The discussion revolves around the expression \( a^n - b^n = 1 \) for positive integers \( a \), \( b \), and \( n \), specifically questioning whether this expression can ever be true when \( a > b \). Participants explore the historical context of this problem and seek to understand if there is a formal proof, particularly in relation to the Catalan conjecture.
Participants generally agree that \( a^n - b^n = 1 \) is not true for \( n > 1 \), but the discussion includes various approaches and reasoning, indicating that multiple views and interpretations remain present.
The discussion does not resolve the formal proof aspect, nor does it clarify all assumptions involved in the reasoning presented. The historical context of the tools used in the proofs is also not fully explored.
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: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?
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.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: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.
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.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.