Another method of proving $x^3+y^3=z^3$ has no integral solutions?

Question

The equation $$ax^2+by^2=z^3$$ has the following parametrization: $$y=q(3ap^2-bq^2)$$ $$x=p(ap^2-3bq^2)$$ $$z=ap^2+bq^2$$ can we deduct from that the Diophantine equation $$x^3+y^3=z^3$$ has no nontrivial solutions by choosing $x=a$ and $y=b$?

Answer 1

That is easy to prove. The proof is rather long but simple.Please disregard any typos.

Case 1: ($p=-1$ and $q=1$)

\begin{eqnarray} x &=& -x (x (-1)^2 - 3 y (1)^2) \\ 2x &=& 3 y \\ \end{eqnarray} Since $\gcd(x,y)=1$ then $x=y=z=0$

\begin{eqnarray} y &=& 1 (3x (-1)^2 - y (1)^2) \\ 2y &=& 3 x \\ \end{eqnarray}which implies as well $$x=y=z=0$$

Case 2: ($p \neq -1$ and $q \neq 1$)

In this case we have: \begin{eqnarray} x &=& \dfrac{3py q^2}{p^3+1 } \\ y &=& \frac{ 3qxp^2}{q^3-1}\\ \end{eqnarray}1) $ x \equiv 0 \pmod 3 $

$\gcd(x,y)=\gcd(x,q)=1$, therefore we can only have $$x=3p$$ $$yq^2=p^3+1$$ Similarly, since $3|x$ and $\gcd(y,x)=\gcd(y,p)=1$, then we must have:$$y=q$$ $$3xp^2=q^3-1$$ By replacing $x$ and $y$ by their respective values,we obtain:

$$ q^3 =p^3+1 $$ $$9p^3=q^3-1$$ Which yields:$$8p^3=0$$ Then, $p=0$ and $q=1$, which contradicts the case we are dealing with $q \neq 1$

2) $ y \equiv 0 \pmod 3 $

$\gcd(x,y)=\gcd(x,q)=1$, therefore we can only have $$x=p$$ $$3yq^2=p^3+1$$ Similarly, since $3|y$ and $\gcd(y,x)=\gcd(y,p)=1$, then we must have:$$y=3q$$ $$xp^2=q^3-1$$ By replacing $x$ and $y$ by their respective values,we obtain:

$$ 9q^3 =p^3+1 $$ $$p^3=q^3-1$$ Which yields:$$8q^3=0$$ Then, $q=0$ and $p=-1$, which contradicts the case we are dealing with $p \neq {-1}$

3) The cases where $x=\pm 1$ or $y=\pm 1$ are easy to demonstrate. In every case, we have a product of 2 numbers which equal to $\pm 1$. Therefore, both must be each either $-1$ or $1$. QED

Answer 2

Your parametrization is almost right, you flipped the second factors for $x$, $y$.

In fact, from $$\sqrt{a}\, x + i \sqrt{b}\, y = (\sqrt{a}\, p + i \sqrt{b}\, q)^3$$

we have with \begin{eqnarray} x &=& p (a p^2 - 3 b q^2) \\ y &=& q( 3 a p^2 - b q^2) \\ z &=& ap^2 + b q^2 \end{eqnarray}

($\pm$ if you wish) the equality

$$a x^2 + b y^2 = (a p^2 + b q^2)^3= z^3$$

While this parametrization provides solutions to the equation $ax^2 + b y^2 = z^3$ it is not clear apriori that all solutions are of this form. Let's just assume furthermore $a=b=1$. Is it clear that all the solutions of the equation $x^2 + y^2 = z^3$ are of the form above, for some $p$ and $q$ ( and $a=b=1$)? In other words, if $N(x+iy)$ is a cube in $\mathbb{Z}$ then $x+iy$ is itself a cube in $\mathbb{Z}[i]$ ?( the converse, explained above, is clear) OK, knowing a bit of arithmetic for the Gaussian integers this can be proved. Is it true in general for $a$, $b$ (positive) integers? It probably has to do with the arithmetic of some quadratic fields. Maybe so.

Assuming it it so, your reasoning is :

From $x^3 + y^3 = z^3$ I get $x \cdot x^2 + y \cdot y^2 = z^3 $ and so for some $p$ and $q$ we have

\begin{eqnarray} x &=& p (x p^2 - 3 y q^2) \\ y &=& q( 3 x p^2 - y q^2) \\ z &=& x p^2 + y q^2 \end{eqnarray}

All right, I suppose you will work with this equation.. Hm...plausible. I vaguely remember seeing something like that in the book of Hardy and Wright..