Difference between like powers proof

[e2m2a]

TL;DR 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?

 

[PeroK]

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.

 

[PeroK]

I was going to refer you to this thread abut a variation of the Catalan conjecture, but I see you were the OP there also!

https://www.physicsforums.com/threads/variation-of-catalan-conjecture.1008090/

 

[nomadreid]

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.

 

[PeroK]

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.

 

[e2m2a]

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.

 

[e2m2a]

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.

 

[e2m2a]

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.