The equation $a^2x^3+y^3=az^3$

Question

An example of equation with a particular numerical value of $a\in\mathbb N$, I have extended to a family of equations that I present here.

Prove the equation $$a^2x^3+y^3=az^3; (x,y,z)\in\mathbb Z^3; a\in\mathbb N; xyz\ne0$$ has no solution for all value of $a\in\mathbb N$.

HINT.- For $a=b^3$ take into account Fermat's Last Theorem for the exposant 3.

Answer 1

Suppose $a\neq 1$

Suppose that $(x,y,z)$ is a solution of equation with $|x|+|y|+|z|$ is minimumm.

So $y=ay_1$. The equation is equivalent to $a^2x^3+a^3y_1^3=az^3$. So $z=az_1$ and the equation is equivalent to $ax^3+a^2y_1^3=a^3z_1^3$. So $x=ax_1$ and then $a^3x_1^3+ay_1^3=a^2z_1^3$. But this is the same $a^2x_1^3+y_1^3=az_1^3$. So $(x_1,y_1,z_1)$ is another solution and $|x_1|+|y_1|+|z_1|<|x|+|y|+|z|$, contradiction.

Answer 2

By FLT for exponent $3$, there are no solutions with $xyz\ne 0$ (non-trivial solutions) if $a$ is a perfect cube. For other $a$ we give a proof by descent (aka induction).

Let $a$ be of the form $a^{3t}a^\ast$ where $a^\ast\gt 1$ is cube-free. If our Diophantine equation has a non-trivial solution with $a$, then it has a non=trivial solution with $a^\ast$. Thus without loss of generality we may assume that there is a prime $p$ such that the highest power of $p$ that divides $a$ is Case 1: $p$ or Case 2: $p^2$.

Case 1: If there is a non-trivial solution, then there is a non-trivial solution with $|x|$ minimal.

Note that $p$ divides $y^3$, so $p$ divides $y$. Let $y=py_1$. Then $p^2$ divides the left-hand side. Since the highest power of $p$ that divides $a$ is $p$, we find that $p$ divides $z$, say $z=pz_1$. But then since $p^3$ does not divide $a^2$, we conclude that $x=px_1$ for some $x_1$. Cancelling the $p^3$ gives us that $a^2x_1^3+y_1^3=az_1^3$, contradicting the choice of $x$.

Case 2: Again we find that $p$ divides $y$. Then $p^3$ divides the left-hand side, so $p$ divides $z$. Since $p^4$ divides both $az^3$ and $a^2x^3$, it follows that $p$ divides $y_1$. Finally, since $p^5$ divides $y^3$ and $az^3$, we conclude that $p$ divides $x$, and we reach the same contradiction as in Case 1.