RE: [math-fun] x^n + y^n = z^n + w^n
Of course, Beal's conjecture (for which there is now a prize of U.S.$100,000 for a proof OR a counterexample) goes as follows: BEAL'S CONJECTURE: If A^K +B^L = C^M , where A, B, C, K, L, M are positive integers, and K, L, M are all greater than 2, then A, B, and C must have a common prime factor. Based on recently-discussed equations for which no examples have ever been found, can Beal's conjecture reasonably be extended? In fact, what is the largest such conjecture that's consistent with what's known? --Dan
dasimov@earthlink.net wrote:
BEAL'S CONJECTURE: If A^K +B^L = C^M , where A, B, C, K, L, M are positive integers, and K, L, M are all greater than 2, then A, B, and C must have a common prime factor.
Based on recently-discussed equations for which no examples have ever been found, can Beal's conjecture reasonably be extended?
Here are my 2 proposals of extension. ----------"SYMMETRICAL EXTENDED" CONJECTURE------------- If A^K + B^L = C^M + D^N where A, B, C, D, K, L, M, N are positive integers, where K, L, M, N are all greater than 4, and where A^K, B^L, C^M, D^N are not twice the same couple, then A, B, C, and D must have a common factor. -------------------------------------------------------- Four remarks on this "symmetrical extended" conjecture: 1) When D=0, it is the Beal's conjecture. For example: 2^5 + 2^5 = 2^6 (common factor 2) 2) "greater than 2" of the Beal's conjecture is changed in "greater than 4" because we know solutions with no common factor and the fourth power, for example this Euler's solution: 158^4 + 59^4 = 134^4 + 133^4 3) "where A^K, B^L, C^M, D^N are not twice the same couple" is an added condition in order to avoid this kind of obvious solution with NO common factor to the four integers: 8^6 + 9^9 = 4^9 + 27^6 4) "A, B, C, and D must have a common factor." Here are some examples: 124^5 + 31^7 = 62^6 + 31^5 (common factor 31) 126^5 + 126^5 = 63^6 + 63^5 (common factor 63) 132^5 + 33^7 = 66^6 + 33^5 (common factor 33) 148^6 + 37^8 = 111^6 + 74^7 (common factor 37) 204^5 + 12^6 = 192^5 + 156^5 (common factor 12) 217^5 + 186^5 = 93^6 + 62^6 (common factor 31) 254^6 + 254^6 = 127^7 + 127^6 (common factor 127) ----------"FULL EXTENDED" CONJECTURE-------------------- If A^K +- B^L +- C^M +- D^N +- ... = 0 where A, B, C, D,... K, L, M, N,... are positive integers, where K, L, M, N,... are all greater than the number of (i^j) elements, and where no subset of elements have a null sum, without changing the signs of the elements then A, B, C, D,... must have a common factor. -------------------------------------------------------- Two supplemental remarks: 5) "where no subset of elements have a null sum, without changing the signs of the elements" is the equivalent of the "not twice the same couple" in the previous extension. It means that 8^6 + 8^6 is allowed... but 8^6 - 8^6 is forbidden! 6) With 4 elements, the other equation, not analyzed with the "symmetrical extended" conjecture, is A^K + B^L + C^M - D^N = 0 Here are some examples with common factors: 2^5 + 2^5 + 2^6 - 2^7 = 0 (common factor 2) 3^5 + 3^5 + 3^5 - 3^6 = 0 (common factor 3) (...) 10^6 + 10^7 + 5^8 - 15^6 = 0 (common factor 5) Perhaps something stupid in these extensions? Something missing or bad exmplained? Or a counterexample (means with no common factor) to these new extended conjectures? Christian.
On the "full extended" conjecture described in my previous email. With 5 elements (implying powers > 5), some examples with common factors: 2^9 - 2^8 - 2^7 - 2^6 - 2^6 = 0 (common factor 2) 12^7 + 10^6 - 8^9 - 6^8 + 4^8 = 0 (common factor 2) 18^6 - 9^8 + 9^7 + 9^7 - 9^6 = 0 (common factor 9) +++ Who will find at least one example WITHOUT common factor? +++ As I wrote, it is needed to have all powers greater than the number of elements. Without this, we can find solutions without common factors, for example with again 5 elements, but powers > 4: 5^5 - 4^6 + 3^6 + 3^5 - 1^n = 0 (no common factor) 7^5 - 6^7 + 4^9 + 3^6 + 2^8 = 0 (no common factor) 13^5 - 5^8 + 5^7 - 3^10 + 2^8 = 0 (no common factor) Christian. -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de Christian Boyer Envoyé : mardi 20 juin 2006 19:01 À : dasimov@earthlink.net; 'math-fun' Objet : RE: [math-fun] x^n + y^n = z^n + w^n dasimov@earthlink.net wrote:
BEAL'S CONJECTURE: If A^K +B^L = C^M , where A, B, C, K, L, M are positive integers, and K, L, M are all greater than 2, then A, B, and C must have a common prime factor.
Based on recently-discussed equations for which no examples have ever been found, can Beal's conjecture reasonably be extended?
Here are my 2 proposals of extension. ----------"SYMMETRICAL EXTENDED" CONJECTURE------------- If A^K + B^L = C^M + D^N where A, B, C, D, K, L, M, N are positive integers, where K, L, M, N are all greater than 4, and where A^K, B^L, C^M, D^N are not twice the same couple, then A, B, C, and D must have a common factor. -------------------------------------------------------- Four remarks on this "symmetrical extended" conjecture: 1) When D=0, it is the Beal's conjecture. For example: 2^5 + 2^5 = 2^6 (common factor 2) 2) "greater than 2" of the Beal's conjecture is changed in "greater than 4" because we know solutions with no common factor and the fourth power, for example this Euler's solution: 158^4 + 59^4 = 134^4 + 133^4 3) "where A^K, B^L, C^M, D^N are not twice the same couple" is an added condition in order to avoid this kind of obvious solution with NO common factor to the four integers: 8^6 + 9^9 = 4^9 + 27^6 4) "A, B, C, and D must have a common factor." Here are some examples: 124^5 + 31^7 = 62^6 + 31^5 (common factor 31) 126^5 + 126^5 = 63^6 + 63^5 (common factor 63) 132^5 + 33^7 = 66^6 + 33^5 (common factor 33) 148^6 + 37^8 = 111^6 + 74^7 (common factor 37) 204^5 + 12^6 = 192^5 + 156^5 (common factor 12) 217^5 + 186^5 = 93^6 + 62^6 (common factor 31) 254^6 + 254^6 = 127^7 + 127^6 (common factor 127) ----------"FULL EXTENDED" CONJECTURE-------------------- If A^K +- B^L +- C^M +- D^N +- ... = 0 where A, B, C, D,... K, L, M, N,... are positive integers, where K, L, M, N,... are all greater than the number of (i^j) elements, and where no subset of elements have a null sum, without changing the signs of the elements then A, B, C, D,... must have a common factor. -------------------------------------------------------- Two supplemental remarks: 5) "where no subset of elements have a null sum, without changing the signs of the elements" is the equivalent of the "not twice the same couple" in the previous extension. It means that 8^6 + 8^6 is allowed... but 8^6 - 8^6 is forbidden! 6) With 4 elements, the other equation, not analyzed with the "symmetrical extended" conjecture, is A^K + B^L + C^M - D^N = 0 Here are some examples with common factors: 2^5 + 2^5 + 2^6 - 2^7 = 0 (common factor 2) 3^5 + 3^5 + 3^5 - 3^6 = 0 (common factor 3) (...) 10^6 + 10^7 + 5^8 - 15^6 = 0 (common factor 5) Perhaps something stupid in these extensions? Something missing or bad exmplained? Or a counterexample (means with no common factor) to these new extended conjectures? Christian. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
My "symmetrical extended" conjecture is still alive. But the "full extended" conjecture (of June 20th) is dead. Here is a very impressive counterexample -probably not the smallest!- found by Jaroslaw Wroblewski: a = 2^21 * 3^43 * 5^26 (46 digits) b = 2^120 * 3^246 * 5^149 (258 digits) c = 2^169 * 3^347 * 5^210 (364 digits) d = 2^439 * 3^902 * 5^546 - 1 (945 digits) e = 2^439 * 3^902 * 5^546 + 1 (945 digits) a^21 + b^11 + c^13 + d^6 - e^6 = 0 Jaroslaw is one of the authors of the best "Prime Generating Polynomials" as reported yesterday by Ed Pegg Jr. My new version of a "full extended" conjecture is now: ----------"FULL EXTENDED" CONJECTURE-------------------- If A^K +- B^L +- C^M +- D^N +- ... = 0 where A, B, C, D,... K, L, M, N,... are positive integers, where K, L, M, N,... are all greater than 2*[(number of i^j elements) - 2], and where no subset of elements have a null sum, without changing the signs of the elements then A, B, C, D,... must have a common factor. -------------------------------------------------------- The unchanged "symmetrical extension" is a subset, because 4 = 2*(4-2). Christian. -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de Christian Boyer Envoyé : mardi 20 juin 2006 19:01 À : dasimov@earthlink.net; 'math-fun' Objet : RE: [math-fun] x^n + y^n = z^n + w^n dasimov@earthlink.net wrote:
BEAL'S CONJECTURE: If A^K +B^L = C^M , where A, B, C, K, L, M are positive integers, and K, L, M are all greater than 2, then A, B, and C must have a common prime factor.
Based on recently-discussed equations for which no examples have ever been found, can Beal's conjecture reasonably be extended?
Here are my 2 proposals of extension. ----------"SYMMETRICAL EXTENDED" CONJECTURE------------- If A^K + B^L = C^M + D^N where A, B, C, D, K, L, M, N are positive integers, where K, L, M, N are all greater than 4, and where A^K, B^L, C^M, D^N are not twice the same couple, then A, B, C, and D must have a common factor. -------------------------------------------------------- Four remarks on this "symmetrical extended" conjecture: 1) When D=0, it is the Beal's conjecture. For example: 2^5 + 2^5 = 2^6 (common factor 2) 2) "greater than 2" of the Beal's conjecture is changed in "greater than 4" because we know solutions with no common factor and the fourth power, for example this Euler's solution: 158^4 + 59^4 = 134^4 + 133^4 3) "where A^K, B^L, C^M, D^N are not twice the same couple" is an added condition in order to avoid this kind of obvious solution with NO common factor to the four integers: 8^6 + 9^9 = 4^9 + 27^6 4) "A, B, C, and D must have a common factor." Here are some examples: 124^5 + 31^7 = 62^6 + 31^5 (common factor 31) 126^5 + 126^5 = 63^6 + 63^5 (common factor 63) 132^5 + 33^7 = 66^6 + 33^5 (common factor 33) 148^6 + 37^8 = 111^6 + 74^7 (common factor 37) 204^5 + 12^6 = 192^5 + 156^5 (common factor 12) 217^5 + 186^5 = 93^6 + 62^6 (common factor 31) 254^6 + 254^6 = 127^7 + 127^6 (common factor 127) ----------"FULL EXTENDED" CONJECTURE-------------------- If A^K +- B^L +- C^M +- D^N +- ... = 0 where A, B, C, D,... K, L, M, N,... are positive integers, where K, L, M, N,... are all greater than the number of (i^j) elements, and where no subset of elements have a null sum, without changing the signs of the elements then A, B, C, D,... must have a common factor. -------------------------------------------------------- Two supplemental remarks: 5) "where no subset of elements have a null sum, without changing the signs of the elements" is the equivalent of the "not twice the same couple" in the previous extension. It means that 8^6 + 8^6 is allowed... but 8^6 - 8^6 is forbidden! 6) With 4 elements, the other equation, not analyzed with the "symmetrical extended" conjecture, is A^K + B^L + C^M - D^N = 0 Here are some examples with common factors: 2^5 + 2^5 + 2^6 - 2^7 = 0 (common factor 2) 3^5 + 3^5 + 3^5 - 3^6 = 0 (common factor 3) (...) 10^6 + 10^7 + 5^8 - 15^6 = 0 (common factor 5) Perhaps something stupid in these extensions? Something missing or bad exmplained? Or a counterexample (means with no common factor) to these new extended conjectures? Christian. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Dear friends, Hmmmm... sorry, sorry, sorry, and again 10, 100, 10^10 times sorry... ...a HUGE bug in a branch of my program forgot a lot of cases. Here is my small counterexample with 5 terms of the new "full extended" conjecture (and of course also of the old version). The smallest power was 6 in the impressive Jaroslaws example, here the smallest power is 7: 6^8 + 2^10 - 5^7 - 3^13 - 2^13 = 0 Please, forget this "full extended" conjecture: I cant indefinitely increase the minimum power of such equations!!! My initial idea was probably stupid. Perhaps no minimum power? Only the "symmetrical extended" conjecture with 4 terms seems still alive, but for how many time? Christian. -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de Christian Boyer Envoyé : mardi 27 juin 2006 09:23 À : 'math-fun' Objet : RE: [math-fun] Extensions of Beal's conjecture My "symmetrical extended" conjecture is still alive. But the "full extended" conjecture (of June 20th) is dead. Here is a very impressive counterexample -probably not the smallest!- found by Jaroslaw Wroblewski: a = 2^21 * 3^43 * 5^26 (46 digits) b = 2^120 * 3^246 * 5^149 (258 digits) c = 2^169 * 3^347 * 5^210 (364 digits) d = 2^439 * 3^902 * 5^546 - 1 (945 digits) e = 2^439 * 3^902 * 5^546 + 1 (945 digits) a^21 + b^11 + c^13 + d^6 - e^6 = 0 Jaroslaw is one of the authors of the best "Prime Generating Polynomials" as reported yesterday by Ed Pegg Jr. My new version of a "full extended" conjecture is now: ----------"FULL EXTENDED" CONJECTURE-------------------- If A^K +- B^L +- C^M +- D^N +- ... = 0 where A, B, C, D,... K, L, M, N,... are positive integers, where K, L, M, N,... are all greater than 2*[(number of i^j elements) - 2], and where no subset of elements have a null sum, without changing the signs of the elements then A, B, C, D,... must have a common factor. -------------------------------------------------------- The unchanged "symmetrical extension" is a subset, because 4 = 2*(4-2). Christian. -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de Christian Boyer Envoyé : mardi 20 juin 2006 19:01 À : dasimov@earthlink.net; 'math-fun' Objet : RE: [math-fun] x^n + y^n = z^n + w^n dasimov@earthlink.net wrote:
BEAL'S CONJECTURE: If A^K +B^L = C^M , where A, B, C, K, L, M are positive integers, and K, L, M are all greater than 2, then A, B, and C must have a common prime factor.
Based on recently-discussed equations for which no examples have ever been found, can Beal's conjecture reasonably be extended?
Here are my 2 proposals of extension. ----------"SYMMETRICAL EXTENDED" CONJECTURE------------- If A^K + B^L = C^M + D^N where A, B, C, D, K, L, M, N are positive integers, where K, L, M, N are all greater than 4, and where A^K, B^L, C^M, D^N are not twice the same couple, then A, B, C, and D must have a common factor. -------------------------------------------------------- Four remarks on this "symmetrical extended" conjecture: 1) When D=0, it is the Beal's conjecture. For example: 2^5 + 2^5 = 2^6 (common factor 2) 2) "greater than 2" of the Beal's conjecture is changed in "greater than 4" because we know solutions with no common factor and the fourth power, for example this Euler's solution: 158^4 + 59^4 = 134^4 + 133^4 3) "where A^K, B^L, C^M, D^N are not twice the same couple" is an added condition in order to avoid this kind of obvious solution with NO common factor to the four integers: 8^6 + 9^9 = 4^9 + 27^6 4) "A, B, C, and D must have a common factor." Here are some examples: 124^5 + 31^7 = 62^6 + 31^5 (common factor 31) 126^5 + 126^5 = 63^6 + 63^5 (common factor 63) 132^5 + 33^7 = 66^6 + 33^5 (common factor 33) 148^6 + 37^8 = 111^6 + 74^7 (common factor 37) 204^5 + 12^6 = 192^5 + 156^5 (common factor 12) 217^5 + 186^5 = 93^6 + 62^6 (common factor 31) 254^6 + 254^6 = 127^7 + 127^6 (common factor 127) ----------"FULL EXTENDED" CONJECTURE-------------------- If A^K +- B^L +- C^M +- D^N +- ... = 0 where A, B, C, D,... K, L, M, N,... are positive integers, where K, L, M, N,... are all greater than the number of (i^j) elements, and where no subset of elements have a null sum, without changing the signs of the elements then A, B, C, D,... must have a common factor. -------------------------------------------------------- Two supplemental remarks: 5) "where no subset of elements have a null sum, without changing the signs of the elements" is the equivalent of the "not twice the same couple" in the previous extension. It means that 8^6 + 8^6 is allowed... but 8^6 - 8^6 is forbidden! 6) With 4 elements, the other equation, not analyzed with the "symmetrical extended" conjecture, is A^K + B^L + C^M - D^N = 0 Here are some examples with common factors: 2^5 + 2^5 + 2^6 - 2^7 = 0 (common factor 2) 3^5 + 3^5 + 3^5 - 3^6 = 0 (common factor 3) (...) 10^6 + 10^7 + 5^8 - 15^6 = 0 (common factor 5) Perhaps something stupid in these extensions? Something missing or bad exmplained? Or a counterexample (means with no common factor) to these new extended conjectures? Christian. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
in the same Catalanian gist as a few years back : (2 + k)/((1/2*k*(3 + k))!*(1/2*(-4 + k + k^2))!) + (2 - k)/((1/2*k*(1 + k))!*(1/2*(-4 + 3*k + k^2))!) + 4/((1/2*(-6 + k + k^2))!*(1 + 1/2*k*(3 + k))!) + k/((1/2*(-6 + k + k^2))!*(1 + 1/2*k*(3 + k))!) == k/((1/2*(-2 + k + k^2))!*(1/2*(-2 + 3*k + k^2))!) go figure! W.
From Wouter: in the same Catalanian gist as a few years back :
(2 + k)/((1/2*k*(3 + k))!*(1/2*(-4 + k + k^2))!) + (2 - k)/((1/2*k*(1 + k))!*(1/2*(-4 + 3*k + k^2))!) + 4/((1/2*(-6 + k + k^2))!*(1 + 1/2*k*(3 + k))!) + k/((1/2*(-6 + k + k^2))!*(1 + 1/2*k*(3 + k))!) == k/((1/2*(-2 + k + k^2))!*(1/2*(-2 + 3*k + k^2))!) go figure! W. (c53) factor(minfactorial((2+k)/((k*(3+k)/2)!*((-4+k+k^2)/2)!) +(2-k)/((k*(1+k)/2)!*((-4+3*k+k^2)/2)!) +4/(((-6+k+k^2)/2)!*(1+k*(3+k)/2)!) +k/(((-6+k+k^2)/2)!*(1+k*(3+k)/2)!) = k/(((-2+k+k^2)/2)!*((-2+3*k+k^2)/2)!))); time= 0 msec. 2 2 2 k + k - 6 (d53) 8 k/((k - 1) (k + 2) (k + k - 4) (k + 3 k - 2) (----------)! 2 2 k + 3 k - 4 2 2 (------------)!) = 8 k/((k - 1) (k + 2) (k + k - 4) (k + 3 k - 2) 2 2 2 k + k - 6 k + 3 k - 4 (----------)! (------------)!) 2 2 --rwg
participants (4)
-
Christian Boyer -
dasimov＠earthlink.net -
R. William Gosper -
wouter meeussen