Diophantine equation of sixth degree [closed]

Question

Show that the following equation admits infinitely many solutions:

$$x^3 + 2 y^6 - 2 z^6 = 1,\qquad \gcd(x,y)=\gcd(x,z)=\gcd(y,z)=1.$$

For example, $(79,5,8)$ is a solution .

Answer 1

Let $F_k$ be the $k$-th Fibonacci number. Cassini's identity is $F_{k-1}F_{k+1}-F_k^2=(-1)^k$, hence $F_{k-1}^2+F_{k-1}F_k-F_k^2=(-1)^k$. Raising this to the third power and reordering terms yields \begin{equation} 2(F_{k-1}^6-F_k^6)+(F_k^2+F_{k-1}F_k-F_{k-1}^2)^3=(-1)^k, \end{equation} which explains infinitely many pairwise coprime integer solutions.

See also sequence A337929 at the OEIS.

Answer 2

There is an Identity shown below:

$(a^2+ab-b^2)^3+(a^2-ab-b^2)^3=2(a^6-b^6)$ ---- (1)

We are given:

$x^3 + 2 y^6 - 2 z^6 = 1$

And we get "OP" equation by taking:

where,

$x=(b^2-ab-a^2)$

$y=a$

$z=b$

& we take, $(a^2-ab-b^2)=1$

For, $(a,b)=(5,3)$ we get $(x,y,z)=(-31,5,3)$

For, $(a,b)=(-2,3)$ we get $(x,y,z)=(11,-2,3)$

For, $(a,b)=(5,-8)$ we get $(x,y,z)=(79,5,-8)$

(The third solution is the one suggested by "OP").