Prove that the equation $2^a - 2^b - 1=3^c$ has no integral solution in integerswith $a,b,c\geq 2$b\geq 3$

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edited Sep 5, 2021 at 0:16

GH from MO

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Any proof why a power of 3 can/can't be equal to (2^a Prove that the equation $2^a - 2^b - 1)1=3^c$ has no solution in integers $a, given a and b are some natural numbers higher than 2?b,c\geq 2$

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edited Sep 4, 2021 at 23:30

Nya

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Any proof why a power of 3 can/can't be equal to (2^a - 2^b - 1), given a and b are some natural numbers higher than 2?

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-b >= 2 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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edited Sep 4, 2021 at 23:28

Nya

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asked Sep 4, 2021 at 23:24

Nya

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Prove that the equation $2^a - 2^b - 1=3^c$ has no integral solution in integerswith $a,b,c\geq 2$b\geq 3$

Any proof why a power of 3 can/can't be equal to (2^a Prove that the equation $2^a - 2^b - 1)1=3^c$ has no solution in integers $a, given a and b are some natural numbers higher than 2?b,c\geq 2$

Any proof why a power of 3 can/can't be equal to (2^a - 2^b - 1), given a and b are some natural numbers higher than 2?

Any proof why a power of 3 can/can't be equal to (2^a - 2^b - 1), given a and b are some natural numbers?

Any proof why a power of 3 can/can't be equal to (2^a - 2^b - 1), given a and b are some natural numbers higher than 2?