Closeness of natural numbers to powers

Jun 2009
83
1
Hi,
informal introduction to the problem is: does in the set of all powers exist arbitrarily long "spaces" not covered by any power?

Strict formulation of the problem is:
let \(\displaystyle p(n):=\min\{|n-a^k|; a,k \in \mathbb{N}, k>1\}\). Then the question is:
Is \(\displaystyle \sup_{n \in N}\{t(n)\}\) finite?
In other words - having set \(\displaystyle M:=\{a^k; a,k \in \mathbb{N}, k>1\}\). Does then for every natural b exist c such that \(\displaystyle M \cap \{c,c+1,c+2,..,c+b-1\}=\emptyset\)?

Possible variations of the problem are:
- we allow "a" to be only prime numbers
- we allow e.g k>2 (this is shape of numbers in Fermat's last theorem)
- instead of a^k we will examine another function...
- ...

Than you for any ideas for proving or disproving this problem (and even the variations).
 

mathbalarka

Math Team
Mar 2012
3,871
86
India, West Bengal
I cannot understand these coded language, but interested in what you want to say, so can you(or anyone) please explain it to me simply? :?
 

CRGreathouse

Forum Staff
Nov 2006
16,046
936
UTC -5
honzik said:
does in the set of all powers exist arbitrarily long "spaces" not covered by any power?
Yes. Up to N there are about N^(1/2) + N^(1/3) + N^(1/5) - N^(1/6) + N^(1/7) - N^(1/10) + ... = sqrt(N) + O(N^(1/3)) powers with exponent 2 or greater. Thus there are gaps of length N/(sqrt(N) + O(N^(1/3))) = sqrt(N) + O(N^(1/3)) with no such powers. \(\displaystyle \lim_N\sqrt N\to+\infty.\)

Short version: Essentially all powers are squares. There are sqrt(N) squares up to N, so there must be gaps of length about N/sqrt(N) which is unbounded.
 
Jun 2009
83
1
Thank you. But how can it be deduced that the mentioned sum of N^(1/i) is sqrt(N) + O(N^(1/3))?
 
Nov 2011
595
16
Humm...that's not a so easy question. I believe like CRGhouse. However what he proposed is definitely not of proof, just an instinctive reason to believe that p(n) might tends to infinity...From this type of reasoning for instance, might would conclude that because the interval between prime behaves as n/ln(n) -> inifnity so there can not be an infinite number of twin primes...which of course is not a derivation..
 

CRGreathouse

Forum Staff
Nov 2006
16,046
936
UTC -5
Dougy said:
Humm...that's not a so easy question. I believe like CRGhouse. However what he proposed is definitely not of proof, just an instinctive reason to believe that p(n) might tends to infinity...From this type of reasoning for instance, might would conclude that because the interval between prime behaves as n/ln(n) -> inifnity so there can not be an infinite number of twin primes...which of course is not a derivation..
No, what I gave was a proof. What you mention is a proof that there are intervals of length at least (x/log x)(1 + o(1)) between primes, not that all intervals have this length! It's an easy application of the pigeonhole principle.
 
Nov 2011
595
16
No, what I gave was a proof
Ok I see the point now. Thanks for the explanation! So basically since the number of powers smaller than N is infinitely smaller than N as N -> infinity, the minimum interval between any n and a power of an integer tends also to infinite. Not as complicated as I thought after all!
 

CRGreathouse

Forum Staff
Nov 2006
16,046
936
UTC -5
Er, yes. As long as you interpret "infinitely smaller" properly (ratio, not difference, goes to infinity).
 
Nov 2011
595
16
Yes, I mean the ratio of the numbers of "powers integers a^k" to N goes to zero...