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

 Math Software Math Software - Mathematica, Matlab, Calculators, Graphing Software

 September 19th, 2018, 08:11 AM #11 Newbie   Joined: Sep 2018 From: tunis Posts: 27 Thanks: 0 (1 - 2m)^3 = (-8) m^3 + 12 m^2 - 6 m + 1 = 3 n^2 + 3 n + 1 how to find n? Last edited by skipjack; September 19th, 2018 at 10:31 AM.
 September 19th, 2018, 10:32 AM #12 Global Moderator   Joined: Dec 2006 Posts: 20,805 Thanks: 2150 The right-hand side is the difference between the cubes of consecutive integers; as this difference is a perfect cube, it must be 1, so n is -1 or 0, and m = 0. Thanks from Ak23
 September 19th, 2018, 03:01 PM #13 Newbie   Joined: Sep 2018 From: tunis Posts: 27 Thanks: 0 ( b +1)^3-b^3=The right-hand side but the difference between consecutive cubes is not a cube. 1^3+(3b^2+3b+1)=3b^2+3b+2 1+ 3*1+3.1+1 = 3 +3 +2 1+ 7 = 8 =2^3 n=-1 et 0 m=0 is it a complete study on this subject? Last edited by skipjack; September 19th, 2018 at 08:15 PM.
September 19th, 2018, 08:26 PM   #14
Global Moderator

Joined: Dec 2006

Posts: 20,805
Thanks: 2150

Quote:
 Originally Posted by Ak23 . . . but the difference between consecutive cubes is not a cube.
The equation states that its right-hand side, 3 n^2 + 3 n + 1, which is the difference between n³ and (n + 1)³, equals its left-hand side, which is (1 - 2m)³. Consecutive cubes differ by a perfect cube only when they are -1 and 0 or 0 and 1.

 September 20th, 2018, 12:56 AM #15 Newbie   Joined: Sep 2018 From: tunis Posts: 27 Thanks: 0 thank you very much I bring here the support of 2n + 1 = x ^ 2? if x is odd we can write it 1-2m x ^ 2 = 4m ^ 2 - 4m + 1 = 2 (2m ^ 2 - 2m) + 1 = 2n + 1 and 4m ^ 2 - 4m + 1 = 2n + 1 (4m ^ 2 - 4m) = 2n (4m ^ 2 - 4m) / 2 = n 2m ^ 2 - 2m = n 2 (m ^ 2 - m) = n From what you showed me. Do we get this with (1-2m)^3 if not show how to do it? Last edited by skipjack; September 20th, 2018 at 03:05 AM. Reason: to correct typos and sign error
 September 20th, 2018, 02:34 AM #16 Newbie   Joined: Sep 2018 From: tunis Posts: 27 Thanks: 0 If x^2 = 2n + 1 = (1 - 2m)^2 sqrt(2n + 1) = |1 - 2m| n = ((-2m + 1)^2 - 1)/2 for any m $\in$ N Last edited by skipjack; September 20th, 2018 at 02:54 AM. Reason: to correct
 September 20th, 2018, 03:00 AM #17 Global Moderator   Joined: Dec 2006 Posts: 20,805 Thanks: 2150 (1 - 2m)³ = (1 - 2m)(1 - 2m)² = (1 - 2m)(1 - 4m + 4m²) = 1 - 6m + 12m² - 8m³ In your last post, I made some corrections. It would make more sense to say "for any m $\small\in$ ℤ" than "for any m $\small\in$ N". Thanks from Ak23
 September 20th, 2018, 03:43 AM #18 Newbie   Joined: Sep 2018 From: tunis Posts: 27 Thanks: 0 1-2m=(3n^2+3n+1)^1/3 there are no solutions because there is the negative root
September 20th, 2018, 03:53 AM   #19
Newbie

Joined: Sep 2018
From: tunis

Posts: 27
Thanks: 0

Quote:
 It would make more sense to say "for any m $\small\in$ ℤ" than "for any m $\small\in$ N".
I did not understand

September 20th, 2018, 04:34 AM   #20
Global Moderator

Joined: Dec 2006

Posts: 20,805
Thanks: 2150

There's no reason to exclude negative values of m.

Quote:
 Originally Posted by Ak23 . . . there are no solutions because there is the negative root
There is no negative cube root of 3n² + 3n + 1.

 Tags show

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post mared Geometry 7 May 16th, 2015 09:50 PM Few_But_Ripe Complex Analysis 1 November 11th, 2011 09:08 AM notnaeem Real Analysis 4 August 16th, 2010 12:32 PM naserellid Algebra 2 August 15th, 2010 02:20 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top