My Math Forum how to show that 2n + 1 = x ^ 2 ?

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 September 20th, 2018, 11:47 AM #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 11:49 AM.
 September 20th, 2018, 01:09 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, 02:27 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, 02:51 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 05:19 PM.
 September 20th, 2018, 05:25 PM #25 Global Moderator   Joined: Dec 2006 Posts: 20,834 Thanks: 2161 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, 07:55 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, 05:50 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, 06:02 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, 02:58 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 03:14 AM.
 September 22nd, 2018, 03:16 AM #30 Global Moderator   Joined: Dec 2006 Posts: 20,834 Thanks: 2161 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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