April 10th, 2017, 06:35 PM  #121 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,591 Thanks: 546 Math Focus: Yet to find out.  
April 12th, 2017, 07:30 AM  #122  
Senior Member Joined: Oct 2011 Posts: 136 Thanks: 1  Quote:
The answer to all questions is guaranteed.  
April 14th, 2017, 06:54 AM  #123 
Senior Member Joined: Oct 2011 Posts: 136 Thanks: 1  Are the axioms of mathematics wrong? WHY a raising of the digit by digit EQUALITY (in basis n) $\displaystyle 2^{nn}==(xn^2+2^{nn})(yn^2+1)$ mod $\displaystyle n^3$, where n is a prime number, n>2, x and y are digits y ≠ 0, [and x + 2y==0 mod n], in the (n1)th power, it turns into an INEQUALITY [and x + 4y==0 mod n]? 
April 19th, 2017, 07:43 AM  #124 
Senior Member Joined: Oct 2011 Posts: 136 Thanks: 1 
nn=n*n=n^2

May 6th, 2017, 11:18 AM  #125 
Senior Member Joined: Oct 2011 Posts: 136 Thanks: 1  Correction. Correction. Unfortunately, in the case of logical inference, I did not notice the exponent in the factors of absentmindedness, and therefore the lemma on power endings is incorrect. It is a pity that none of the many thousands of readers have indicated an oversight. Fortunately, the error was immediately eliminated, and with a significant simplification of the proof, but the publication will have to start anew. 
May 6th, 2017, 01:25 PM  #126 
Senior Member Joined: Oct 2011 Posts: 136 Thanks: 1  The authentic proof of the Fermat's Last Theorem In Memory of my MOTHER All calculations are done with numbers in base n, a prime number greater than 2. The notations that are used in the proofs: $\displaystyle A'$ – the first digit from the end of the number $\displaystyle A$; $\displaystyle A_{[k]}$ is the kdigit ending of the number A (i.e. $\displaystyle A_{[k]} = A$ mod n$\displaystyle ^k)$; $\displaystyle nn=n*n=n^2$. Here the well known properties of Fermat’s equality for natural and coprime numbers $\displaystyle A,B,C$: 1°) $\displaystyle A^n=C^nB^n$ [$\displaystyle =(CB)P$] //and $\displaystyle B^n=C^nA^n$ [$\displaystyle =(CA)Q$], C$\displaystyle ^n=A^n+B^n$ [$\displaystyle =(A+B)R$]//. From here 1a°) $\displaystyle (CB)P+(CA)Q(A+B)R=0$, where we denote with the letters $\displaystyle a, b, c$ the greatest common divisors, respectively, of the pairs of numbers $\displaystyle (A, CB), (B, CA), (C, A+B).$ Then, 2°) if $\displaystyle (ABC)'≠0,$ then $\displaystyle CB=a^n, P=p^n, A=ap$; $\displaystyle CA=b^n, Q=q^n, B=bq$; $\displaystyle A+B=c^n, R=r^n, C=cr$; 2a°) but if, for example, $\displaystyle B_{[k]}=0$, but $\displaystyle B_{[k+1]}≠0$, then $\displaystyle (CA)_{[kn1]}=0$, where $\displaystyle kn1>k$ (what is important in 82°); 3°) the number $\displaystyle U=A+BC=un^k$, where k>1. From here we find that ($\displaystyle A+B)(CB)(CA)=2U$ and (if $\displaystyle k=2$) 3a1°) $\displaystyle A_{[2]}=(a^n)_{[2]}=(a'^n)_{[2]}$, $\displaystyle B_{[2]}=(b^n)_{[2]}=(a'^n)_{[2]}$, $\displaystyle C_{[2]}=(c^n)_{[2]}=(a'^n)_{[2]}$; consequently (see 5°), 3b1°) $\displaystyle (A^n)_{[3]}=(a'^{nn})_{[3]}$, $\displaystyle (B^n)_{[3]}=(b'^{nn})_{[3]}$; $\displaystyle (C^n)_{[3]}=(c'^{nn})_{[3]}$; consequently (see 1°), 3c1°) $\displaystyle (a'^{nn})_{[3]}+(b'^{nn})_{[3]}(c'^{nn})_{[3]}=((A^n)_{[3]}+(B^n)_{[3]}(C^n)_{[3]})_{[3]}=0$. 4°) The digit $\displaystyle A^n_{(k+1)}$ is uniquely determined by the ending of $\displaystyle A_{[k]}$ (a simple consequence of the binomial theorem). 5°) Lemma. Every prime divisor of the factor R of the binomial A$\displaystyle ^{n^k}+B^{n^k}=(A^{n^{k1}}+B^{n^{k1}})R$, where $\displaystyle k>1$, the natural numbers A and B are mutually prime and the number A+B is not a multiple of a prime $\displaystyle n>2$, can be presented as: $\displaystyle m=dn^k+1$. ([url=http://www.mathforum.ru/forum/read/1/20535/page/63/]) ++++ And now the proof itself FLT. It consists of an endless sequence of cycles in which the exponent k (3°), starting with the value 2, increases in 1. Thus, we consider the equality 3c1° on threedigit endings: 6°) $\displaystyle (a'^{nn})_{[3]}=((c'^{nn})_{[3]}(b'^{nn})_{[3]})_{[3]}=[((c'^{nn})_{[3]}(b'^{nn})_{[3]})P_{[3]}]_{[3]}$, where: a) according to Lemma 5°, each prime factor of the number P ends with 01, and b) each prime cofactor of the number P is in degree n. And consequently (see 4°) $\displaystyle P_{[3]}=001$. Similarly and $\displaystyle Q_{[3]}=R_{[3]}=001$. And now from the equality 1a°, we have: $\displaystyle [(CB)+(CA)(A+B)]_{[3]}=0$. Where 72°) the number $\displaystyle U=A+BC=un^3$, so NOW k=3, and we compose the source data for the next cycle (increasing k by 1): 3a2°) $\displaystyle A_{[3]}=(a^{nn})_{[3]}=(a'^{nn})_{[3]}$, B$\displaystyle _{[3]}=(b^{nn})_{[3]}=(a'^{nn})_{[3]}$, $\displaystyle C_{[3]}=(c^{nn})_{[3]}=(a'^{nn})_{[3]}$; consequently (see 4°), 3b2°) $\displaystyle (A^n)_{[4]}=(a'^{nnn})_{[4]}$, $\displaystyle (B^n)_{[4]}=(b'^{nnn})_{[4]}$; $\displaystyle (C^n)_{[4]}=(c'^{nnn})_{[4]}$; consequently (see 1°), 3c2°) $\displaystyle ((A^n)_{[4]}+(B^n)_{[4]}(C^n)_{[4]})_{[4]}=(a'^{nnn})_{[4]}+(b'^{nnn})_{[4]}(c'^{nnn})_{[4]}=0$. [And if, for example, $\displaystyle B_{[2]}=0$, then $\displaystyle (CA)_{[kn1]}=0$ and from 1a° we find that $\displaystyle 2B_{[3]}=0$ и $\displaystyle U_{[3]}=0$.] Then we repeat the reasoning of 6°7° with obtaining k=4 and go to the next cycle. And so on to infinity. Finally, the end of the numbers A, B, C take the following form: 8°) $\displaystyle A_{[k+1]}=(a'^{n^k})_{[k+1]}$, $\displaystyle B_{[k+1]}=(b'^{n^k})_{[k+1]}$, $\displaystyle C_{[k+1]}=(c'^{n^k})_{[k+1]}$, where k tends to infinity, that indicates the impossibility of the equality of 1° and of the truth of the FLT. ============== Victor Sorokine. Mezos. 5.5.2017 =============== Control text in Word, see: Вестник Мечты  Главная страница 
June 4th, 2017, 06:16 AM  #127 
Senior Member Joined: Oct 2011 Posts: 136 Thanks: 1  Fermat's Last Theorem. Proof for school children
Proof of the Fermat's theorem (1 phrase) From the known properties of the Fermat’s equality follows (without calculations): If the second digits of all the prime factors of the numbers A, B, and C are reduced to zero, then the new reduced numbers A°, B°, C° become infinitely large. The question is: is this the truth of the FLT? No one answered this question. And what will you say? (Вестник Мечты  Главная страница ). +++++++++++++++++++++++++++++++++++++++++++++++++ Fermat's Last Theorem. Proof for school children In Memory of my MOTHER The essence of the contradiction. In Fermat's equality, after decreasing the second digits in prime factors of the numbers to zero, the new reduced numbers become infinitely large. Therefore, by adopting the simplest properties of Fermat's equality for truth, the proof of FLT can be considered school. All calculations are done with numbers in base n, a prime number greater than 2. The notations that are used in the proofs: , , – the first, the second, the kth digit from the end of the number ; is the kdigit ending of the number (i.e. ); . Here the well known properties of Fermat’s equality for natural and coprime numbers : 1°) [] //and [], [//. From here 1a°) , where we denote with the letters the greatest common divisors, respectively, of the pairs of numbers , , . Then, 2°) if ≠0, then , , ; , , ; , , ; 3°) the number , where , from here ; 3a°) but if, for example, , but ≠0, then , where , and in the the equality 3b°) (see 3°) the number . At the start (that is, in the first cycle  see below the beginning of the proof), for 4a1°) , , ; consequently (see 5°), 4b1°) , ; ; consequently (see 1° and 2°), 4c1°) , from here (from the expansion formulas and 2°) 4d1°) et . 5°) The digit is uniquely determined by the ending of (a simple consequence of the binomial theorem). That is, the end , etc, do not depend on the digit ! (The decisive lemma; perhaps it should be considered as the Fermat's Middle Theorem.) And now the proof itself FLT. It consists of an endless sequence of cycles in which the exponent k (3°), starting with the value 2, increases in 1. It is easy to see (see 5°) that the equalities 4d° are CONTRADICT, since in their left parts the second digits and in the bases are absent, and in the right  PRESENT. But under the condition: this contradiction is eliminated. And then . Similarly, . And now from the equalities 1a° and 3b° we have: 6°) . Where (see 3°) 7) the number , so NOW , and we compose the source data 4a°4d° for the next cycle (increasing in the formulas 41° the number k and indices by 1). [And if, for example, , then , and from 1a° we find that и .] The next cycle is analyzed in a completely similar way. And so on to infinity. Finally, the end of the numbers take the following form: 8°) , where k tends to infinity. And if we restore the values of the second digits in the factors then the infinite values of the numbers only increase, that indicates the impossibility of the equality of 1° and of the truth of the FLT. ============== Victor Sorokine. Mezos. May 11, 2017 /P.S. This is the second, simpler proof of the WTF. The first is dated 5.5.2017./ Last edited by victorsorokin; June 4th, 2017 at 06:20 AM. 
June 4th, 2017, 07:38 AM  #128 
Senior Member Joined: Aug 2012 Posts: 1,888 Thanks: 525 
Can you walk us through a simple example for, say, n = 3?

June 5th, 2017, 07:48 AM  #129 
Senior Member Joined: May 2013 Posts: 115 Thanks: 10 
Starting reading your text, I have questions & objections Question. Looks like you give two definitions of $\displaystyle A_{[K]}$ The second one isn't clear.And what do you mean by $\displaystyle A_{[k]}=A$? Objection. You're right when you say e.g.$\displaystyle CB=a^n$.But then you claim a is the gcd of A and CB.If this is important to your work, you need to prove it,cause it's not "well known" at all Before I read any further I must have your reply. NOTE 1:A proof of infinetely many steps is not acceptable.That's why we have induction NOTE 2:I've been working for a while,on the possibility that $\displaystyle n^DA+BC$,where D is "dangerously big".So far,I only collect tears NOTE 3.Writing in a hazy way,and posting based on previous texts that have been found wrong, keep people from reading your work. 
June 6th, 2017, 04:45 AM  #130  
Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24  Quote:
I think ANY more simple proof will involve the infinite descent, but here the problem seems to me another one, but I'm too busy with my errors... in what I'm sure is a more usefull way to solve this problem (someone is trying to retranslate in simple known modular algebra). I'm still waiting for $n=2$ and $n=3$ examples too: one works, why the other not ? Why we can then say "than for any bigger $n$ it doesen't works too..." ? You prove / show / discover that property ? I Think I answer to this problems, for so I claim I've 3 similar way to prove FLT, based on the power of integers property and the fact that the derivate is a curve from $n>2$, so we will fell in an infinite descent for each $n>2$.... Last edited by complicatemodulus; June 6th, 2017 at 04:48 AM.  

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