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

Question

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.

Answer 1

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

Answer 2

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 $m,n,k$.