B Difference between powers of numbers equaling 1

[e2m2a]

TL;DR Summary

The Difference of Powers of Numbers What conditions must exist for the difference to be one?

If you take the cube of 2, you get 8. If you take the square of 3, you get 9. That is, 3 squared minus 2 cube equals 1. Are there any other examples of this? Where the difference between two powered numbers is equal to one? And is there some kind of theorem that says when this is possible? Can't find anything on the subject on the internet.

 

[fresh_42]

So you are asking for quadruple $(a,b,n,m)$ such that $a^n-b^m=1$. Essential information is missing, namely, where we are allowed to select the numbers from, especially whether it is $\mathbb{N}^4,\mathbb{Z}^4$ or $\mathbb{Z}^2\times \mathbb{N}^2.$ And do we require $n,m > 1$?

 

[e2m2a]

Not sure about the symbols but let me keep it simple. First, I am interested in the case where a and b are integers and n and m are integers. Then I am interested the case where a and b are integers and n and m can be any rational number.

 

[Office_Shredder]

Fresh alluded to in his post an infinitude of trivial solutions of the form $a^b - (a^b-1)^1=1$. So you probably really want to know if there are more examples where both powers are forced to be larger than 1 (in the integer case at least, I guess the rational powers could be smaller than 1).

 

[PeroK]

Summary:: The Difference of Powers of Numbers What conditions must exist for the difference to be one?

If you take the cube of 2, you get 8. If you take the square of 3, you get 9. That is, 3 squared minus 2 cube equals 1. Are there any other examples of this? Where the difference between two powered numbers is equal to one? And is there some kind of theorem that says when this is possible? Can't find anything on the subject on the internet.

I can't find any others! Are there any?

 

[PeroK]

Not sure about the symbols but let me keep it simple. First, I am interested in the case where a and b are integers and n and m are integers. Then I am interested the case where a and b are integers and n and m can be any rational number.

It only makes sense to look for whole number solutions. E.g., for any $a^n$, we have $b = (a^n+1)^2$ and $m = \frac 1 2$.

 

[PeroK]

Can't find anything on the subject on the internet.

You didn't look very hard

https://www.mathpages.com/home/kmath104/kmath104.htm

 

[PeroK]

https://en.wikipedia.org/wiki/Catalan's_conjecture

 

[e2m2a]

You didn't look very hard

https://www.mathpages.com/home/kmath104/kmath104.htm

Wow! Thanks. This is what I was looking for.

 

[e2m2a]

Ok. I need to be more specific. The Catalan conjecture deals with consecutive numbers and specifically with 8 and 9. I am interested in the following: a^n - 2^m = 1. Where n and m are rational numbers and a is an integer. Any research done on this case? Any theorems?

 

[PeroK]

Ok. I need to be more specific. The Catalan conjecture deals with consecutive numbers and specifically with 8 and 9. I am interested in the following: a^n - 2^m = 1. Where n and m are rational numbers and a is an integer. Any research done on this case? Any theorems?

Tha Catalan conjecture is for all integers. See post #6 re rational solutions.

 

[mfb]

Ok. I need to be more specific. The Catalan conjecture deals with consecutive numbers and specifically with 8 and 9. I am interested in the following: a^n - 2^m = 1. Where n and m are rational numbers and a is an integer. Any research done on this case? Any theorems?

This is just a special case of the conjecture, where you force one of the powers to be a power of 2. Doesn't change the result. If there is just one solution of the more general question then there is only one solution to this more restrictive problem, the one you posted already.
Non-integer n,m and adding zero and one just adds possible trivial cases (like 2^1 - 2^0 = 1 or 9^1/2 - 2^1 = 1 and so on).