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September 15th, 2017, 02:00 AM  #1 
Senior Member Joined: Oct 2011 Posts: 135 Thanks: 1  Fermat's theorem. Proof by 2 operations Fermat's theorem. Proof by 2 operations The essence of the contradiction. The hypothetical Fermat's equality is contradictory between the second digits of the factors of the number $А$. All calculations are done with numbers in base n, a prime number greater than 2. The notations that are used in the proofs: $A'$, $A''$ – the first, the second digit from the end of the number $A$; $A_2$ is the twodigit ending of the number A (i.e. $A_2=A$ mod $n^2$); Consider the Fermat’s equality in the base case (its properties 2°3° are proved here: The Proof of Fermat's Last Theorem for the Base Case, viXra.org ePrint archive, viXra:1707.0410) for coprimes positive $A$, $B$, $C$; $A'≠0$, prime $n$, $n>2$: 1°) $A^n=C^nB^n$ [$=(CB)P$], where (as is known) 2°) $A'≠0$, $CB=a^n$, $P=p^n$, $A=ap$, $p'=1$, $a'≠0$, $(a^n)'=a'$, $(a'^{n1})'=1$ (Fermat's small theorem); 3°) $(A+BC)_2=0$, from here 3a°) $(ap)_2=(a^n)_2$ and therefore 3b°) $p_2=(a^{n1})_2$. 4°) If $p''=0$, then we multiply term by term equality 1° by such $g^{nn}$, that $p''≠0$. Properties 2°3° are preserved, and we leave the notation of the numbers as before. And now the proof itself FLT. We represent the endings $a_2$ and $p_2$ in the form: $a2=(xn+a'^n)_2$ and $p_2=p''n+1$, where x and y are digits. First we substitute these values of the endings in the lefthand side of the equality 3a°: 5°) $[(xn+a'^n)(p''n+1)]_2=(a'^n)_2$, from here 5a°) $(a'np''n+xn)_2=0$, or (see 2°) $a'p''+x=0$ (mod n). Now, we substitute the value of $a_2$ in the righthand side of the equality 3b°: 6°) $(xn+a'n)^{n1}_2=[(n1)xna'^{n2}+1]2=(nxa'^{n2}+1)_2=(nxa'^{n1}/a'+1)_2$. From 3b° we have: 6a°) $xa'^{n1}/a'+p''=0$ (mod $n$), or $xa'^{n1}+a'p''=0$ (mod $n$), or $x+a'p''=0$ (mod $n$), It follows from 5a° and 6a° that $x=y=0$, which contradicts to 2° and 4°. From this contradiction follows the truth of the FLT. (Mezos, France. 4 September 2017) 
September 16th, 2017, 03:44 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,731 Thanks: 707 
What? That is not the way "proof by contradiction" works! "Proof by contradiction" starts by assuming that what is to be proved is not true then arrives at two statements that contradict one another. Since two contradictory statements cannot both be derived from a true statement, it follows that the original assumption, that what s to be proved is not true, must be false. And therefore that what was to be proved is true. Here, you don't do that. You assert 2° and 4° which you say have already been proved (I will grant you that) and then prove 5a° and 6a° which you say contradicts 2° and 4°. But you never assert that "Fermat's last theorem" is false to derive those. If you have done everything correctly, then you have disproved Fermat's last theorem! Oh, wait. I just realized this was posted by Victor Sorokin. What a waste of my time! 
September 17th, 2017, 01:30 AM  #3  
Senior Member Joined: Oct 2011 Posts: 135 Thanks: 1  Quote:
 
September 17th, 2017, 04:11 AM  #4 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,354 Thanks: 464 Math Focus: Yet to find out.  
September 17th, 2017, 04:28 AM  #5 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,500 Thanks: 757  
September 23rd, 2017, 01:38 PM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,731 Thanks: 707 
Ah! It took me a moment to figure out what "method from the prive" meant since I did not know what "prive" meant. Now, thanks to romsek, I realize it is a "method from the privy"! All is clear now.

September 23rd, 2017, 01:47 PM  #7 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,500 Thanks: 757 
we aim to please. ... especially in the privy 

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