From: mert0236@sable.ox.ac.uk (Thomas Womack) Newsgroups: sci.math Subject: Re: factorization to 19 numbers? Date: 2 Dec 1996 13:17:59 GMT Bob Silverman (numtheor@tiac.net) wrote: : Mikko haapanen wrote: : Every sufficiently large integer is the sum of at most 16 4th powers. : I am not sure if the exact bound is known, i.e. the largest integer that : is not the sum of 16 4th powers. I played with this problem for cubes, a little. From numerical evidence, it looks like every sufficiently large number is the sum of 5 cubes; there are 2 numbers (23,239) which are sums of 9 cubes, 15 which are sums of 8, 121 (largest <2^24 is 8042) which are sums of 7, and 3922 (below 2^24, but there are none between 2^21 and 2^24) which are sums of 6. Has anyone proved these, or any other bounds. I know that you can't have every number the sum of 3 cubes [work mod 9, a^3 = 0,1 or -1], but I thought the best proved result was at most 7 cubes. -- Tom Dort, wo man Buecher verbrennt, verbrennt man am Ende auch Menschen (Heinrich Heine) ============================================================================== From: cohen@math.u-bordeaux.fr (Henri Cohen) Newsgroups: sci.math Subject: Re: factorization to 19 numbers? Date: 3 Dec 1996 17:48:12 GMT The fact that g(4)=19, i.e. that EVERY (not only sufficiently large) positive integer is the sum of 19 fourth powers was proved 12 years ago by Balasubramanian, Deshouillers and Dress. From memory, it involved first proving that every sufficiently large (with an EXPLICIT bound around 10^350) integer was sum of 19 fourth powers, using very delicate estimates of integrals (if you don't do it delicately, you get something more like 10^2000 or more). Then, using quite a large amount of computer calculation done cleverly (of course it is out of the question to test 10^350 numbers!) they were able to prove that all numbers up to 10^360 or so are sum of 19 fourth powers. The result g(5)=37 is due to Chen, and all other g(k) are (theoretically) known (g(k)=2^k-2+floor((3/2)^k) if a certain condition, almost certainly always fulfilled, is true), see Hardy and Wright. For G(3), one still does not know how to improve the result G(3)<=7 which is more than half a century old, although extensive numerical experiments seem to show that G(3)=4, the largest number not a sum of 4 cubes being N=7373170279850, a result due to B. Landreau 1996 (note: this is based and MUCH MORE than a simple computation up to some limit, but of course is a conjectural result). Henri Cohen ============================================================================== From: numtheor@tiac.net (Bob Silverman) Newsgroups: sci.math Subject: Re: factorization to 19 numbers? Date: Tue, 03 Dec 1996 03:18:16 GMT mert0236@sable.ox.ac.uk (Thomas Womack) wrote: >Bob Silverman (numtheor@tiac.net) wrote: >: Mikko haapanen wrote: >: Every sufficiently large integer is the sum of at most 16 4th powers. >: I am not sure if the exact bound is known, i.e. the largest integer that >: is not the sum of 16 4th powers. >I played with this problem for cubes, a little. From numerical evidence, >it looks like every sufficiently large number is the sum of 5 cubes; >there are 2 numbers (23,239) which are sums of 9 cubes, 15 which are >sums of 8, 121 (largest <2^24 is 8042) which are sums of 7, and >3922 (below 2^24, but there are none between 2^21 and 2^24) which are >sums of 6. >Has anyone proved these, or any other bounds. I know that you can't have >every number the sum of 3 cubes [work mod 9, a^3 = 0,1 or -1], but I >thought the best proved result was at most 7 cubes. The Waring problem for cubes is still open. Recent computations (1982) suggest G(3) = 4, not 5. You probably didn't look high enough. All integers except 23 and 239 are the sum of 8 cubes. (1939, Dickson) ==============================================================================