Prove that there exists infinite solutions for this Diophantine equation

Question

Prove that the Diophantine equation $$x^4+y^3=z^2$$ has infinitely many primitive solutions.

My work:

We can re-write the equation as $$y^3=z^2-x^4$$ or $$y^3=(z+x^2)(z-x^2)$$

Now if $|z|>x^2$ then $b$ has to be positive otherwise $b$ will be negative. How to approach after that$?$ What to do if we have to prove infinite solutions of a equation$?$

Any help is greatly appreciated.

Answer 1

Just so we are on the same page we note that "OP" requested solution for, $(x^4+y^3=z^2)$ & @Dietrich Burde" provided link for equation, $(m^2+n^3=w^4)$ which are both not the same. Anyway no harm done. There is parametric solution for the below eqn:

$x^4+y^3=z^2$

$x=p^2q(2k-3)$

$y=p^3q^2$

$z=p^4q^2(k-1)(5k-7)$

Fo, $k=3$ we have,

$735^4+8575^3=960400^2$

Answer 2

Infinitely many relatively prime solutions are listed here in section $4.1$, for the generalised Fermat equation $$ x^p+y^q=z^r $$ for the case $$ \frac{1}{p}+\frac{1}{q}+\frac{1}{r}>1, $$ the so-called spherical case, where we may assume that
$(p,q,r)=(2,2,r), (2,q,2), (2,3,3), (2,3,4), (2,4,3), (2,3,5)$. Here we have $$ (p,q,r)=(2,3,4). $$ A parametrisation for this case is given in the above article as \begin{align*} x & = 4ts(s^2 − 3t^2)(s^4 + 6t^2s^2 + 81t^4)(3s^4 + 2t^2s^2 + 3t^4),\\ y & = ±(s^2 + 3t^2)(s^4 − 18t^2s^2 + 9t^4),\\ z & = (s^4 −2t^2s^2 +9t^4)(s^4 +30t^2s^2 +9t^4) \end{align*} for $s\not\equiv t\bmod 2$ and $3\nmid s$, and also for the other cases.

Answer 3

Part (2) of my answer:

I tried editing my answer but it would not go through.

The missing info is:

$p=k^2-2$

$q=3k-4$

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