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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.

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    $\begingroup$ Try $x=2^m$ and $y=2^n$ for suitable $m$ and $n$. $\endgroup$ – Lord Shark the Unknown Jan 30 at 20:31
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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 $x=2^7, y=2^5, z=2^4$

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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 $m,n,k$.

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