How to solve this Diophantine equation?

Question

Can anyone say how one can find solutions to the Diophantine equation $$x^3+y^4=z^2$$ in General? Only a few triples of numbers have been found, and most likely this equation has infinitely many solutions.

Examples of triples: $(6,5,29),(2,1,3),(9,6,45)$...

Answer 1

This is a case of the generalized Fermat equation $$ x^p+y^q=z^r. $$ For $(p,q,r)=(3,4,2)$ we have $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}>1$, which is the spherical case. Here we have infinitely many integer solutions for this triple. The solutions are given by a finite set of polynomial parametrisations of the equation, see the following paper:

F. Beukers, The diophantine equation $Ax^p + By^q = Cz^r$, Duke Math.J. 91(1998), 61-88.

Further Reference: The generalized Fermat equation.

Answer 2

Here is one simple parameterization. We have,

$$x^4 +(y^2-1)^3 = (y^3+3y)^2$$

given the Pell equation $x^2-3y^2 =1$.

Answer 3

Above equation shown below has parameterization:

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

The below parameterization has no restriction such as the

Pell equation condition demonstrated by Tito Piezas.

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

$y=(p)(q)^2(k-1)$

$z=(p)^2(q)^4(2k-3)$

where, $p=(k-2)$ and $q=(k^2-2)$

For $k=3$ we get : $(-343)^3+(98)^4=(7203)^2$

Answer 4

"OP" asked for parametric solution for $(x,y,z)=(6,5,29)$ in $x^3+y^4=z^2$

Solution is:

$x=3p^3(8k^2-40k+50)$

$y=p^2(20k^2-104k+135)$

$z=p^4(2k-5)^2(116k^2-540k+621)$

Where, $p=(4k^2-27)$

For $k=(13/5)$, we get after removing common factors:

$6^3+5^4=29^2$

Answer 5

"OP" enquired about integer coefficent's for the parametric

solution for the equation $(x^2+y^4=z^2)$. "OP" just needs

to substitute $k=(m/n)$ in the parametrization & the resulting

solution after removing common factors is given below.

$x=6(u^3)(v^2)$

$y=(u^2)(v)(10m-27n)$

$z=(u^4)(v^2)(116m^2-540mn+621n^2)$

And $u=(4m^2-27n^2)$ & $v=(2m-5n)$

For $(m,n)=(13,5)$ we get:

$6^3+5^4=29^2$