36) \;\text{ or } (12,-36)$$

Translate this back in terms of $x,y,z$, this means the integral solutions of

$$y^6 - 4x^3 + 3z^2 = 0,\quad y \ne 0$$ all has the form

$$x = y^2\quad\text{ and }\quad z = \pm y^3$$

If one further impose the condition that $x,y,z$ are pairwise co-prime positive integers, this leaves us one and only one possibilities. Namely, $x = y = z = 1$.

Method 2 - Fermat Last Theorem

Let $x,y,z$ be any pairwise co-prime positive integer solution to the equation $$y^6 - 4x^3 + 3z^2 = 0$$ Let $A = y^3+z$, $B = y^3-z$ and $C = 2xy$, we have $$A^3 + B^3 = (y^3 + z)^3 + (y^3 - z)^3 = 2y^3(y^6 + 3z^2) = (2xy)^3 = C^3$$ In order not to contradict with Fermat's last Theorem, we need $B = 0$. This means $y^3 = z$. Together with the requirement $\gcd(y,z) = 1$, we find $y = z = 1$ and so does $x$.

• Is there a simple explanation for why this problem is more tractable than Fermat's Last Theorem? Feb 12 '15 at 17:17
• @AlexR. FLT covers polynomial with arbitrary large degree. We don't have any general tool to deal with them. This one can be reduced to an elliptic curve in an elementary manner. In fact, that is true for all equations of the form $ax^3 + by^6 + cz^2 = 0$. Since rational points on an elliptic curve forms a group, we simply have more tools to use! Feb 12 '15 at 18:08
• Just about every problem is more tractable than FLT. In this case, of course, we only need FLT for n=3, which has a proof that even I can understand (the one in Hardy and Wright which I read in high school). Jun 17 '15 at 2:47