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: $1+511=512$.

  • $\begingroup$ Are you allowing negative numbers among the solutions? (It's reasonably clear you don't consider $(1,0,1)$ to be a solution!) $\endgroup$ – Barry Cipra Oct 22 '14 at 14:15
  • $\begingroup$ No, it's just about positive numbers. $\endgroup$ – Pim Oct 22 '14 at 14:16
  • $\begingroup$ I edited the title to make that clear $\endgroup$ – Barry Cipra Oct 22 '14 at 14:19
  • $\begingroup$ 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). $\endgroup$ – Bart Michels Oct 25 '14 at 21:13
  • 3
    $\begingroup$ Take that $1+511=512$ and multiply through by $511^k$ for a clever choice of $k$. $\endgroup$ – Gerry Myerson Dec 6 '14 at 5:53

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.