Fermat's Last Theorem. Proof of Pierre Fermat
In Memory of my MOTHER
The proof is done only for the basic case, to which are easily reduced almost all the other cases. All calculations are done with numbers in base n, a prime number greater than 4.
The notations that are used in the proof:
A', A'', A''', A_(k) – the first, the second, the third, the k-th digit from the end of the number A;
A_[k] is the k-digit ending of the number A (i.e. A_[k] = A mod n^k).
00°) Let us show that if A^n=C^n-B^n=(C-B)P, where A'>1, then for any arbitrarily big k, the endings of the numbers (C-B)_[k+1] and P_[k+1] are the endings of the numbers A'^{n^k} and 1^{n^k}.
A KEY tool in the proof: after increasing the degree of the equality A^n=(C-B)P into (n-1)-th degree the equality becomes false when we compare the digits.
Here the elementary properties of Fermat’s equality for the base case:
0°) A'>1, B'>1, C'>1.
If B'=0 /or C'=0/, then the numbers C-A /or A+B/ in the formula 3° are not taken into account.
1°) A^n=C^n-B^n [=(C-B)P] //and B^n=C^n-A^n [=(C-A)Q], C^n=A^n+B^n [=(A+B)R]//.
1a°) (C-B)P+(C-A)Q-(A+B)R=0,
where we denote with the letters a, b, c the greatest common divisors, respectively, of the pairs of numbers (A, C-B), (B, C-A), (C, A+B).
Then,
2°) if (ABC)'=/=0, then C-B=a^n, P=pn and A=a^p,
//similarly for integers B and C: C-A=b^n, Q=q^n and B=bq; A+B=c^n, R=r^n and C=cr//;
3°) the number U=A+B-C=un^{k+1}, where k>0.
From here we find that (C-B)+(C-A)-(A+B)=0 (mod n^2) [note: if B'=0 /or C'=0/, then the number C-A /or A+B/ =0 (mod n^{kn-2}) and not taken into account HERE].
3a°) ap-a^n=bq-b^n=c^n-cr=0,
A-a^n=B-b^n=c^n-C=0 and
A-a'^n=B-b'^n=c'^n-C=0 (mod n^2) and
3b°) A_[2]={a^n}_[2]={a'^n}_[2], B_[2]={b^n}_[2]={a'^n}_[2], C_[2]={c^n}_[2]={a'^n}_[2].
4°) P'=Q'=R'=p'=q'=r'=1 and consequently P_[2]=Q_[2]=R_[2]=01.
5°) The digit (A^n)_(k+1) is uniquely determined by the ending of A_[k].
Here it is the Proof of the FLT per se.
On the basis of the property we found previously p'=q'=r'=1, consists of an endless sequence of cycles in which the exponent of the degree k increases by 1:
I. 6a°) P_[2]=(5°)=(p'^{n^1})_[2]=(p'^{n^k})_[2]=(4°)=01, k_I=1. =>
7a°) A_[2]=(3b°)=(a'^{n^1})_[2] =>
8a°) (A^n)_[3]=(5°)=(a'^{n^2})_[3]=(1°, 2°)=({a^n}_[3]*{p^n}_[3])_[3] => (A KEY Lemma, see Appendix*)) =>
=> (a^n)_[3]=(a'^{n^2})_[3]; (p^n)_[3]=(5°)=(p'^{n^2})_[3]=001 [=(q'^{n^2})_[3]=(r'^{n^2})_[3]], => (1a°)=>
9a°) U=A+B-C=un^3 [=un^{k+1}] =>
10a°) A_[3]=(B-C)_[3]=(a^n)_[3]=(a'^{n^2})_[3] =(a'^{n^k})_[3] => k_II=2 =>
II. 6b°) P_[3]= (p^'{n^2})_[3]=001, k_II=2, and then the same as in loop I: => A_[3]=(a'^{n^2})_[3] => (A^n)_[4]=..., where with each loop the index k in nk increases by 1, and thus INFINITELY.
If, for example, B ends with k+1 zeros, then the expression (C-A)Q in 1a° has (k+1)n zeros at the end, which is much longer than in U_[3] we are interested in and thus can be discarded. As a result of the equality 1a°, we find that the number B ends with k+2 zeros and not with k+1 zeros. Then we go to the calculations in the next cycle.
That's basically is the incredible proof.
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*) The KEY Lemma.
If (a^n)_[k^+1]=(a'^{n^k})_[k+1], (p^n)_[k+1]=1, {(a^n)*(p^n)}_[k+2]={a'^(n^{k+1})}_[k+2],
where k>0 and 1<a'<n,
then a_[k+2] = {a'^(n^{k+1})}_[k+2] and p_[k+2] = (p'^(n^{k+1})}_[k+2]=1.
Since the proof for any k is exactly the same, we will present the proof of the Lemma only for k=1 (this is the starting case in the proof of FLT), i.e.
11°) If (a^n)_[2]=(a'^{n^1})_[2],(p^n)_[2]=(1'^{n^1})_[2]=01, {(a^n)*(p^n)}_[3]=(a'^{n^2})_[3],
where 1<a'<n,
then
a_[3] = (a'^{n^2})_[3] and p_[3] = (p'^{n^2})_[3]=1.
Proof.
When a_[3] = (a'^{n^2})_[3] and p_[3] = (p'^{n^2})_[3]=1,
then the equality (a'^{n^2})_[3]=[{(a^n)_[3]}*{(p^n)_[3]}]_[3] is an identity, and remains an identity even after the increase of the equality into (n-1)-th degree.
Let us assume now that a[3] and p[3] have different values of third digits, i.e.:
a°_[3] =x00+(a'^{n^2})_[3] and p°_[3] = y00+1. (Here x and y are DIGITS.)
12°) From the equality [(x00+(a'^{n^2})_[3])(y00+1)]'''=(a'^{n^2})'''
we easily find that
(x+a'y)'=0 and x=(-a'y)'.
Now let's elevate the equation 12° into (n-1)-th degree:
[=001].
The three-digit ending of the number y01 in (n-1)-th degree is equal to (n-1)y+1 with the third digit [(n-1)y]' [here x and y are digits!].
The three-digit end of the number a°_[3]=x00+(d^{nn})_[3] in (n-1)-th degree is equal to
[(n-1)x00([a'^{nn}]')^{n-2}+1]_[3],
where (a'^{nn})'=a' (see Little Theorem),
and the third digit of the number a°^{n-1} is equal to (n-1)x(a')^{n-2},
or [(n-1)x(a')^{n-1}]/a',
where (a'^{n-1})'=1.
And now we arrive to the conclusion that the third digit of the product
[(a°^n)^{n-1}]*[(p°^n)^{n-1}] will be equal to:
(n-1)x/a'+a'(n-1)y [=0],
and thus:
x=-a'a'y,
where a'=/=1 (!).
Here lies a contradiction with the prviously found property of the n-th degree, where x=-a'y.
Thus, we must recognize that the real values are:
A_[3] = a'^{n^2}_[3] and p_[3] = p'^{n^2}_[3]=1.
Victor Sorokine. Mezos. 2 March 2017
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The original text, in Word, can be found on this site: http://rm.pp.net.ua/