# How to solve this Diophantine equation?

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

## 5 Answers

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.

Here is one simple parameterization. We have,

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

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

• @AlexD $r=2$ in this question. – Dietrich Burde Dec 29 '18 at 23:34
• Thanks, @DietrichBurde. Let me try again: A solution to the problem here is a tuple of 3 numbers. But the parameterization here only relates $x$ and $y$. What about $z$? – Alex D Dec 30 '18 at 11:21
• @AlexD We have $X^4+Y^3=Z^2$ and $X$,$Y$,$Z$ depend on $x$ and $y$ only. So no $z$ in here, only two parameters $x,y$ here for an equation in three variables $X,Y,Z$. – Dietrich Burde Dec 30 '18 at 11:47
• Well, @DietrichBurde, you can find out more, but how can you find a general formula (solve this equation in general form), expressed in arbitrary numbers (for example) m, n, k? For example, find the top three numbers (6, 5, 29)? – Yan Dashkow Dec 30 '18 at 11:55
• @YanDashkow We find them using one of the parametrizations, see Beukers article. – Dietrich Burde Dec 30 '18 at 13:31

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 : $$(-343)^3+(98)^4=(7203)^2$$

• I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)? – Yan Dashkow Dec 30 '18 at 10:55

"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:

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

• Wait, why can't k be whole? – Yan Dashkow Dec 30 '18 at 19:30

"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:

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