B Difference between like powers proof

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The expression a^n - b^n = 1 is never true for positive integers a, b, and n when a > b and n > 1. A proof using binomial expansion shows that the minimum difference between a^n and b^n occurs when a = b + 1, resulting in a difference greater than 1. Historical context indicates that the tools for such proofs have existed since ancient times, although modern formulations clarify the impossibility of the expression. Observations of specific cases further confirm that the differences increase as n and b grow. Therefore, it is established that a^n - b^n = 1 cannot hold under the given conditions.
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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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e2m2a said:
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.
 
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PeroK said:
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.
 
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.
 
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PeroK said:
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.
 
PeroK said:
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.
 
PeroK said:
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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