From: rusin@math.niu.edu (Dave Rusin) Newsgroups: sci.math.research Subject: Re: [new posting] A problem similar to FLT. Date: 8 Dec 1998 14:59:54 GMT Min-Ho Ahn wrote: >As David Rusin pointed out, I made a mistake in the previous posting. I'm not sure this message made it clear that all I did was to point out well-known counterexamples to Euler's conjecture generalizing the Fermat conjecture. The poster is asking for small numbers of nth powers which add to another nth power. Roughly speaking, nothing is known for certain except that FLT has been proven, and comparatively small counterexamples to a couple of generalizations have been found. Here is a summary I have used before: Two nonzero squares can sum to another. Three nonzero cubes can sum to another, but not two. Three nonzero fourth powers can sum to another, but not two. Four nonzero fifth powers can sum to another, but not two; It has not been decided whether or not three fifth powers can sum to another. What is the minimum number of positive sixth powers whose sum is another sixth power? It's more than two, certainly, but really nothing else is known. I would not be surprised to hear that five sixth powers can sum to another, but in fact no one has ever even found SIX sixth powers whose sum is another sixth power (and some researchers are keenly interested in this.) What should be a comparably difficult problem, that of finding two equal sums of three sixth powers, admits the easy solution 3,19,22; 10,15,23 found by Rao in 1934. Less is known for higher exponents. As far as I know there are no examples known of fewer than n-1 nth powers summing to another nth power. For even powers (or for odd powers with a positivity requirement added) one obtains a variety of different problems asking for solutions to Sum x_i^n = Sum y_i^n, possibly with different numbers of summands on the two sides. One can ask for parametric solutions (e.g. it is not yet known for sure that there is not a parametric solution to x^4+y^4+z^4=w^4 -- there _are_ infinitely many distinct integer solutions). One can ask for numbers multiply expressible as a sum of k n-th powers, or a count of the number of representations as a sum of n-th powers, or a count of the numbers in a given range which can be expressed as a sum of n-th powers, or a count of the number of [positive] n-th powers needed to express every [sufficiently large] integer as a sum of this many n-th powers. One can ask for sets of integers the sum of whose n-th powers are equal for more than one n. And so on. There's plenty of conjecture to keep us busy for years, but both proofs and counterexamples are hard to come by. For more Additive Number Theory, see e.g. index/11PXX.html dave ============================================================================== From: rusin@math.niu.edu (Dave Rusin) To: Tim Chow Subject: Re: [new posting] A problem similar to FLT. Date: 8 Dec 1998 23:00:54 GMT [est...] Newsgroups: sci.math.research You wrote: >Let f(n) be the smallest integer s >= 2 such that there exists an nth power >that is the sum of s nth powers of positive integers. Then FLT tells us >that f(n) > 2 for n > 2. Certainly f(n) <= 2^n, since we may simply add up >2^n copies of "1^n" to get 2^n. Are there any general results about f(n) >beyond this? It seems that the methods used for Waring's problem should be >able to get better upper bounds, although the inference does not seem to be >immediate. Waring's problem is related, but I'd have to say the relationship is tenuous. Let's see, you want Let f(n) be the smallest integer s >= 2 such that there exists an nth power that is the sum of s nth powers of positive integers. and it's taken 350 yr to show f(n) > 2 for all n > 2. Waring's problem deals with Let g(n) be the smallest integer s >= 2 such that every positive integer is the sum of s nth powers of positive integers. So f(n) >= g(n). Not so good. We known g(n) exists (Hilbert) and barring a number-theoretic miracle, g(n) = 2^n + [(3/2)^n] - 2. Thus for n=1,2,3,4,5 we have g(n) = 1,4,9,19,37. On the other hand, f(n) <= 2^n, as you noted, but in fact f(n)=1,2,3,3, [3 or 4]. The gap is widening! Now, of course, you can do better. Take, say, Let G(n) be the smallest integer s >= 2 such that every sufficiently large integer is the sum of s nth powers of positive integers. Then f(n) <= G(n) <= g(n). The values of G are not known: they begin G(n) = 1, 4, [4,5,6, or 7], 16. I know G(5) <= 18, but not much else. There's also a trivial lower bound: There are only N^(1/n) nth powers in [1, N], and so if k < n you can certainly not make N - O(1) different sums with k of these powers; you must have G(n) => n for all n. If I were forced to make a conjecture, I might think f(n) <= n, although no one knows six 6th powers which sum to another 6th power. But that would give a pretty set of inequalities 3 <= f(n) <= n <= G(n) <= g(n) = [presumed] 2^n + [(3/2)^n] - 2 If these conjectures are at all close to true, they suggest the Waring's problem link is sort of a red herring: I suppose G(n) is pretty far from n. By the way, you can play the game a little differently. In FLT one might at first assume all terms are positive, but it's no loss to consider positive or negative powers which sum to another. Or to allow sums _or differences_ of powers. This makes for a different Euler-type conjecture: We _do_ know of five sixth powers whose _signed sum_ is another 6th power. It also puts another spin on the Waring's problem game. Let h(n) be the smallest integer s >= 2 such that every (positive) integer is the sum or difference of s nth powers of positive integers. and similarly H(n) for "all sufficiently large". Then h(2) = 2, and it is possible that it is possible that all integers not congruent to +-4 mod 9 are expressible as sums of _three_ cubes (although it is still not known if it is possible to write 30 as a sum of three signed cubes). So perhaps h(n) is much less than g(n). One more twist: Let k(n) be the smallest integer s >= 2 such that every positive integer is the sum of s nth powers of positive rationals. (and you can define more numbers with "sufficiently large" and "or differences" in your sentences). Then f(n) <= k(n) <= g(n). I don't know about the values of k but they're less than g in general. We have k(2)=4, k(3)=3 and k(5)<=6; I don't know about k(4). These low numbers might lead one to think we could write _every_ positive integer as a sum of about six 6th powers of rationals; in fact we don't know how to do that for _any_ integer in a nontrivial way! Ayway, it would seem this is the variant of Waring's problem most appropriate for your insight. dave ============================================================================== Date: Wed, 9 Dec 1998 10:18:02 -0500 (EST) From: "Timothy Y. Chow" To: Dave Rusin Subject: Re: [new posting] A problem similar to FLT. > Now, of course, you can do better. Take, say, > > Let G(n) be the smallest integer s >= 2 such that every sufficiently > large integer is the sum of s nth powers of positive integers. > > Then f(n) <= G(n) <= g(n). The values of G are not known: they begin > G(n) = 1, 4, [4,5,6, or 7], 16. I know G(5) <= 18, but not much else. > There's also a trivial lower bound: There are only N^(1/n) nth powers > in [1, N], and so if k < n you can certainly not make N - O(1) different > sums with k of these powers; you must have G(n) => n for all n. Well, the letter G is usually used for the following quantity: the smallest integer s such that every sufficiently large integer is the sum of s nth powers of *nonnegative* integers. Then it is known that G(n) <= n (log n + log log n + O(1)), proved by Wooley in his Annals of Math paper, "Large improvements in Waring's problem." The problem is that this result does not tell us anything immediately about G as you've defined it, because an nth power is trivially the sum of one nth power. It seems that one might be able to get around this, but this would require understanding the methods well enough to be able to tweak them. > If I were forced to make a conjecture, I might think f(n) <= n, although > no one knows six 6th powers which sum to another 6th power. But that would > give a pretty set of inequalities > 3 <= f(n) <= n <= G(n) <= g(n) = [presumed] 2^n + [(3/2)^n] - 2 > If these conjectures are at all close to true, they suggest the Waring's > problem link is sort of a red herring: I suppose G(n) is pretty far from n. But n log n is much better than 2^n. Tim. ============================================================================== From: Dave Rusin Date: Wed, 9 Dec 1998 09:28:52 -0600 (CST) To: tchow Subject: Re: [new posting] A problem similar to FLT. Cc: rusin@math.niu.edu Yes, thanks. I got a little sloppy about which function was which. I don't really know what conclusion to draw though; I guess we come to expect from Waring's-problem-type research that "usually" numbers are sums of "something like" n n-th powers; with a few more nth powers thrown in, that representation wouldn't be unique, in general, and so we expect to be able to find nth powers which are sums of about that many other nth powers. That's pretty vague (possibly reassuring, I suppose), but I think you'll not get too much more from trying to shoehorn the Euler conjecture into the Waring framework. I'd put a lot more money on algebraic geometry - type solutions a la Bremner to show that there exist small sums of nth powers which are another nth power, and a lot on sieve - type solutions to show almost everything is a sum of enough nth powers. By the way, if you have a favorite reference for additive number theory let me know; I think it would be appropriate on the corresponding web page here. dave ============================================================================== From: "Tsukamoto Tenma" <1@2.3> Newsgroups: sci.math Subject: Proof/Attempted Proof of Euler's conjecture for n=6 Date: Tue, 14 Mar 2006 17:23:26 +0800 The following is my proof of x^6+y^6+z^6=w^6 not having nontrival integral solutions. I think it's correct, but I hope you can help me check if there are flaws in it. Assume there exists such a nontrival integral solution (x,y,z,w). Then it corresponds to a solution (p,q,r,s) of p^3+q^3=r^3+s^3 such that the absolute value of each of p,q,r,s is a perfect square. The general solution to p^3+q^3=r^3+s^3, by J. Binet, (as cited by here: http://www.geocities.com/titus_piezas/Equalsums.pdf, I hope I can get an access of Binet's original paper) is: (multiply p,q,r,s by a constant if necessary) p=1-(a-3b)(a^2+3b^2) q=(a+3b)(a^2+3b^2)-1 r=(a+3b)-(a^2+3b^2)^2 s=(a^2+3b^2)^2-(a-3b) Check the "trival" cases of (-1<=a<=1 or -1<=b<=1), and there are no solutions that produce a nontrival solution of x^6+y^6+z^6=w^6, so consider only the "nontrival" cases below. Note that -rs=(a^2+3b^2)^2-2a(a^2+3b^2)+a^2-9b^2 has its absolute value being a perfect square. Outside of "trival cases", (a^2+3b^2-a)>=14, (a^2+3b^2-a-2)^2<-rs<(a^2+3b^2-a)^2, and this implies (a^2+3b^2-a-1)^2=-rs=(a^2+3b^2)^2-2a(a^2+3b^2)+a^2-9b^2, which simplified as 3b^2=2a^2-2a-1 and note that |b|<|a|. Then s=(a^2+3b^2)^2-(a-3b). If a=3b, 3b^2=2a^2-2a-1=2(9b^2)-2(3b)-1 doesn't have any integral solutions. But |a+3b|<|a^2+3b^2|, so (a^2+3b^2-1)^2, Tsukamoto Tenma <1@2.3> wrote: > The following is my proof of x^6+y^6+z^6=w^6 not having nontrival integral > solutions. This is very interesting. I think for sixth powers we do know how to make 7 sixth powers sum to another sixth power[*]; and we know that 2 sixth powers cannot sum to another sixth power. I don't recall hearing before that it was proven that 3 sixth powers cannot sum to another sixth power, so if this proof is correct, that narrows the unknown range to the questions of whether 4, 5, or 6 sixth powers can sum to another sixth power. (Without any real evidence at all I might guess that 4 isn't enough, 6 is, and 5 probably is. Simcha Brudno keeps asking about this.) > I think it's correct, but I hope you can help me check if there > are flaws in it. > > Assume there exists such a nontrival integral solution (x,y,z,w). Then it > corresponds to a solution (p,q,r,s) of p^3+q^3=r^3+s^3 such that the > absolute value of each of p,q,r,s is a perfect square. The general solution > to p^3+q^3=r^3+s^3, by J. Binet, (as cited by here: > http://www.geocities.com/titus_piezas/Equalsums.pdf, I hope I can get an > access of Binet's original paper) is: (multiply p,q,r,s by a constant if > necessary) > p=1-(a-3b)(a^2+3b^2) > q=(a+3b)(a^2+3b^2)-1 > r=(a+3b)-(a^2+3b^2)^2 > s=(a^2+3b^2)^2-(a-3b) This is indeed a solution; I'll take your word for it that this parameterizes all solutions. (You have to be a little careful -- does this parameterize the set of all _ordered_ quadruples (p,q,r,s)? Or the set of possible values of the set-of-sets {{p,q},{r,s}} ?) > Check the "trival" cases of (-1<=a<=1 or -1<=b<=1), and there are no > solutions that produce a nontrival solution of x^6+y^6+z^6=w^6, so consider > only the "nontrival" cases below. > Note that -rs=(a^2+3b^2)^2-2a(a^2+3b^2)+a^2-9b^2 has its absolute value > being a perfect square. Typo here: -rs=(a^2+3b^2)^4 - 2a(a^2+3b^2)^2 + a^2-9b^2 . Note that this equals ((a^2 + 3b^2)^2 - a)^2 - (3b)^2, so that it's almost immediate what the nearest square is. (You don't need to separate out the trivial cases.) > Outside of "trival cases", (a^2+3b^2-a)>=14, > (a^2+3b^2-a-2)^2<-rs<(a^2+3b^2-a)^2, Typo persists, but the idea is OK. Certainly -rs < ((a^2+3b^2)^2 - a)^2. Rather than look at the other inequality you can actually show -rs is larger than the very next square down: (-rs) - ((a^2+3b^2)^2 - a - 1)^2 = 12a^2b^2 + 18(b^2-1/4)^2+(2a^4-2a-17/8) Clearly the right side is positive for any real a and b unless a is between the two real roots of the quartic, which are near -.75 and +1.23. So when a and b are integers, positivity is assured unless a=0 or a=1; but even then, the value of the quartic is -17/8, while the quartic in b has a value larger than 17/8 unless b is between -.77 and +.77, so again using integrality it is only the case b=0 that could be a problem. So we have shown that for _integers_ a and b we have (-rs) - ((a^2+3b^2)^2 - a - 1)^2 > 0 unless b=0 and a = 0 or 1. Those two cases are degenerate anyway (two or four of p,q,r,s are zero). > and this implies > (a^2+3b^2-a-1)^2=-rs=(a^2+3b^2)^2-2a(a^2+3b^2)+a^2-9b^2, which simplified as > 3b^2=2a^2-2a-1 and note that |b|<|a|. Then s=(a^2+3b^2)^2-(a-3b). If a=3b, > 3b^2=2a^2-2a-1=2(9b^2)-2(3b)-1 doesn't have any integral solutions. But > |a+3b|<|a^2+3b^2|, so (a^2+3b^2-1)^2 perfect square. So the proof looks good to me, with typos corrected and maybe the inequalities trimmed down a little. Good job! dave [*] 1077^6 + 894^6 + 702^6 + 474^6 + 402^6 + 234^6 + 74^6 = 1141^6 ==============================================================================