-3
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I was looking for a natural power of 3 that could be written like

Binary format:

11..(N times)..11011..(M times)..11

Example: 1111110111111111111111 (...isn't a power of 3)

Or could also be written like

3^x = 2^a - 2^b - 1

(x is arbitrary, "a" and "b" are natural numbers, a = N-M-1, b = M, and the single zero in binary format is a must)

But couldn't find any, so I thought there might be some proof that there's no such numbers (altho that would contradict intuition) or maybe it can be proven that there might be such numbers?

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5
  • $\begingroup$ $3^1=2^3-2^2-1$ $\endgroup$
    – Andreas Blass
    Commented Sep 4, 2021 at 23:28
  • $\begingroup$ A bit less trivially, $3^3=2^5-2^2-1$. $\endgroup$
    – Andreas Blass
    Commented Sep 4, 2021 at 23:29
  • $\begingroup$ Thank you, I forgot to clarify that the single zero in binary format is a must $\endgroup$
    – Nya
    Commented Sep 4, 2021 at 23:29
  • $\begingroup$ Oh, 3^3 = 2^5 - 2^2 - 1 perfectly matches, I got my question answered, thank you very much $\endgroup$
    – Nya
    Commented Sep 4, 2021 at 23:34
  • $\begingroup$ Is it suppose to be $a,b>2$? $\endgroup$
    – markvs
    Commented Sep 5, 2021 at 0:00

1 Answer 1

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The equation $2^a-2^b-1=3^c$ has no integral solution with $a,b\geq 3$. Indeed, in this case the left-hand side is $\equiv 7\pmod{8}$, while the right-hand side is either $\equiv 1\pmod{8}$ or $\equiv 3\pmod{8}$.

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  • 2
    $\begingroup$ $b=1$ is clearly impossible, so we are left with $b=2$, $2^a-5=3^c$, which has (at least) the two solutions $8-5=3$ and $32-5=27$ (as noted by Andreas). Are there any more? $\endgroup$
    – Gerry Myerson
    Commented Sep 5, 2021 at 0:56
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    $\begingroup$ @GerryMyerson: Those are the only two solutions, by Bennett's theorem: personal.math.ubc.ca/~bennett/B-CJM-Pillai.pdf $\endgroup$
    – GH from MO
    Commented Sep 5, 2021 at 1:21

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