
Number Theory Number Theory Math Forum 
 LinkBack  Thread Tools  Display Modes 
March 5th, 2018, 06:34 PM  #11 
Senior Member Joined: May 2016 From: USA Posts: 922 Thanks: 368 
Admittedly off topic. I have long wondered what attracts people to things like Fermat's Last Theorem. Part, I suspect, is an unwarranted intuition that things easy to express are easy to prove. Part is a simple desire to solve a puzzle; I have to believe that desire motivated Wiles and the many skilled mathematicians who had sought a proof before him. And for many decades, I am confident that people did not understand the actual complexity of the problem. Now that a proof has been found and the complexities involved in finding a proof have been delineated, the continuing interest baffles me. "Wiles found a proof and x years later I did too" seems a rather puerile goal. Or am I missing something? 
March 19th, 2018, 06:31 AM  #12 
Member Joined: Aug 2015 From: Chiddingfold, Surrey Posts: 47 Thanks: 2 Math Focus: Number theory, Applied maths 
See below for interesting results of further extensive research. Comments welcome. ODD POWERS AND FLT. Although it is known that Fermatâ€™s last theorem is true for all powers higher than two, an attempt to disprove this with odd powers, might be to show that there is at least one set of three integers A>B>C where all of the following statements are true : 1. An MOD (B +C) =0 2. Bn MOD (AC) = 0 3. CnMod(AB) = 0 4. (Bn+Cn) MOD A =0 5. (AnBn) MOD C= 0 6. (An Cn) MOD B = 0 7. A, B & C all different 8. A, B & C relatively prime 9. A.B.C MOD 2 =0 10. When 2n+1 is prime then A.B.C MOD (2n +1) = 0 (Sophie Germain, but under Nom de plume until 1807) 11. A,B.C not in arithmetical progression (Bottari (1907), Goldziher (1913), Milhajinee (1952), Rameswar Rao (1969) Alternatively if it can be shown that for all possible sets of A,B & C, at least one of these statements cannot be true, then we would have a new proof of the theorem. However a comprehensive computer aided examination of all 35,820,200 possible sets of A,B & C from 3,2,1 to 600,699,698 and powers from 3 to 21 have shown that all the above statements are true in the number of triads as follows: Power 3 20,522 Power 5 13,659 Powers 7, 11, 17, 19 54,749 (not statement 10) Power 9 8,245 Power 11 6,892 Power 15 5,207 Power 21 3,702 (The sets of triads where 2p+1 is not prime are identical , and of which, the others subsets.) For comparison the number or primitive Pythagoras triples up to A,B, C = 593,465,368 is only 95. When all statements other than 1 are true, (statement 11 with A,B,C = 5,4,3) So how can that be since FLT is true? The answer is that there is one more statement that is true which confirms that FLT is true (Apart from A^n  B^n  C^n >< 0). I leave it to the reader to discover that other statement. Clue : check ONE more odd prime power. Last edited by magicterry; March 19th, 2018 at 06:38 AM. Reason: update 
March 19th, 2018, 02:16 PM  #13 
Senior Member Joined: Feb 2010 Posts: 643 Thanks: 111  
March 19th, 2018, 02:45 PM  #14 
Senior Member Joined: Sep 2015 From: USA Posts: 1,790 Thanks: 922  
March 19th, 2018, 03:21 PM  #15  
Senior Member Joined: Aug 2012 Posts: 1,777 Thanks: 482  Quote:
Today I learned! What do you think is crazy? That Trump became president? Or that the only viable options available to the American people were Trump and Hillary? Or that the DNC rigged its own process to nominate the only person in the country who could manage to lose to Trump? I'd go with all of the above.  

Tags 
attempt, find, flt, proof, simple 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Collatz Conjecture Unrevised Proof Attempt  relaxursoul  Number Theory  0  November 14th, 2017 12:20 PM 
harmonic sum attempt thing of a jig  tejolson  Real Analysis  1  March 17th, 2013 07:54 PM 
My attempt at some limit calculations  Math4dummy  Calculus  3  February 22nd, 2012 08:56 PM 
??(x+1)/(x^2+4x+8) dx [Attempt to solve]  Riazy  Calculus  2  January 21st, 2011 08:45 AM 
Integral $x*e^[x^(2)] * dx. With an attempt to solve it.  Riazy  Calculus  1  January 15th, 2011 05:23 PM 