$x^2+y^3 = z^4$ for positive integers

Question

How can I solve this diophatine equation :

$$x^2+y^3=z^4$$ for $(x, y, z) \in \mathbb{Z}_{>0}$

I tried to look on wolfram alpha yet it seems like there aren't any solutions...

Answer 1

$$X^{n}+Y^{n+1}=Z^{n+2}$$

Solution always can be written, for example.

$$X=(c^2-b^2)^{(n+2)}b^{(n^2+2n-1)}c^{(n+1)^2}$$

$$Y=(c^2-b^2)^{(n+1)}b^{(n-1)(n+2)}c^{n(n+1)}$$

$$Z=(c^2-b^2)^{n}b^{(n-1)(n+1)}c^{(n^2+1)}$$

If you make this change. $X^2+Y^3=Z^4$

$$p=tz(2zk^2+t)$$

$$s=tzk^2(2zk^2-t)$$

The result of such decision.

$$X=sp^3$$

$$Y=2tzk^2p^2$$

$$Z=kp^2$$

Where the number $t,z,k$ - integers and set us. You may need after you get the numbers, divided by the common divisor.

Answer 2

This is a special case of the generalized Fermat equation $$x^p+y^q=z^r$$ For $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}\le 1$ it has only finitely many coprime integer solutions, as has been proved by Darmon and Granville. For $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}>1$ however, there are infinitely many coprime nonzero integer solutions, given by a finite set of 2-parameter families, see Beukers, The Diophantine equation $Ax^p + By^q = Cz^r$, Duke Math. J. 91 (1998), 61-88. The explicit parameterizations (with proofs) can be found in Chapter 14 of Cohen's book Number Theory, Vol. II.