From: Tenie Remmel Newsgroups: sci.math Subject: Sums of cubes, 4th-powers, etc. Date: Sat, 14 Dec 1996 12:04:54 -0800 It has been proved that every number is a sum of 9 cubes, and 19 4th-powers. But what about 5th-powers, I think it's 37, but I don't have a proof (however, 223 needs 37 5th-powers, so it can't be less). And not all numbers need 9 cubes. Actually, I think every sufficiently large number is a sum of: 5 cubes, 16 4th-powers (maybe 15), 10 5th-powers. In fact, 1290740 seems to be the biggest number that needs 6 cubes. This I tested to 200 million. Similarly, all numbers above 13792 seem to be a sum of 16 4th-powers, and all numbers above 51033617 a sum of 10 5th-powers. 1) Are there any real (i.e. proven) results here? 2) Why do you need more 4th-powers than 5th-powers?! This is very strange. 3) Any lower bounds? It can't be N-1 Nth-powers, since there are only a maximum of (K^(1/N))^(N-1) sums of N-1 Nth powers below K and this being K^((N-1)/N) is always less than K. P.S. Testing these to 200 million isn't hard, it only takes 1-4 hours with a Pentium-90. Runs out of memory if you go any higher, so I couldn't go to a billion. -- Tenie Remmel tjr19@mail.idt.net ============================================================================== From: saouter@irisa.fr (Saouter Yannick) Newsgroups: sci.math Subject: Re: Sums of cubes, 4th-powers, etc. Date: 15 Dec 1996 14:55:54 GMT In article <32B30866.FDD@mail.idt.net>, Tenie Remmel writes: > It has been proved that every number is a sum of 9 cubes, > and 19 4th-powers. > > But what about 5th-powers, I think it's 37, but I don't have > a proof (however, 223 needs 37 5th-powers, so it can't be less). > > And not all numbers need 9 cubes. Actually, I think every > sufficiently large number is a sum of: > > 5 cubes, > 16 4th-powers (maybe 15), > 10 5th-powers. > > In fact, 1290740 seems to be the biggest number that needs 6 cubes. > This I tested to 200 million. Similarly, all numbers above 13792 > seem to be a sum of 16 4th-powers, and all numbers above 51033617 > a sum of 10 5th-powers. > > 1) Are there any real (i.e. proven) results here? Let $g(k)$ be the least number of positive k-th powers necessary to write any positive numbers. Then the expression of $g(k)$ is exactly known since 1944 as far as I know. > > 2) Why do you need more 4th-powers than 5th-powers?! > This is very strange. 16 biquadrates are needed and 37 quintic powers are needed. There is however open problems e.g. : - Show that any positive number but a finite number ones can be expressed as sum of 4 cubes. - Show that any integer (positive or not) can be expressed as the sum of 3 cubes, positive or not.