20200712, 04:24  #34 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2·3^{3}·173 Posts 
Formal note from a moderator:
Do not create any more threads that cover the same subject. Stop posting the same content repeatedly. This is your only warning. 
20200712, 04:39  #35 
Jun 2020
10111_{2} Posts 
I'm sorry!
I'm so sorry! I don't know how to supplement or modify the original article.
Last fiddled with by Zcyyu on 20200712 at 04:53 
20200712, 05:28  #36 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2×3^{3}×173 Posts 
You have a limited time from when you make a post to modify it. If you need to make corrections after that you might quote just the part that needs fixing and then indicate that the you are posting a change. And then state the update.

20201105, 23:42  #37  
Jun 2020
17_{16} Posts 
Quote:
congruent number order of group The upper bound estimate 80451910 Have certain that it is a group of 32 orsders 32 99015854 Have certain that it is a group of 32 orsders 32 124245334 Have certain that it is a group of 32 orsders 32 144362005 Have certain that it is a group of 32 orsders, 128,The order could be > 32 251741886 Have certain that it is a group of 32 orsders 32 253375590 Have certain that it is a group of 32 orsders 32 277872990 Have certain that it is a group of 32 orsders 32 433300054 Have certain that it is a group of 32 orsders 32 565004406 Have certain that it is a group of 32 orsders 32 840075621 Have certain that it is a group of 32 orsders 32 904789886 Have certain that it is a group of 32 orsders 32 936966030 Have certain that it is a group of 32 orsders 32 Last fiddled with by Zcyyu on 20201105 at 23:46 Reason: A new discovery 

20210115, 10:16  #38  
Jun 2020
23 Posts 
we have new found about congruent number
Quote:
Since 157 is a prime number with modulo 8 and 5, there must be a solution of type (1.1.157.1) scheme That is, quaternion quadratic equations Q1) x^2+y^2=157*z^2 (1) Q4) w^2x^2=2*y^2 (2) The general solution of equation (2) is x =  u ^ 22 * v ^ 2 , y = 2uv, w = u * u + 2 * v * v Substituting the expression of X, y for (1), 157 * Z ^ 2 = u ^ 4 + 4 * V ^ 4 (3) When u = 356411, v = 571761, equation (3) holds in the range of positive integers z=40,75981,25052 x=52,67710,95761 y=40,75981,25052 w=52,67710,95761 last st=x^2=52,67710,95761^2 s+t=w^2=78,08714,68723^2 s=157*z^2=157*5,31566,61805^2 t=y^2 =40,75981,25052^2 Last fiddled with by Zcyyu on 20210115 at 10:21 Reason: I hope teachers and friends like it 

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