Difference between like powers proof

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  • #1
e2m2a
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Is the difference between like powers never equal to 1?
This may seem like a trivial question but I don't know if there is a formal proof for this. Is the following expression never true? a^n-b^n =1, where a >b, a,b,n are positive integer numbers. Was this known since ancient times? Or is there a modern proof for this?
 

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  • #2
PeroK
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Summary: Is the difference between like powers never equal to 1?

This may seem like a trivial question but I don't know if there is a formal proof for this. Is the following expression never true? a^n-b^n =1, where a >b, a,b,n are positive integer numbers. Was this known since ancient times? Or is there a modern proof for this?
If ##a > b##, then there is a minimum difference between ##a^n## and ##b^n##. If we fix ##b##, then the minimim difference is when ##a = b+1##. And:
$$(b+1)^n - b^n = 1 + nb + \binom n 2 b^2 + \dots nb^{n-1} > 1$$Assuming ##n > 1##, of course.
 
  • #4
nomadreid
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If ##a > b##, then there is a minimum difference between ##a^n## and ##b^n##. If we fix ##b##, then the minimim difference is when ##a = b+1##. And:
$$(b+1)^n - b^n = 1 + nb + \binom n 2 b^2 + \dots nb^{n-1} > 1$$Assuming ##n > 1##, of course.
The OP also wanted to know whether the proof was modern. Since PeroK's proof is based on a binomial expansion, then the OP can look at https://en.wikipedia.org/wiki/Binomial_theorem#History , which at least shows how far back the tools for this proof existed.
 
  • #5
PeroK
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Although I posted a proof, it's clear from looking at the first few cases that ##a^n - b^n = 1## is impossible for ##n > 1##:
$$1, 4, 9, 16, \dots$$$$1, 8, 27, 64 \dots$$$$1, 16, 81, 256 \dots$$And the differences are clearly only getting larger as ##n## and ##b## increase.
 
  • #6
e2m2a
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If ##a > b##, then there is a minimum difference between ##a^n## and ##b^n##. If we fix ##b##, then the minimim difference is when ##a = b+1##. And:
$$(b+1)^n - b^n = 1 + nb + \binom n 2 b^2 + \dots nb^{n-1} > 1$$Assuming ##n > 1##, of course.
Ok, thanks for the reply.
 
  • #7
e2m2a
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Although I posted a proof, it's clear from looking at the first few cases that ##a^n - b^n = 1## is impossible for ##n > 1##:
$$1, 4, 9, 16, \dots$$$$1, 8, 27, 64 \dots$$$$1, 16, 81, 256 \dots$$And the differences are clearly only getting larger as ##n## and ##b## increase.
Thanks for the response.
 
  • #8
e2m2a
354
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Although I posted a proof, it's clear from looking at the first few cases that ##a^n - b^n = 1## is impossible for ##n > 1##:
$$1, 4, 9, 16, \dots$$$$1, 8, 27, 64 \dots$$$$1, 16, 81, 256 \dots$$And the differences are clearly only getting larger as ##n## and ##b## increase.
Thanks for answering.
 

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