March 31st, 2017, 01:26 PM  #1 
Newbie Joined: May 2016 From: Pretoria Posts: 7 Thanks: 0  Fermat's last theorem
FERMATâ€™S LAST THEOREM an + bn = cn No three positive integers a, b and c satisfy the equation for value of n greater than 2 21power3 + 35power3 = 44power3 9261 + 42875 = 85184 Divide by 90 9261/90 42875/90 85184/90 equals 102.90 476.38 946.48 Round to 103 476 946 Times 90 103*90 476*90 946*90 Equals 9270 42840 85140 9261 42875 85184 Minus big from small ( stay within 1 to 45 ) 9 + 35 = 44 44 = 44 Lourie 31 March 2017 
March 31st, 2017, 01:48 PM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,797 Thanks: 715 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Try again. Dan Last edited by topsquark; March 31st, 2017 at 01:50 PM.  
April 1st, 2017, 12:04 AM  #3 
Newbie Joined: May 2016 From: Pretoria Posts: 7 Thanks: 0 
Let us put it this way. No three positive integers a,b, and c satisfy the equation of [ an+bn=cn ] for value of n bigger than 2. meaning: an + bn can not equal cn according to the theorem? Last edited by Lourie; April 1st, 2017 at 12:27 AM. 
April 1st, 2017, 12:37 AM  #4 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,597 Thanks: 546 Math Focus: Yet to find out.  

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