Prove there are infinitely many (x, y, z) positive integers satisfying $x^5 + y^7 = z^9$

Prove there are infinitely many (x, y, z) positive integers satisfying $$x^5 + y^7 = z^9$$

I have reduced the problem to finding only one solution $$(x_0,y_0,z_0)$$ and then using the fact that there are infinitely many solutions of the form $$(x_0*k^{63},y_0*k^{45},z_0*k^{35})$$

I have tried even making a program to check for high enough numbers. I've seen methods on here of 'guessing' the form of the solutions, but for other exponents, based on modulos.

any help can suffice.

• Try $x=2^m$ and $y=2^n$ for suitable $m$ and $n$. – Lord Shark the Unknown Jan 30 at 20:31

Lets look for $$2^n$$ formula that might work:

$$(2^k)^5+(2^l)^7=(2^n)^9$$

so we know $$2^t+2^t=2^{t+1}$$

Than we look for $$5 k=7l=9n-1$$

$$k=\frac{9n-1}{5}, l=\frac{9n-1}{7}$$

we are looking for int, so you want the fractions to be integers.

It means $$7|9n-1$$ and $$5|9n-1$$ which is $$35|9n-1$$

Just from a look, you can tell $$n=4$$ will work

So $$x=2^7, y=2^5, z=2^4$$

This is a special case of the generalized Fermat equation $$x^p+y^q=z^r$$ for $$(p,q,r)=(5,7,9)$$. This case is hyperbolic, because $$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}<1.$$ For the primitive solutions, there are only finitely many, and for specific cases no non-trivial integer solutions:

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

Actually, Beal's conjecture says that there are no positive coprime integer solutions for $$(p,q,r)=(5,7,9)$$.

If we drop the assumption of $$gcd(x,y,z)=1$$ we trivially have infinitely many solutions, e.g., with powers of $$2$$:

$$(2^{7m})^5+(2^{5n})^7=(2^{k})^9,$$ for suitable $$m,n,k$$.