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