From: Noam D. Elkies Subject: Re: Sum of cubes conjecture Date: 17 Apr 1999 22:26:34 GMT Newsgroups: rec.puzzles,sci.math Keywords: Numbers expressible as sums of two cubes in multiple ways From article <3719171E.10B3@library.com>: > I wrote a brute force computer program to test positive integers >starting at 1 and increasing by 1 looking for positive integers that >could be expressed as the sum of cubes of positive integers in more than >one way. The program found 1729 and 13 other numbers. It has long been known that there are infinitely many primitive solutions of A^3+B^3=C^3+D^3 ("primitive" i.e. "with A,B,C,D having no common factor >1"). > In all cases, at least one of A,B was composite. Also, at least >one of C,D was composite. I conjecture that this is always the case. >Is this true or false? False. >Why? Because, to give the two smallest counterexamples, 31^3 + 1867^3 = 397^3 + 1861^3 and 61^3 + 1823^3 = 1049^3 + 1699^3 with each of A,B,C,D prime in both cases. These examples are easy to find by exhaustive search, once one is looking for prime solutions -- primes are rare enough that asking all of A,B,C,D to be prime in the output of an unrestricted search is a needle-in-a-haystack problem. [The standard way to do such a search is to list all p^3+q^3 for distinct primes p,q up to say 2500, then sort the resulting file and look for duplicates. If space is at a premium one can use the method described by D.J.Bernstein at http://pobox.com/~djb/papers/sortedsums.dvi .] One heuristically expects lots of solutions: the number of quadruples (A,B,C,D) of primes less than M is asymptotically proportional to (M/log(M))^4 by the Prime Number Theory, and one expects that A^3+B^3=C^3+D^3 will hold about one time in M^3 for such numbers, for a total number of solutions proportional to M / (log M)^4. This increases without bound as M grows. > Also, can a positive integer be expressed as the sum of cubes of >positive integers in more than 2 ways? Presumably you mean "as the sum of *two* cubes", else the problem is trivial. (How much is one and one and one and one and one and...?) The answer to your question is also yes; indeed there exists an integer expressible as a sum of two cubes in at least a xillion ways, for any finite value of a xillion. That is to be sure a harder theorem, but has been known in some form since Diophantus. See the relevant chapter of Dickson's History of the Theory of Numbers for more details. Or, if you're familiar with the arithmetic of elliptic curves, simply start with a curve of the form x^3+y^3=m with a point P of infinite order -- say m=7, P: (x,y) = (2,-1) -- and note that its first xillion multiples in the group law are distinct rational solutions to x^3+y^3=m; clearing denominators we find an integer D^3 * m expressible as the sum of two integer cubes in at least a xillion ways. It is not yet known if for each N there is an integer expressible as a sum of two *relatively prime* cubes in at least N ways. --Noam D. Elkies (Dept. of Math., Harvard University) [change "mathematics" in e-addr. to the abbr. in the above line] ============================================================================== From: djb@koobera.math.uic.edu (D. J. Bernstein) Subject: Re: Sum of cubes conjecture Date: 24 Apr 1999 23:28:18 GMT Newsgroups: rec.puzzles,sci.math David Bernier wrote: > What about the problem of expressing a (strictly) positive > integer N as a sum of two (strictly) positive cubes in three > or more different ways? Same method; just look for multiples of P in the first quadrant. > This suggests O(log(K)) numbers N two strictly positive cubes in three or more ways, and a finite > number of N for the 4 or more ways question. But those ways are not independent! If N has many factors then it has many chances to be written as a sum of cubes. For example, the number 99984532154060736 has 4480 divisors, and can be written in 3 ways as a sum of positive cubes. My table of solutions shows (modulo typos) digits in N 8 9 10 11 12 13 14 15 16 17 ------------------------------------------------- #{N: #ways=3} 1 7 18 52 158 415 933 2231 5173 11745 #{N: #ways=4} 1 6 20 45 111 #{N: #ways=5} 1 ---Dan ============================================================================== From: Jim Ferry Subject: Re: Sum of cubes conjecture Date: Mon, 26 Apr 1999 18:28:39 -0500 Newsgroups: rec.puzzles,sci.math Andrew Nikitin wrote: > > > Also, can a positive integer be expressed as the sum of cubes of > >positive integers in more than 2 ways? > > The positive answer to this question was already posted, but it doesn't show > minimal such number. The least known to me is > > 166^3 + 435^3 = 227^3 + 422^3 = 254^3 + 413^3 = 87539319 I got these from Sloane's On-Line Encyclopedia of Integer Sequences 2 ways 1729,4104,13832,20683,32832,39312,40033,46683,64232,65728, 110656,110808,134379,149389,165464,171288,195841,216027,216125, 262656,314496,320264,327763,373464,402597,439101,443889,513000, 513856 3 ways 87539319,119824488,143604279,175959000,327763000,700314552, 804360375,958595904,1148834232,1407672000,1840667192,1915865217, 2363561613,2622104000,3080802816,3235261176,3499524728,3623721192, 3877315533 4 ways (involving at least one sum of coprime cubes) 95773976104625,159380205560856,2983266899506341,3924747381450168, 4145402010642984,10998043552638016,13796337654911448 18170126765973000 | Jim Ferry | Center for Simulation | +------------------------------------+ of Advanced Rockets | | http://www.uiuc.edu/ph/www/jferry/ +------------------------+ | jferry@expunge_this_field.uiuc.edu | University of Illinois | ============================================================================== [Correction received in email -- djr] Near the end of the text, one can read that the least positive integer which is the sum of two cubes in three different ways is : 166^3 + 435^3 = 227^3 + 422^3 = 254^3 + 413^3 = 87539319 This is the right number, but the decomposition is false. The right version is : 167^3 + 436^3 = 228^3 + 423^3 = 255^3 + 414^3 = 87539319 =============================================================================