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

Solution 1:

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.