Fermat's Last Theorem. Elementary proof

[uncool]

Victor: would you mind posting, in one post, your entire proof?

=Uncool-

[Victor Sorokine]

Victor: would you mind posting, in one post, your entire proof?

=Uncool-

Yes!

==========

 

Refinement of one old idea with the desired contradiction: the numbers A, B, C are infinite.

 

Designations:

[math]a_i[/math] - i-th digit from the end in the number [math]a[/math];

[math]a_{(i)}[/math] - the i-digit ending of the number [math]a[/math].

[math]a_{[i+1]}[/math] - the head part of the number [math]a[/math], being finished by number [math]i+1[/math], or the number [math]a-a_{(i)}[/math].

 

Let us examine the numbers A, B, C in the numeration system on the prime base [math]n>3[/math].

 

1°) let us assume that the equality [math]A^n+B^n=C^n[/math] is possible and where, as is known,

 

2°) [math]U=A+B-C>0[/math], where [math]U_i=(A_i+B_i-C_i+[u_{(i-1)}]_i)[/math] (where [[math]U_{(i-1)}]_i=0[/math] or 1).

 

Since the second case (ABC divided by n) proves, actually, just as the first (ABC not divided by n), then let us concentrate attention at the first case.

 

 

Proof of the FLT (first case)

 

Let us introduce the numbers

 

3°) [math]v=A_{(i)}+B_{(i)}-C_{(i)}[/math] and [math]V_{(i)}=A_{(i)}^n+B_{(i)}^n-C_{(i)}^n[/math],

 

and also [math]W_{[i+1]}=V_{(i+1)}-V_{(i)}[/math].

 

Obviously, [math]V_{(1)}[/math] it is not equal to zero (since the number-digits [math]A_1[/math], [math]B_1[/math], [math]C_1[/math] do not contain the dividers of the form [math]m=pn+1[/math]$ and therefore they do not satisfy Fermat’s equality).

 

Therefore there is this smallest rank [math]t+1[/math], for which the number [math]V_{t+1}=n-d[/math] is not equal to zero ([math]V_1=0[/math] according to little Fermat's theorem).

 

Consequently in order to ensure equality [math]V_{t+1}=0[/math]$ the digit [math]U_t[/math] (and no previous [math]U_i[/math]!) must increase by value [math]d[/math].

 

However, according to the Lemma (its proof will be represented separately) an increase of the digit [math]U_t[/math] by the number-digit d leads not only to the equality of digit [math]v_{t+1}=0[/math], but also to the appearance of positive [math]W_{[t+1]}[/math], thus and to not the equality to zero sums [math]V_{(t+1)}[/math].

Consequently, for the transformation [math]V_{(t+1)}[/math] into the zero Fermat’s equality we ARE AGAIN forced to reduce to zero digit, now already [math]V_{s}[/math] in the rank above tn, also, in this case furthermore with the advent of a positive digit, now already in the rank above [math]sn[/math].

 

And so to infinity.

 

In the second case (for example, A divided by n) the number U takes the form: [math]U=An^{kn-k}+B-C[/math], and the prime number

 

3-2°) [math]v_{(kn+1)}=(An^{kn-k})_{(kn+1)}+B_{(kn+1)}-C_{(kn+1)}[/math]. However, into the formula of the number [math]V_{(i)}=A_{(i)}^n+B_{(i)}^n-C_{(i)}^n[/math] coefficient [math]n^{kn-k}[/math] is not included.

 

* * *

 

The plain and elegant proof of Lemma will be represented after “running-in” of the first part of the proof.

 

Thank!

[Victor Sorokine]

Logic of the rank proof

 

 

1. Preliminarily the number U=A+B-C is reduced to the form, which consists of some “nines”, without considering zero at the end. This makes it possible to easily prove Lemma.

 

2. In the terms of the number U we leave such digital ends that the based on them Fermat’s equality deliberately would not be carried out.

 

3. But now we attempt “to make” incorrect Fermat’s equality, beginning from the smallest rank with the error - rank t+1. This “repairing” can be carried out only uniquely - with the aid of the change in the number U (respectively and in the numbers A, B, C) of the digits of the previous rank t.

 

4. In this case we have only one possibility: in THE EMPTY place in the ranks t of the current number U to transfer digits of the rank t of the number U, which is been the solution of the hypothetical of Fermat’s equations.

 

5. However, together with this, as asserts Lemma, appears positive number in the rank s, which exceeds THE GREATEST rank in the sum of degrees, based on the number U (and its terms). And it is this is what important: inclusively to influence the digit s is impossible with the aid of the digits of the number U in the ranks from 1 to t, since these digits are fixed and undertaken from the solution of Fermat’s equation.

 

6. And to reset to zero s-th digit in the reconstructed Fermat’s equality is possible only in an only manner: to introduce into the number U the digits of s-1-th rank.

 

7. But in this case AGAIN appears positive number in the reconstructed Fermat’s equality - in the rank, which lies BEYOND the limits of the reconstructed Fermat’s equality.

 

And so to infinity.

 

It seems that the FLT is proven (with an accuracy to lemmas).

[Victor Sorokine]

Lemma. Attempt of the proof

 

If the number [math]U=n^t-n^k[/math] (i.e. consists of digits n-1, or “nines”, with zero at the end), where U=A+B-C>0, if ABC not is divided by n; [math]U=n^{kn-k}A+B-C>0[/math], if A is multiple n; A+B>C>A>B>0; k is a constant; t as as desired is great; the number A, B, C are integers and maximum rank of the number [math]D=A^n+B^n-C^n[/math] (where prime n>2) is equal s,

than with an increase of the rank of the number U by 1 the rank of the number s also increases.

 

Let us examine proof only for n=3, since the generalization for any prime n completely obviously and consists, actually, of the replacement of the number 3 by n.

 

With the condition A+B>C>A>B it is possible unique values of the numbers, entering in U:

 

A=222222222222222…10

+

B=222222222222222…01

-

C=222222222222222…11

===================

U=222222222222222…00 (how conveniently twos, but with k zero at the end).

 

During the replacement of any digit 2 in the number C, for example, by 1 it is necessary in the same rank to replace by digit 1 digit 2 in the number A or C, which leads to the disturbance of the condition A+B>C>A>B.

 

So whether this? If “yes”, then lemma and FLT are true.

[Victor Sorokine]

Elementary proof of Fermat's last theorum (editing)

 

Let us examine the numbers A, B, C in the numeration system on the prime base n>3.

let us assume that the equality

 

1°) [math]A^n+B^n=C^n[/math] is possible and where, as is known,

 

2°) [math]U=A+B-C=un^k>0[/math], where: A, B, C are integers, [math]A+B>C>A>A>U[/math] and [math]u>1[/math].

 

With the aid of the multiplication of the equality 1° by the appropriate number [math]g^n[/math] let us lead the number U to the form (this possibility it was repeatedly shown by the author earlier):

 

3°) [math]U=n^t-n^k[/math].

 

Now one of the possible solutions of the equation of 2° is such:

 

A_1=999999999999999…10

+

B_1=999999999999999…01

-

C_1=999999999999999…11

====================

U = 999999999999999…00 (here before the sufficiently large number of “nines”, or digits n-1, they stand k of zeros).

 

In this case it is easy to see that the number [math]D_1=A_1^n+B_1^n-C_1^n>0[/math], and with any attempt to replace at least one of “the nines” by another digit with the retention of the conditions 2° and 3° it leads to AN INCREASE of the number D and, thus, the equality D=0 IS IMPOSSIBLE. Thus FLT is proven.

 

P.S. With n=2 the number u=1 and proof loses force.

[Victor Sorokine]

For the sly nut there is a bolt with the screw!

 

Elementary proof of Fermat's last theorum (by refined method of correction)

 

Designations:

[math]a_i[/math] – i-th digit from the end in the number a;

“9” – digit n-1 in the numeration system on the base n.

[math]a_{(i)}[/math] – the i-digit ending of the number a.

 

Let us assume that with n>3 and unequal integers A, B, C equality (with smallest C)

 

1°) [math]A^n+B^n=C^n[/math] is possible and where, as is known, in the case of 1 (ABC not multiply n)

 

2°) [math]U=A+B-C=un^k>0[/math] (where u not is multiple n) with the highest rank t ([math]U_t>0[/math]).

 

With the aid of the multiplication of the equality 1° by the appropriate number [math]g^n[/math] let us lead the number A+B to the form:

 

3°) [math]A+B=n^t-1[/math].

 

We find from the equality 2° that the digit

 

4°) [math]U_t=A_t+B_t-C_t=[/math]“9”[math]-C_t[/math] or “9”[math]+1-C_t[/math] (where 1 transfer from the previous sum).

 

And since [math]U_t[/math] is not equal to zero, the transfer of digit 1 into the rank [math]U_{t+1}[/math] is absent. But since all digits [math]U_{t+i}[/math], where [math]i=1,2, . . . [/math] are equal to zero, is the sum

 

5°) [math]A_{t+i}+B_{t+i}-C_{t+i}=0[/math], where [math]i=1, 2, . . . [/math],

 

in this case among the digits [math]A_{t+i}[/math], [math]B_{t+i}[/math], [math]C_{t+i}[/math] deliberately there is positive ones, since [math]U<B<A<C[/math] and [math]A+B>2U[/math].

 

(But if there are no such digits, then [math]A^n+B^n-C^n>0[/math], which was shown three days ago.)

 

It thus, follows from the equality

 

6°) [math]V=(A_{(t)})^n+(B_{(t)})^n-(C_{(t)})^n>0[/math] it follows that [math]V_s=d[/math] (where [math]s>t+1[/math]) is not deliberately equal to zero.

 

Consequently in order to ensure the equality [math]V_s=0[/math], it is necessary in the place of the digits [math]A_{s-1}[/math], [math]B_{s-1}[/math], [math]C_{s-1}[/math] with the condition [math]A_{s-1}+B_{s-1}-C_{s-1}=0[/math] to place other digits – with the condition

[math]A_{s-1}+B_{s-1}-C_{s-1}=d[/math]. However, this IMPOSSIBLY, since in this case the rank of the number U must be equal not to t, but s [>t].

 

Case of 2 proves in perfect analogy, if we replace formula for U by the formula, in which the significant part of the number, multiple n, is multiplied by [math]n^{kn-k}[/math], in this case with the erection of this number into the degree it is used in the previous form.

 

Thus truth of the FLT is proven.

[Victor Sorokine]

Elementary proof of Fermat's last theorum

 

 

Let us assume that for the prime n>2 and relatively prime integers A, B, C

 

1°) [math]A^n+B^n=C^n[/math], where, as is known, last k [k>1] digits of the number

 

2°) [math]U=A+B-C[/math] in the numeration system on the foundation of n are zeros.

 

The case I. ABC is not divisible by n.

 

With the aid of the multiplication of the equality 1° by the appropriate number [math]g^{nn}[/math] we convert the last two digits of the number A+B in n-1, after which on the next-to-last digits of the numbers A, B, C we obtain the rigorous equality

 

3°) [math]A_2+B_2=C_2[/math].

 

Let us assume that not all digits [math]A_2, B_2, C_2[/math] are equal to zero.

 

Let us introduce the numbers A', B', C', obtained from the numbers A, B, C with the aid of the resetting to zero of the second digits. Obviously, the [math]{A', B', C'}[/math] is not the solution of Fermat’s equation and the number

 

4°) [math]D=A'^n+B'^n-C'^n[/math] is not equal to zero.

 

Let [math]D_ t=d[/math] be a positive digit of the smallest rank. Let us attempt it to reset to zero. For this to the sum of the digits [math]A'_{t-2}[/math], [math]B'_{t-2}[/math], [math]C'_{t-2}[/math] it is necessary to add positive digit n-d. However, this is impossible. Actually,

 

If t-2 is not a next-to-last rank (i.e. 2), then we are forced to change digits beyond the limits of the second rank and in this case not to correct the digit of the second rank.

But if t-2=2, then we cannot restore the original values of the numbers A, B, C for that reason, what both sums – [math]A_2+B_2-C_2[/math] and [math]A'_2+B'_2-C'_2[/math] – are equal to zero and addition to the second sum of the number n-d cannot change sum [math]A_2+B_2-C_2[/math] [=0].

 

Thus, contradiction is resolved uniquely: all three numbers – [math]A_2, B_2, C_2[/math] – are equal to zero.

 

5°) Let us note that the strictness of the equality 3° is not disrupted from the multiplication of the equality 1° by each number [math]g^{nn}[/math], where [math]g=1,2,… n-1[/math]. In this case the previous conclusions about the equality of all second digits to zero preserve their force. But THE CONTRADICTION follows from this:

from one side, accordingly 4°, the second from the end the digit in the sum [math]1^n+2^n+…+(n-1)^n[/math] is [math](n-1)/2[/math], while with another, as shows direct summing up, it is equal to zero.

 

The case II. ABC it is multiple n. Analogous, but more complex arithmetically proof will be represented after the consideration of the first case.

 

It is easy to show that in the Case II two-digit ends of the significant parts of all three numbers A, B, C are equal and the change by 1 next-to-last digit of the significant part in the number U leads to the change by 1 ([math]kn+4[/math])-th digit in the number [math]A^n+B^n-C^n[/math].

[khaled]

I agree that this could be a lot clearer but I can't help wondering something.

 

If you start by saying

 

"Elementary proof of FLT

 

 

Let us assume that for natural A, B, C

 

1°) A^n+B^n=C^n [or E=A^n+B^n-C^n=0], where prime n>2,

 

"

 

doesn't that mean you have assumed the truth of Fermat's last theorem?

 

in case he wanted to prove the Fermat's Last theorem,

 

inductively, he has to move from one side to another, by assuming the initial state is true ...

 

but in case there is only the FLT in that form, numerical substitution is not a proof ...

 

----------------------

 

moreover, what he is trying to do is not a proof to the FLT,

 

he's just testing the FLT at some conditions ...

Edited February 11, 2011 by khaled

[Victor Sorokine]

That same Pierre Fermat [victory]

 

Many years ago I proved the Theorem (it published on the forum «dxdy»):

 

If two-digit endig [math]A_{(2)}[/math] of the number A, recorded in the numeration system in the prime base [math]n>2[/math], there is a two-digit endig [math](g^n)_{(2)}[/math] of certain degree [math]g^n[/math], then the s-digit number A is a s-digit ending of a certain degree [math]A'^n[/math].

 

With the aid of this theorem easily proves the FLT.

 

Actually, as is well known, all significant parts of the numbers A, B, C satisfy theorem condition. Using this, let us accept for the length (i.e. for the number of the digits) of the numbers A, B, C the length (it is possible doubled – that, however, not compulsorily) of the number [math]C^n[/math], equal to s. (Understandable that before the numbers A, B, C it will stand long number of zeros.) And now, according to theorem, the numbers A, B, C can be written down in the form:

 

1°) [math]A=a^n+n^sA'[/math]; [math]B=b^n+n^sB'[/math]; [math]C=c^n+n^sC'[/math].

 

After substituting these values into the Fermat’s equation, on the s-digit endig of degrees we obtain second (and identity!) Fermat’s equation:

 

2°) [math]a^{nn}+b^{nn}=c^{nn}[/math].

 

If one of the numbers A, B, C is multiple by [math]n^k[/math], then this in no way prevents the appearance of second equality.

 

Further conversion, similar 1°, we make with the numbers a, b, c and after their substitution into the equality of 2° we obtain tertiary Fermat’s equation – already of the degree of [math]n^3[/math].

 

And so on, to the equation of power [math]n^t[/math], when the root of the degree [math]n^t[/math] at least of one of the numbers A, B, C will not become FRACTIONAL.

 

Thus, Fermat's last theorem is proven.

[Victor Sorokine]

Modification of the proof of the FLT, based on the Theorem:

 

If two-digit ending [math]A_ {(2)}[/math] of the number A, recorded in the base prime n>2, there is a two-digit ending [math](g^n)_{(2)}[/math] of a certain degree [math]g^n[/math], the s-digit number A is a s-digit ending of a certain degree [math]a'^n[/math].

 

* * *

 

It is known that in the base case (prime n>2, integers A, B, C are coprime) among the numbers A-B, C+B, C+A are such, that

1) two-digit endings of both terms are the two-digit endings of some n-th degrees,

2) both terms are not multiple n and

3) their sum is the sum of n-th degrees.

 

Let us assume that this number is A-B and, which means,

 

[math]A=C-B=a^n[/math], [math]B=C-A=b^n[/math],

 

[math]A_{(2)}=(a'^n)_{(2)}=(a^n)_{(2)}[/math], [math]B_{(2)}=(b'^n)_{(2)}=(b^n)_{(2)})[/math], therefore,

 

[math]A_{(s)}=(a'^n)_{(s)}=a'^n-pn^s[/math], [math]B_{(s)}=(b'^n)_{(s)}=b'^n-qn^s[/math], (here s is the length of the number A)

 

[math]A-B=(a'^n-b'^n)-(p-q)n^s=a^n-b^n[/math], from where

 

[math]a'^n-b'^n=a^n-b^n[/math].

 

Quick calculation shows that

[math]a'^n=A=a^n, b'^n=B=b^n[/math], from where [math]A+B-C=0[/math], and Fermat’s equality is impossible.

[Victor Sorokine]

Two key moments of the proof

 

Simple elementary proof of the FLT is based on two theorems:

 

1. If the two-digit end of the number A, not multiple prime n>2, are the two-digit end of any number [math]g^n[/math], then the number itself A is the two-digit end of any number [math]G^n[/math].

The simple proof of this theorem is based on the fundamental fact of the theory of numeration on the prime base n: all last digits in the multiplication table [math]g*i[/math], where g is positive number and i takes the value of each digit from 0 to n -1, ARE DIFFERENT.

 

2. Since one of the combined numbers A-B, C+B, C+A is the sum of degrees.

For example: [math]A-B=a^n-b^n[/math] (*), where the length of the numbers [math]a^n[/math] and [math]b^n[/math] does not exceed sn digits, but the sn-digits ends of the numbers A and B (i.e., numbers themselves A and B!) there are ends of some degrees [math]D^n[/math] and [math]E^n[/math], then bases in the pairs D and a and E and b at the length s of numbers ARE EQUAL, otherwise, as it is easy to show, equality * is absent.

But since [math]a^n=C-B[/math], then from the equality [math]A=a^n[/math] follows equality [math]A+B-C=0[/math], but Fermat’s equality IS IMPOSSIBLE with this condition.

 

Here, strictly, all proof of Fermat's last theorum.

[Victor Sorokine]

Is examined simple proof of the FLT, which consists of three simple algebraic formulas and page of reasonings on the level of arithmetic for the first class, based on one simple lemma

 

1. If in the prime base of n>2 number [math]a^n[/math] finishes by digit 1, then the second significant (following nonzero) range is 1;

 

and more complex (years 20 ago by me as if proven):

 

2. The number [math]2^n-1[/math] is not divided by [math]n^2[/math].

 

= = =

 

Elementary proof of the Fermat's last theorem

 

Designation: [math]a_1[/math] - last digit of the number a.

 

Let us examine the first 1st case: abc is not divided by n.

 

Let us take equivalent Fermat’s equality in the prime base n

 

1°) [math]a^n+b^n-c^n=(c-b)P+(c-a)Q-(a+b)R=0[/math], or [math]AP+BQ-CR=0[/math], where

 

2°) [math]a+b-c=U=un^k[/math] and [math]A+B-C=-2U=-2un^k[/math] (k>1),

 

3°) P, Q, R are degrees and [math]P_1=Q_1=R_1=1[/math],

 

4°) (k+1)-digit ending of the number b is transformed into 1 (with the retention of the property of 3°).

 

(Continuation it follows.)

[Victor Sorokine]

Proof of the FLT in the case 1.

 

Proof of the FLT in the case 1.

 

Let us deduct from the number

 

6°) [math]D=-a^n+b^n+c^n=-(c^n-b^n)+(c^n-a^n)+(a^n+b^n)=-AP+BQ+CR[/math] [[math]=2b^n[/math]] the number

 

7°) [math]d=-A+B+C [=(c-b)+(c-a)+(a+b)=2b] [/math], multiplied by the number [math]b^{(n-1)} [/math] [[math]=E[/math]]:

 

8°) [[math]F=[/math]] [math]-A(P-E)+B(Q-E)+C(R-E)=0[/math]. [KEY].

 

And now, after the rejection of common zero ending in three terms, we obtain for the last digits of cofactors the following equation:

 

9°) [math]-xA_1+y*1+zC_1=0[/math], where

 

[math]A_1+1-C_1=0[/math] (see 1°), and the number

 

x, y, z (taking into account lemma 1) have values of 1 or 0 and one of them is deliberately equal to 1.

 

And we see that the equation 9° has only unique solution:

 

10°) [math]A_1=1, x=y=1, z=0[/math], which contradicts lemma 2.

 

Case 2: a (either b or c) is divided by n proves analogously, but without the complication 10°.

 

Thus, for the final proof of the FLT to us is not sufficient only a proof of lemma 2, to the examination of which we must pass.

[Victor Sorokine]

Completing correction.

 

In reality the proof contains 6 special subcases (fortunately, easily surmountable) - on 3 in each of two base cases:

 

[math]A_1+B_1=n, A_1+C_1=n, B_1+C_1=n, A_1-B_1=0, C_1-B_1=0, C_1-A_1=0[/math].

 

Difficulty easily is removed with the aid of the selection of the corresponding “working” number (a, b or c) and the corresponding to it pair of the formulas of 6° and 7°.

[Victor Sorokine]

Fermat's last theorum

 

Description of conditions.

 

Let us examine canonical Fermat’s equality in the base with prime n> 2:

 

1°) [math]a^n+b^n-c^n=(c-b)P+(c-a)Q-(a+b)R=0[/math], or [math]AP+BQ-CR=0[/math], where

 

2°) on the last digits [math]x_1[/math] there are the equalityes [math]a_1=A_1, b_1=B_1, c_1=C_1, [/math]

 

3°) [math]a+b-c=U=un^k[/math] and [math]A+B-C=-2U=-2un^k (k>0) [/math],

 

number-expression [math]A+B-C[/math] has three combined with it numbers:

 

4°) [math]d(a)=A-B+C[/math] [[math]=2a[/math]], [math]d(b)=-A+B+C[/math] [[math]=2b[/math]], [math]d©=A+B+C[/math] [[math]=2c[/math]].

 

Accordingly the number [math]AP+BQ-CR[/math] also has three combined with it numbers:

 

5°) [math]D(a)=AP-BQ+CR[/math] [[math]=2a^n[/math]], [math]D(b)=-AP+BQ+CR[/math] [[math]=2b^n[/math]], [math]D©=AP+BQ+CR[/math] [[math]=2c^n[/math]].

 

The numbers a, b, c into 4° and 5° let us name workers.

 

Subtraction from the numbers [math]D(a), D(b), D© [/math] of the numbers [math]d(a), d(b), d© [/math], multiplied respectively by [math]E(a) [/math] [[math]=a^{n-1}[/math]], [math]E(b) [/math] [[math]=b^{n-1}[/math]], [math]E© [/math] [[math]=c^{n-1}[/math]], generates three characteristic equations, equivalent to the Fermat’s equality:

 

6a°) [F(a)=] [math]A(P-E)-B(Q-E)+C(R-E)=0[/math],

6b°) [F(b)=] [math]-A(P-E)+B(Q-E)+C(R-E)=0[/math],

6c°) [F(c )=] [math]A(P-E)+B(Q-E)+C(R-E)=0[/math]. (©)

 

Proof of the FLT is decomposed into 2 cases.

 

Case 1: any of the numbers A, B, C, A' [[math]=c+b[/math]], B' [[math]=c+a[/math]], C' [[math]=a-b[/math]] not is multiple n.

For the proof of the FLT in this case it is possible to use any characteristic equation 6°.

 

Case of 2: one (and only one!) from the numbers A, B, C, A' [[math]=c+b[/math]], B' [[math]=c+a[/math]], C' [[math]=a-b[/math]] is multiple n. For the proof of the FLT in this case it is not possible to use the characteristic equation 6°, in which the signs before the terms with the same letters either both coincide or both they do not coincide with the signs of the terms a, b, c in [math]A, B, C, A', B', C'[/math]. So, for A do not befit 6a°, for B – 6b°, C – 6c°, A' – 6b° and 6c°, B' – 6a° and 6c°, C' – 6a° and 6b°. Thus, for each of the numbers [math]A, B, C, A', B', C'[/math], i.e. for each case, there is a characteristic equation from 6°, which satisfies the requirement indicated.

 

 

Proof of Fermat's last theorum

 

Since all proofs in all cases are completely analogous, we will prove only the case, when C', or a-b, is divided by n, with the characteristic equation 6c° and with the working number c. For this with the aid of the multiplication of initial Fermat’s equality 1° by the appropriate number [math]G^{nn}[/math] we convert (k+1)-digit ending of the number c into 1 : From this they do not change propertyes of the numbers P and Q: they finish to digit 1 and are n-th degrees. It is easy to see that now the second significant digits (not of identical rank) in the numbers P and Q is 1. And after the rejection of total zero endings of equal length in three terms in the equation 6c°, we obtain for the last digits of cofactors the following equation:

 

7°) [math]xA_1+yB_1+z*1=0[/math], where [math]A_1+B_1-1=n[/math] (cf. 1°) и [math]A_1-B_1=0[/math], and numbers

 

x, y, z have values of 1 or 0 and one of them is deliberately equal to 1. And it is easy to see that the equation of 7° DOES NOT HAVE OF SOLUTION.

 

Theorem is proven.

[Victor Sorokine]

Entertainment

 

Proof one additional case: [math]c=n^k*c'[/math], where c' it is not divided by n.

 

From two possible characteristic equations let us take 6b° with the working number b.

 

It is very easy to show that either the number P or in the number Q (k+1)-th digit is equal to 1.

 

After the conversion of the kn-digit ending of the number b into 1 we obtain in the initial equality 1° such values of the last digits: [math]a_1=n-1, b_1=1, c_1=0[/math].

 

But last digits in the characteristic equality 6b° generate the equation:

 

[math]-x*(n-1)+y*1+z*0=(0 /or/ n)[/math], where x and y are equal to 0 or 1 and one of them is deliberately equal to 1.

 

But neither number -(n-1)+1 nor number -(n-1) nor number 1 is equal to zero.

 

Consequently, initial Fermat’s equation integral solution does not have.

[Victor Sorokine]

VTF for the junior schoolboys. Basic case

 

Basic case. The numbers abc, a+b, a-b, c-b, c+b, c-a, +a are not divided by n.

Let for the convenience the number a+b-c is divided by [math]n^2[/math] and is not divided by [math]n^3[/math] (i.e. k=2).

 

Consequently, the numbers P, Q, R do not finish simultaneously with 001 – otherwise the number a+b-c is divided by [math]n^3[/math].

 

* * *

 

The following combinations of the three-digit endings of the numbers P, Q, R are thus, possible:

1) P=.101, Q=.101, R=101; 2) P=.101, Q=.101, R=001; 3) P=.101, Q=.001, R=.001.

Remaining versions are obtained by the transposition of the numbers P, Q, R.

 

Any of the numbers a, b, c – let for the certainty the number a – it is possible to take as the working number and with the aid of the multiplication Fermat’s equality by the appropriate number [math]g^{nn}[/math] to convert the three-digit ending of the working number a into 001. Then three-digit ending of the number [math]a^{n-1}[/math], or E, will also be equal to 001.

 

Now let us take any of characteristic equalities – for example, 6c° A(P-E)+B(Q-E)+C(R-E)=0. And since all three numbers P, Q, R finish by digit 1, are possible the following combinations of three-digit endings in the bracketed expressions:

1) P-E=.100, Q-E=.100, R-E=100; 2) P-E=.100, Q-E=.100, R-E=000; 3) P-E=.100, Q-E=.000, R-E=.000.

 

Consequently, expression in each bracket in the equality 6c° we can reduce by two zeros, after leaving in them only the third (from the end) digits, which now become the latter. And now instead of the brackets (P-E), (Q-E), (R-E) in 6c° we can place one of the collections:

1) 1, 1, 1; 2) 1, 1, 0; 3) 1, 0, 0 with any order of the entering in them numbers.

 

Let us designate last dogots in the numbers a, b, c through a', b', c'.

 

If in the collection there are two zeros, then the equality of 6c° is converted into one of the equalities: a'=0, b'=0, c'=0, where all single-valued numbers a', b', c' are not equal to zero. Therefore, the equality 6c° impossibly.

 

If in the collection there are no zero, then the equality 6c° is converted into equality 2c'=0, where the number c' is not equal to zero. Consequently, the equality 6c° is impossible in this case.

 

And finally if in the collection there are two zeros, then or a'+b', or a'+c', or b'+c' finishes by zero, and this will be already the second case.

 

For understanding of the proof in the second case it is necessary to have eightyear formation and to know the formulation of Fermat's Little theorem.

[Dave]

Victor, at this point I think we can agree this isn't really mathematics and has been dragged on for a long time now. Therefore, I'm moving this thread to speculations. Please do not attempt to start any new threads on this topic in the mathematics forum; they will either be locked or deleted on sight.

[Incendia]

I remember watching a horizon documentary about a mathematician who proved this theorem. You are too late Victor. Theorem has been proved.

Edited June 12, 2011 by Incendia