Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

It's well-known that every natural number can be written as a sum of 4 squares of integers.

Has there been any recent progress about the similar problem for the cubes, 4-th powers and so on? I believe this was proven to be representable using some N — that depends on the power — and what's the story about it?

share|improve this question
add comment

4 Answers

up vote 5 down vote accepted

Arguably, the problem "what is the least g=g(k) such that every integer can be written as the sum of g k-th powers" is less interesting than the version that ignores the random stuff that's happening with a finite number of special cases.

Namely, the "real" question should be "what is the least G=G(k) such that for some N, every integer greater than N can be represented as the sum of G k-th powers".

For example, every number is the sum of 19 4th powers, but every number greater than 13,792 is actually the sum of just 16 4th powers. The "16" was known for quite some time; the verification that 13,792 is the last offender is quite recent (I found the value on Wikipedia, btw).

Evaluating G(k) is harder than evaluating g(k), and most of the actual values are still not known. I don't think there was tremendous recent progress on this front, although there certainly is progress on things like bounds, number of representations, etc.

You should look at Wooley's survey here (I haven't read it yet).

share|improve this answer
    
Indeed, 16 is more interesting, especially the probability theory says the bound could be 5. –  Ilya Nikokoshev Oct 14 '09 at 4:55
add comment

I don't know, but I attended a talk by John Conway on this sort of stuff at Berkeley last year. Unfortunately I didn't take notes, but below is the abstract from the talk.

Nearly Universal Quadric Forms

I’m rather proud of having proved the “Fifteen Theorem,” according to which one can check that a positive definite quadratic form with integral matrix represents all positive integers provided only that it represents those up to 15. I’m almost as proud of having made the “290 Conjecture,” now proved by Bhargava and Hanke, that performs the similar service for integer valued forms. Why? – because together these two theorems end more than 200 years of work on universal quadratic forms, the first theorem in that subject being Lagrange’s 1770 Four Squares Theorem.

It’s now time to try to do the same for near-universality of three-dimensional forms, which subject was begun by Legendre’s much subtler Three Squares Theorem of 1798.

share|improve this answer
add comment

This is called "Waring's problem."

share|improve this answer
add comment

Yes! This is a classical problem with a lot of history. From Wikipedia:

In number theory, Waring's problem, proposed in 1770 by Edward Waring, asks whether for every natural number k there exists an associated positive integer s such that every natural number is the sum of at most s kth powers of natural numbers (for example, every number is the sum of at most 4 squares, or 9 cubes, or 19 fourth powers, etc.). The affirmative answer, known as the Hilbert–Waring theorem, was provided by Hilbert in 1909.1 Waring's problem has its own Mathematics Subject Classification, 11P05, "Waring's problem and variants."

You might also like the 15 theorem.

share|improve this answer
    
Every number can be written as [1 first power], "4 squares, or 9 cubes, or 19 fourth powers, etc.". As mentioned above, all sufficiently large numbers can be written as 16 fourth powers. Assuming that the 9 cubes is optimal in this sense (i.e., there are arbitrarily large numbers requiring all 9 cubes), we get an interesting sequence 1,4,9,16.... Is this just coincidence or can every sufficiently large natural number be expressed as k^2 kth powers? (I know NOTHING about this field, just found the question interesting...) –  Jason DeVito Nov 19 '09 at 1:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.