Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math

Topic: Can all numbers be represented by (a^n) +/- (b^m) {where n and m are >1} ?
Replies: 0  

Advanced Search

Back to Topic List Back to Topic List  
karpmage

Posts: 3
From: New Zealand
Registered: 11/17/12
Can all numbers be represented by (a^n) +/- (b^m) {where n and m are >1} ?
Posted: Nov 17, 2012 11:12 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

Can all of the natural numbers be expressed by (a^n) plus or minus (b^m)? Where n and m are equal or greater than two, and are whole numbers, and a and b are whole numbers. I've noticed that there is no obvious solution for numbers of the form (2^n-2), excluding 2. i.e. for 6, 14, 30, 62, etc... and I was wondering, if these numbers can't in fact be expressed in the way above, whether there are any other numbers that also cannot be expressed this way.

It'd be great if someone could:
a) provide a solution for numbers of the form (2^n-2) {although the number 30 may not necessarily disprove the other numbers, as the numbers could be seen as Mersenne Primes times two)
b) prove that every number can be expressed this way through theory
c) provide reasoning that the numbers (2^n-2)cannot be expressed this way

Hopefully I made myself clear. This problem's been troubling me for some time; would be great to get it off my chest.



Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.