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 September 20th, 2018, 12:47 PM #21 Newbie   Joined: Sep 2018 From: tunis Posts: 27 Thanks: 0 (1 - 2m)³ = (1 - 2m)(1 - 2m)² =(1 - 2m)(2n+1) =-4mn+2n-2m+1 for 4mn+2n-2m=-1 m=2n+1/4n+2 n=2m-1/-4m+2 replacing m and n by its value can we find the solution? Last edited by Ak23; September 20th, 2018 at 12:49 PM. September 20th, 2018, 02:09 PM #22 Newbie   Joined: Sep 2018 From: tunis Posts: 27 Thanks: 0 I did all the calculation that gives -4mn+2n-2m+1 (-4)*(1/2)*((-12)/3)+2*((-12)/3)-2*(1/2)+1=0 m=1/2 n=(-12)/3 September 20th, 2018, 03:27 PM #23 Newbie   Joined: Sep 2018 From: tunis Posts: 27 Thanks: 0 (1 - 2m)³ (1-2*(1/2))^3=0 3n^2+3n+1 3*((-12)/3)^2+3*((-12)/3)+1=37 interprétation -4mn+2n-2m+1 admits for solution 0 and 37 it is the minimum 0 for up to 37 0 1 7 19 37 7 and 19 are not cubes the 1 yes which gives 64--27=37 3n^2+3n+1=1 = (1 - 2m)³=1 n=0 n=-1 m=0 September 20th, 2018, 03:51 PM #24 Newbie   Joined: Sep 2018 From: tunis Posts: 27 Thanks: 0 I did the test of the theorem a ^ 3 + b ^ 3 = c ^ 3 based on 3n ^ 2 + 3n + 1 algorithm generating the difference between the consecutive cubes the outgoing resultant is put under the cubic root if the whole spell is true. If it interests you, I post algo. En following your method. I understood what it is. Last edited by skipjack; September 20th, 2018 at 06:19 PM. September 20th, 2018, 06:25 PM #25 Global Moderator   Joined: Dec 2006 Posts: 21,113 Thanks: 2327 It''s known that a³ + b³ = c³ has no solution in positive integers, which implies that at least one of the cubes on the left-hand side is zero if the remaining cubes are integers. Thanks from Ak23 September 21st, 2018, 08:55 AM #26 Newbie   Joined: Sep 2018 From: tunis Posts: 27 Thanks: 0 a³+b³+c³=d³ 3³+4³+5³=6³ but we can find the solution with 3 cubes. September 21st, 2018, 06:50 PM #27 Newbie   Joined: Sep 2018 From: tunis Posts: 27 Thanks: 0 if a+b=c aⁿ+bⁿ=cⁿ it's false for any number. c×aⁿ⁻¹+c×bⁿ⁻¹=cⁿ 2×5²+3×5²=5³ 50 + 75 =125 September 21st, 2018, 07:02 PM   #28
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Quote:
 Originally Posted by Ak23 if a+b=c aⁿ+bⁿ=cⁿ it's false for any number.
If a = 0 and b = c, it's true for any values of c and n. September 22nd, 2018, 03:58 AM #29 Newbie   Joined: Sep 2018 From: tunis Posts: 27 Thanks: 0 In proof of Wallis theorem, you have: A + B = C ABC ≠ 0 A,B and C are relatively primes B is divisible by 32 A = 3 mod 4 = 3 C = 1 mod 4 = 1 B = 32 n is prime ≤5 Aⁿ×Bⁿ×Cⁿ = (A×B×C)ⁿ solving this with elliptic curves A + B ≠ C, y²=x(x-3)(x+32) y²=x³+29x²-96x 3+32=1 Last edited by skipjack; September 22nd, 2018 at 04:14 AM. September 22nd, 2018, 04:16 AM #30 Global Moderator   Joined: Dec 2006 Posts: 21,113 Thanks: 2327 What theorem of Wallis are you referring to? Is this related to your previous posts or are you digressing into a new topic? Tags show Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mared Geometry 7 May 16th, 2015 10:50 PM Few_But_Ripe Complex Analysis 1 November 11th, 2011 10:08 AM notnaeem Real Analysis 4 August 16th, 2010 01:32 PM naserellid Algebra 2 August 15th, 2010 03:20 AM

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