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+2n2m+1 for 4mn+2n2m=1 m=2n+1/4n+2 n=2m1/4m+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+2n2m+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 (12*(1/2))^3=0 3n^2+3n+1 3*((12)/3)^2+3*((12)/3)+1=37 interprétation 4mn+2n2m+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 6427=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 lefthand side is zero if the remaining cubes are integers.

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 
Global Moderator Joined: Dec 2006 Posts: 21,113 Thanks: 2327  
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(x3)(x+32) y²=x³+29x²96x 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?


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