[math-fun] System of 2 diophantine equations
I am looking for at least one integer solution of this system: a^3 - d^3 = b^3 - e^3 = c^3 - f^3 a^3 - g^3 = b^3 - h^3 = c^3 - i^3 Quite easy to find near solutions, for example: 165^3 - 72^3 = 178^3 - 115^3 = 162^3 - 51^3 165^3 - 618^3 = 178^3 - 619^3 = 162^3 - 235788435 Unfortunately 235788435 is not a cube... It seems that there is no solution with integers < 300000^3. Any idea? Christian.
Thanks to Edwin Clark for his remark: looking for one solution using DISTINCT positive integers. Christian. -----Message d'origine----- De : math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] De la part de Christian Boyer Envoyé : samedi 14 juin 2008 11:18 À : 'math-fun' Objet : [math-fun] System of 2 diophantine equations I am looking for at least one integer solution of this system: a^3 - d^3 = b^3 - e^3 = c^3 - f^3 a^3 - g^3 = b^3 - h^3 = c^3 - i^3 Quite easy to find near solutions, for example: 165^3 - 72^3 = 178^3 - 115^3 = 162^3 - 51^3 165^3 - 618^3 = 178^3 - 619^3 = 162^3 - 235788435 Unfortunately 235788435 is not a cube... It seems that there is no solution with integers < 300000^3. Any idea? Christian. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
There are some corollaries; perhaps these could be used to work backwards, building up a solution from smaller partials: d^3 - g^3 = e^3 - h^3 = f^3 - i^3, and the system d^3 - e^3 = g^3 - h^3, d^3 - f^3 = g^3 - i^3, which is a smaller version of your goal. ------------ A New Kind of Industrial Mathematics Consider a related subproblem: providing interesting sets of integers with prescribed density. We have primes with density falloff 1/logN, squares with density falloff N^-1/2, cubes with N^-2/3, powers of 2 with falloff logN/N, products of two primes with falloff loglogN/logN, etc. We can make combinations of these to create guesses like "every odd number is the sum of a prime and a power of 2" or "every even number is the sum of a prime and Mersenne-prime exponent". We can tune these to be truer by adding another thin set like 2^2^N, or noticing modular restrictions (primes are mostly odd), or saying N>1000, or "almost all N" (meaning exceptions are thin, -> 0%). We could introduce meta-sets, such as "numbers that are NOT the sum of three squares". With very little investment of fact, a whole bushel of conjecture! (borrowing from Mark Twain). I can identify one application of this sort of pseudo-theorem: Sometimes, as part of a larger algorithm, I'll need to say "There are lots of primes with properties A,B, & C; search until you find one." Or, I'll cheerfully assume the existence of a small primitive root. For engineering applications, this is accepted behavior, even though the rigor is absent. There's an obvious tendency to regard these assertions as mere annoyances, since we have no hope of proof for most of them. Any refutation will usually be based on trivial arguments, such as a numerical counterexample, or a simple mod 12 argument. Usually the refutations can be tuned away, frustrating the refuter. I'll argue for a more "industrial" approach: We can generate millions of these guesses mechanically, classify them mechanically, run the usual-suspect numerical tests, and mark the survivors as "provisionally true". We could generate some implications automatically, such as Goldbach-even2 -> Goldbach-odd3. The idea is to create a database somehwat like the sequence database or the real-number database. When confronted with a question, we could consult the database, hoping for a match or near-match that could cast light on our question. We could apply data-mining methods to look for generalizations that are actually interesting. This applies modern technology to the old idea of warehousing results that might be useful in the next century. Rich --------- Quoting Christian Boyer <cboyer@club-internet.fr>:
I am looking for at least one integer solution of this system: a^3 - d^3 = b^3 - e^3 = c^3 - f^3 a^3 - g^3 = b^3 - h^3 = c^3 - i^3
Quite easy to find near solutions, for example: 165^3 - 72^3 = 178^3 - 115^3 = 162^3 - 51^3 165^3 - 618^3 = 178^3 - 619^3 = 162^3 - 235788435 Unfortunately 235788435 is not a cube... It seems that there is no solution with integers < 300000^3.
Any idea?
Christian.
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What methods have you tried? My idea: make a list of numbers that are a difference of cubes in 2 or more ways, (is there a quick test for this, or do we have to slowly build up a big hash table?) then look for a,b,c such that a^3 - b^3 is a difference of cubes in 2 ways, same with a^3 - c^3, b^3 - c^3. --Joshua Zucker On Sat, Jun 14, 2008 at 2:17 AM, Christian Boyer <cboyer@club-internet.fr> wrote:
I am looking for at least one integer solution of this system: a^3 - d^3 = b^3 - e^3 = c^3 - f^3 a^3 - g^3 = b^3 - h^3 = c^3 - i^3
Quite easy to find near solutions, for example: 165^3 - 72^3 = 178^3 - 115^3 = 162^3 - 51^3 165^3 - 618^3 = 178^3 - 619^3 = 162^3 - 235788435 Unfortunately 235788435 is not a cube... It seems that there is no solution with integers < 300000^3.
Any idea?
Christian.
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Integers < 300,000^3 was a computation done by Frank Rubin in 2006. Latest news on this problem: using tables built during his computation on Cabtaxi(10) http://cboyer.club.fr/Taxicab.htm, Uwe Hollerbach worked on his list of 10,597,218 numbers < 9.5 * 10^20 that are differences of cubes in two or more ways. If his computation is correct, the system a^3 - d^3 = b^3 - e^3 = c^3 - f^3 = S a^3 - g^3 = b^3 - h^3 = c^3 - i^3 = S' has no solution with S and S' < 9.5 * 10^20. Deceptive! But not a proof that the problem is impossible. Christian. -----Message d'origine----- De : math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] De la part de Joshua Zucker Envoyé : dimanche 15 juin 2008 06:44 À : math-fun Objet : Re: [math-fun] System of 2 diophantine equations What methods have you tried? My idea: make a list of numbers that are a difference of cubes in 2 or more ways, (is there a quick test for this, or do we have to slowly build up a big hash table?) then look for a,b,c such that a^3 - b^3 is a difference of cubes in 2 ways, same with a^3 - c^3, b^3 - c^3. --Joshua Zucker On Sat, Jun 14, 2008 at 2:17 AM, Christian Boyer <cboyer@club-internet.fr> wrote:
I am looking for at least one integer solution of this system: a^3 - d^3 = b^3 - e^3 = c^3 - f^3 a^3 - g^3 = b^3 - h^3 = c^3 - i^3
Quite easy to find near solutions, for example: 165^3 - 72^3 = 178^3 - 115^3 = 162^3 - 51^3 165^3 - 618^3 = 178^3 - 619^3 = 162^3 - 235788435 Unfortunately 235788435 is not a cube... It seems that there is no solution with integers < 300000^3.
Any idea?
Christian.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Ooops, sorry. Please read: "list of 10,597,218 numbers < 9.5 * 10^20 that are differences of cubes in THREE or more ways" -----Message d'origine----- De : Christian Boyer [mailto:cboyer@club-internet.fr] Envoyé : dimanche 15 juin 2008 15:13 À : 'math-fun' Objet : RE: [math-fun] System of 2 diophantine equations Integers < 300,000^3 was a computation done by Frank Rubin in 2006. Latest news on this problem: using tables built during his computation on Cabtaxi(10) http://cboyer.club.fr/Taxicab.htm, Uwe Hollerbach worked on his list of 10,597,218 numbers < 9.5 * 10^20 that are differences of cubes in two or more ways. If his computation is correct, the system a^3 - d^3 = b^3 - e^3 = c^3 - f^3 = S a^3 - g^3 = b^3 - h^3 = c^3 - i^3 = S' has no solution with S and S' < 9.5 * 10^20. Deceptive! But not a proof that the problem is impossible. Christian. -----Message d'origine----- De : math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] De la part de Joshua Zucker Envoyé : dimanche 15 juin 2008 06:44 À : math-fun Objet : Re: [math-fun] System of 2 diophantine equations What methods have you tried? My idea: make a list of numbers that are a difference of cubes in 2 or more ways, (is there a quick test for this, or do we have to slowly build up a big hash table?) then look for a,b,c such that a^3 - b^3 is a difference of cubes in 2 ways, same with a^3 - c^3, b^3 - c^3. --Joshua Zucker On Sat, Jun 14, 2008 at 2:17 AM, Christian Boyer <cboyer@club-internet.fr> wrote:
I am looking for at least one integer solution of this system: a^3 - d^3 = b^3 - e^3 = c^3 - f^3 a^3 - g^3 = b^3 - h^3 = c^3 - i^3
Quite easy to find near solutions, for example: 165^3 - 72^3 = 178^3 - 115^3 = 162^3 - 51^3 165^3 - 618^3 = 178^3 - 619^3 = 162^3 - 235788435 Unfortunately 235788435 is not a cube... It seems that there is no solution with integers < 300000^3.
Any idea?
Christian.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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