From: rcw9@tutor.open.ac.uk (Richard Walker) To: rusin@math.niu.edu Date: Tue, 5 May 1998 17:22:11 +0100 Subject: Re: Representation by 5-th powers - question you wrote: >In article you >write: >>Does anyone have any information on the following problems, or know >>where information can be found please? > >Did you get any responses? Yours the first but my enquiry was only posted yesterday or day before. >I suppose you know Chen has g(5)=37, but >that doesn't appear to be one of the questions you asked. Yes, that's right. These questions are related to Waring's problem, but different. >Is it even obvious that there _is_ an integer k in >>#1 What is the least number k such that every rational can be >expressed >>as the sum of not more than k 5-th powers of rationals? The following identity shows that k does exist. w^5-(w-3)^5-(w-4)^5+(w-5)^5+w^5-(w+3)^5-(w+4)^5+(w+5)^5=2880w If then we set w=y/2880, we have a representation of y as eight rational 5-th powers. We can get an infinity of similar expressions by replacing (3,4,5) by other Pythagorean triples. When it comes to #2, representation by -positive- rationals I don't think it is at all obvious (at least not to me!) that a k exists. My hunch is that it does. Many thanks for the reply and the references. Best wishes, Richard ============================================================================== From: Dave Rusin Date: Tue, 5 May 1998 12:31:10 -0500 (CDT) To: rcw9@tutor.open.ac.uk Subject: Re: Representation by 5-th powers - question >When it comes to #2, representation by -positive- rationals I don't >think it is at all obvious (at least not to me!) that a k exists. My >hunch is that it does. I realized after I wrote you earlier that this is trivial, e.g. to represent 3/7 as a sum of positive rational fifth powers, simply write 3*7^4 as a sum of positive integral fifth power, and divide through by 7. So (3/7)= two (5/7)^5 's + three (3/7)^5 's + seven (2/7)^5 's. ============================================================================== From: rcw9@tutor.open.ac.uk (Richard Walker) To: rusin@math.niu.edu Date: Wed, 6 May 1998 05:49:16 +0100 Subject: Representation by 5-th powers Richard Walker wrote: >>When it comes to #2, representation by -positive- rationals I don't >>think it is at all obvious (at least not to me!) that a k exists. My >>hunch is that it does. You replied: I realized after I wrote you earlier that this is trivial, e.g. to represent 3/7 as a sum of positive rational fifth powers, simply write 3*7^4 as a sum of positive integral fifth power, and divide through by 7. So (3/7)= two (5/7)^5 's + three (3/7)^5 's + seven (2/7)^5 's. Good point, thanks. I should have seen that, but I didn't. So k obviously (!) exists and is <=37. The same construction shows that every positive rational is the sum of at most four rational squares and nine rational cubes, etc. But in the case of cubes Hardy and Wright showed that only -three- positive rational cubes are in fact needed. So for example 23/1 would need nine cubes by the above construction but can be represented as follows: (35303/12768)^3+(2760/17689)^3+(2087677/1698144)^3=23/1 and there are infinitely many such solutions. It is easily seen that the three rational cubes cannot in general be reduced to two (and in fact x^3+y^3=23 does not have rational solutions.) However I think that to represent 23 as a sum of -squares- of rationals still requires -four- squares. Suppose eg 23=x^2+y^2+z^2, x, y, z rational. Then 23*m^2=X^2+Y^2+Z^2 for some integers m, X ,Y, Z. But if m=2^p*(2n+1), then 23*m^2=4^p*23*(2*n+1)^2=4^p*(8*k+7), which is not representable by three squares - is the working correct? So it is -easier- to represent 23 by rational cubes than by rational squares, which is perhaps unexpected and different from what we anticipate from Waring's problem? I was intrigued by this, which I think is quite surprising, and wondered if there are comparable results for higher powers and what work has been done on the subject. It seems to me that it should be possible to represent every positive rational as the sum of much fewer than 37 fifth powers of positive rationals, but I haven't got any leads, and numerical investigations are hard. Oh, it has just occurred to me (pursuing your observation above) that if we write a/b as (a*t)/(b*t) then we can choose t so that (a*t)*(b*t)^4 is sufficiently large to ensure that it can be represented by no more than 18 5-th powers, since all numbers beyond a certain point require only 18 5-th powers, which result I believe is due to Brudern 1990. So that's an immediate improvement but I would guess still far from the best result, since the same reasoning only gets us to 7 for rational cubes. I hope most of this makes some kind of sense, I wrote it -very- late at night. Thanks again for your response, it's difficult to get any feedback on this sort of stuff and I really would be very grateful for any more thoughts you might come up with. Best wishes, Richard ============================================================================== From: Dave Rusin Date: Wed, 6 May 1998 07:57:13 -0500 (CDT) To: rcw9@tutor.open.ac.uk Subject: Re: Representation by 5-th powers Your points are well taken. I'm surprised this topic is better known, since in general, arithmetic algebraic geometry does better rationally than integrally. I'll ask around to see if anyone here has any better leads but I don't think there's very much in print. dave ============================================================================== From: Dave Rusin Date: Wed, 6 May 1998 08:10:38 -0500 (CDT) To: rcw9@tutor.open.ac.uk Subject: Re: Representation by 5-th powers Best match from MathSciNet found. Searching also reminded me that Andrew Bremner has done a lot of algebraic geometry related to sums of powers; he's a good correspondent, too, so you might want to email him. 90g:11041 11D85 Choudhry, Ajai Representation of every rational number as an algebraic sum of fifth powers of rational numbers. (English) Enseign. Math. (2) 35 (1989), no. 1-2, 19--20. Based on an identity, this note proves that every rational number is representable as the sum of at most six fifth powers of rational numbers. Reviewed by Hong Bing Yu � Copyright American Mathematical Society 1990, 1998 dave ============================================================================== From: rcw9@tutor.open.ac.uk (Richard Walker) To: rusin@math.niu.edu Date: Thu, 7 May 1998 03:15:22 +0100 Subject: Re: Representation by 5-th powers You wrote: >Best match from MathSciNet found. Searching also reminded me that >Andrew Bremner has done a lot of algebraic geometry related to sums of >powers; he's a good correspondent, too, so you might want to email >him. > 90g:11041 11D85 > Choudhry, Ajai > Representation of every rational number as an algebraic sum of fifth > powers of rational numbers. (English) > Enseign. Math. (2) 35 (1989), no. 1-2, 19--20. > Based on an identity, this note proves that every rational number is > representable as the sum of at most six fifth powers of rational > numbers. > Reviewed by Hong Bing Yu > ) Copyright American Mathematical Society 1990, 1998 >dave Many thanks for all this. I have put in for a copy of the reference, which looks to be a very close match. I am almost sure that it wil turn out to be what I just worked out [literally in the last couple of days]. But at least I have been working on the right lines! I needed a point of order 4 on x^5+y^5+z^5+u^5+v^5+w^5 which then gives the result above, and in fact I finally saw how to do it from a bit in a paper by [coincidence!] Andrew Bremner. I had been trying to get down to six 5-th powers for several weeks. I will be surprised if the identity referred to in the abstract isn't the same one I got. I feel that six can be reduced to five but I can't see the way at present. I will certainly take up your suggestion of contacting Andrew Brenmer. If you come across anything else relevant please do let me know. If I make any significant progress I'll update you. Kind regards, Richard ============================================================================== From: rcw9@tutor.open.ac.uk (Richard Walker) To: rusin@math.niu.edu Date: Thu, 7 May 1998 03:15:22 +0100 Subject: Re: Representation by 5-th powers You wrote: >Best match from MathSciNet found. Searching also reminded me that >Andrew Bremner has done a lot of algebraic geometry related to sums of >powers; he's a good correspondent, too, so you might want to email >him. > 90g:11041 11D85 > Choudhry, Ajai > Representation of every rational number as an algebraic sum of fifth > powers of rational numbers. (English) > Enseign. Math. (2) 35 (1989), no. 1-2, 19--20. > Based on an identity, this note proves that every rational number is > representable as the sum of at most six fifth powers of rational > numbers. > Reviewed by Hong Bing Yu > ) Copyright American Mathematical Society 1990, 1998 >dave Many thanks for all this. I have put in for a copy of the reference, which looks to be a very close match. I am almost sure that it wil turn out to be what I just worked out [literally in the last couple of days]. But at least I have been working on the right lines! I needed a point of order 4 on x^5+y^5+z^5+u^5+v^5+w^5 which then gives the result above, and in fact I finally saw how to do it from a bit in a paper by [coincidence!] Andrew Bremner. I had been trying to get down to six 5-th powers for several weeks. I will be surprised if the identity referred to in the abstract isn't the same one I got. I feel that six can be reduced to five but I can't see the way at present. I will certainly take up your suggestion of contacting Andrew Brenmer. If you come across anything else relevant please do let me know. If I make any significant progress I'll update you. Kind regards, Richard ============================================================================== Date: Thu, 7 May 1998 13:40:48 -0500 (CDT) From: Neil Dummigan To: Dave Rusin Subject: Re: powers (fwd) ---------- Forwarded message ---------- Date: Thu, 7 May 1998 03:47:00 -0400 (EDT) From: Trevor Wooley To: dummigan@math.niu.edu Subject: Re: powers Dear Neil, yes indeed, using rational powers is more efficient. I do not know much on this problem for higher powers. For cubes you need only 4, maybe only 3, I seem to remember a paper by Landau(?) in a conference proceedings from the 60's, but there must be other references. No sign of our reprints yet ... I am chasing thenmm up. Best wishes, Trevor. ==============================================================================