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 September 15th, 2017, 03:00 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 two-digit 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 e-Print archive, viXra:1707.0410) for co-primes positive $A$, $B$, $C$; $A'≠0$, prime $n$, $n>2$: 1°) $A^n=C^n-B^n$ [$=(C-B)P$], where (as is known) 2°) $A'≠0$, $C-B=a^n$, $P=p^n$, $A=ap$, $p'=1$, $a'≠0$, $(a^n)'=a'$, $(a'^{n-1})'=1$ (Fermat's small theorem); 3°) $(A+B-C)_2=0$, from here 3a°) $(ap)_2=(a^n)_2$ and therefore 3b°) $p_2=(a^{n-1})_2$. 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 left-hand 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 right-hand side of the equality 3b°: 6°) $(xn+a'n)^{n-1}_2=[(n-1)xna'^{n-2}+1]2=(-nxa'^{n-2}+1)_2=(-nxa'^{n-1}/a'+1)_2$. From 3b° we have: 6a°) $-xa'^{n-1}/a'+p''=0$ (mod $n$), or $-xa'^{n-1}+a'p''=0$ (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, 04:44 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,875 Thanks: 766 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, 02:30 AM   #3
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 Originally Posted by Country Boy 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!
The proof is carried out by the method from the prive. You do not understand?

September 17th, 2017, 05:11 AM   #4
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 Originally Posted by victorsorokin The proof is carried out by the method from the prive. You do not understand?
Designer sunglasses?

September 17th, 2017, 05:28 AM   #5
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 Originally Posted by victorsorokin The proof is carried out by the method from the privy. You do not understand?
ftfy

 September 23rd, 2017, 02:38 PM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,875 Thanks: 766 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, 02:47 PM #7 Senior Member     Joined: Sep 2015 From: USA Posts: 1,653 Thanks: 840 we aim to please. ... especially in the privy

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