Positive integer solutions to $x^4+y^7=z^9$

Question

A while ago, a maths teacher gave me this problem: find solutions to $x^4+y^7=z^9$ with $x,y,z>0$. I found $(2^{56})^4+(2^{32})^7=(2^{25})^9$. In general, if $k=8+9l$ then $(2^{7k},2^{4k},2^{\frac{28k+1}{9}})$ is a solution. Then the teacher asked me two questions: are there any other solutions, and are there any other solutions where at least one of $(x,y,z)$ is odd.

My progress for the second question: it's easy to see that if at least one is odd, exactly two are odd. If $x$ and $z$ are odd, $1+0\equiv(2k+1)^9 (\text{mod }16)$, and similar equations exist for $(x,y)$ or $(y,z)$ are odd. Also, I've bruteforced this equation for $y\leq15000$ and $z<y$, but I wasn't able to find any solutions.

Any thoughts?

EDIT: of course, for any number n, we can find solutions such that $x,y,z\equiv 0 \text{ (mod n)}$, by multiplying $x^4+y^7=z^9$ with $n^{\text{lcm}(4,7,9)}$. Also, my teacher gave the hint about the question about odd solutions: $1+511=512$.

Are you allowing negative numbers among the solutions? (It's reasonably clear you don't consider $(1,0,1)$ to be a solution!) — Barry Cipra, Oct 22 '14 at 14:15
What makes this question likely to be hard is the fact that there are solutions, meaning that considering the equation modulo some numbers can never exclude further solutions, unless one starts to investigate deeper arithmetic information about the numbers (which I don't expect given the complexity of the equation). — Bart Michels, Oct 25 '14 at 21:13
$1/4 + 1/7 + 1/9 < 1$. In their 1994 paper, Darmon and Granville has shown that if $1/p + 1/q + 1/r < 1$ then for $A, B, C$ nonzero integers, $Ax^p + Bx^q = Cx^z$ has finitely many primitive solutions. So the equation has only finitely many primitive solution as a consequence. Here's a heuristic reason for this. — Balarka Sen, Dec 03 '14 at 18:31
Take that $1+511=512$ and multiply through by $511^k$ for a clever choice of $k$. — Gerry Myerson, Dec 06 '14 at 05:53

score 12 · Accepted Answer · edited Jun 04 '20 at 07:35

Note that $(x^4:y^7:z^9)$ is the simple ratio $(48:1:49)$. I basically started from that and looked for $N$ such that $48N,N,49N$ are respectively 4th, 7th, and 9th powers. This is basically what Gerry Myerson suggested in spelling out your teacher's hint, except that $48+1=49$ yields a somewhat simpler solution than $1+511=512$. (What makes these ratios work is that $48/24$ and $512/29$ are odd integers.)

Thus, a solution with $y,z$ odd is:

$$ (x,y,z) = (2 \! \cdot \! 3^{16} \! \cdot \! 7^{49}, 3^9 \! \cdot \! 7^{28}, 3^7 \! \cdot \! 7^{22}) $$

There are similar solutions with $x,z$ or $x,y$ odd.

Positive integer solutions to $x^4+y^7=z^9$

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