CC BY
28
Section 3. Mathematics
Section 3. Mathematics
Drushinin Victor Vladimirovich, National Research Nuclear University “MEPHI" Sarov Physics and Technology Institute Department of Mathematics E-mail: Sarov, vvdr@newmail.ru
Generalization of Fermat’s last theorem
Abstract: We consider the one-power sums of numbers, such as the sum of the three numbers of Fermat’s theorem for equality or inequality them to zero. We formulated several theories on their properties. We give a new simple proof of Fermat’s last theorem.
Keywords: small and Fermat’s last theorem, one-power sums of numbers.
Fermat’s last theorem (FLT) is inequality x1 + xП — x 1 Ф 0 for natural and xk and n > 3 . This theorem is formulated almost four hundred years ago in 1637, but the first time one fully proved in 1995 by the British mathematician Andrew Wiles [1]. The proof is very bulky, is about one hundred villages A4. Wilds uses elliptic curves of the form y = x3 + ax2 + bx + c and the equivalence of these features for modular functions. In this article we present a simple method of proving FLT and generalize this inequality or equal to the sum of more complex one-power variables. In the future, we use only natural numbers and zero.
We introduce a one-power sums numbers of the form
F (m,n ) = Ъ{-1ТХ1, (1)
k=l
and call them sums sums of Fermat. In FLT there are m = 3 ,а1 = a2 = 2,a3 = 1 . Immediately take (x1, x 2,..., xm) = 1, i. e. all numbers in (1) have a greatest common divisor equal to “1”. (1) always has positive and negative summands. Let us ask ourselves when F(m,n) Ф 0 and when F(m,n) = 0. At first we turn to the table. № 1. In this table, there are the last digits of numbers x = A1 -10 + x0 in different power n, i. e. xn. We see that last digits are repeated with a period of «4»: both xn and xn+4k have the same last digits. We select in F(m,n) = 10 • Fl (m,n) + F0 (m,n) the last digit
F0 (m,n).
Table 1. - The value of the last digit of number (Д10 + x0) in different powers
n; x „ 1 2 3 4 5 6 7 8 9 0
2 1 4 9 6 5 6 9 4 1 0
3 1 8 7 4 5 6 3 2 9 0
4 1 6 1 6 5 6 1 6 1 0
5 1 2 3 4 5 6 7 8 9 0
6 1 4 9 6 5 6 9 4 1 0
7 1 8 7 4 5 6 3 2 9 0
8 1 6 1 6 5 6 1 6 1 0
9 1 2 3 4 5 6 7 8 9 0
10 1 4 9 6 5 6 9 4 1 0
For example, F(3,1) = 6 + 5 - 9 = 2,F0 (3,1) = 2; F(3,2) = 36 + 25 -81 = -30,F0 (3,2) = 0. Tabl. 1 also indicates at repeatability oflast digit of Fermat’s sums with a period of «4»
F0 (m,n) = F0 (m,n + 4k), (2)
if both F(m,n) and F(n,n + 4k) are positive or both
negative.Inthecaseofthe change of sign
F0 (m,n ) = 10 - F0 (m,n + 4k). (3)
Example. F (3,1) = 6 + 5 - 9 = 2;F0(3,1) = 2; F(3,5 ) = = 65 + 55 - 95 =-48148;F0 (3,5) = 8 . Here, using (3), the changing of the last digit happen due to the change of the sign before F (m,n) .Thereafter can be formulated
18
Generalization of Fermat’s last theorem
Theorem D1. If the last digit of the first four Fermat’s sums { (m,l);F0 (m,2);F0 (m,3);F0 (m,4))* 0 , then all values F(m,n) Ф 0. This meansthat The sum mat form a group of the extent to which there are four irreducible representations.
If {(W)^ (m2)uF0 (m,3)uFc (m,4)) = 0 ,
then a question about significance of F(m,n) requires further consideration.
Next, we will use two small Fermat’s theorem (SFT). SFT1: (ap - a):p; (p - a) always is a multiple of the prime number p. SFT2: (p 1 -l):p, where (a,p) = 1; -1) is a multiple of p, if a and p are coprime. Using Table. № 1 these theorems, we added the following formulas. If p = 4k + 5 , then (ap - a)(l0-p). Since the the last digits ap and a identical and their difference is divided by « 10». If (a,p) = 1, then ( -l):5*p . Examples: (213 -2) = 10-13-63,
(( -1) = 5 - 07-771. Also, there is (aik+n -an))0) for any n and k.
Theorem D2. If F(m,l) = 0 , then all F(m,p)p . Moreover F(m,p) can be zero if m > 3.
Using SFT1, we will prove this theorem.
m
F (m,p) = F (m,p)- F(m, 1) = ]Г (- 1 )'*( op -x( = p ■ M.
k=l
Examples. F(3,1) = 6 + 5-11 = 0;F(3,3) = 63 + 53 --113 =-3-330. F (8,1) = 1 + 2 + 2 - 3 + 4 + 8 - 7 - 7 = 0;
F(8,3) = 13 + 23 + 23 - 33 + 43 + 83 -73 -73 = -23 -3 • 5.
m
Theorem D3. If all (xk, p ) = 1 and X(-l) = Cm
k=1
does not equal to a zero or are not multiple of , then
x m A
F(m,p - l) = X(-l) Xp-1 ф 0 . The proof based on
k=1
SFT2.
m
F(m,p- 1)=H-1)“‘ 3 = [SFT2] = p-M + Cm * 0.
k=1
The meaning of this theorem that if the sum of the three (Pythagorean triple) or more squares with different signs is zero, and C 3 not a multiple of «3», then at least one summand is a multiple «3». Quadratic identity 4a2ß2 + (a2 -ß2) -(a2 +ß2) = 0, giving Pythagorean triples have either the first or the second summand a multiple of «3». In the Druzhinin’s article [2], there was given the formula for an arbitrary number of squares that form Pythagorean chain. Example. 32 + 42 +122 + 32 + 42 +122 + 842 + 36122 + 65268842 +
+ 213001139016122 - 213001139016132 = 0 . In this sum there is C 7 = 6, and only the second and last summands are not multiples of «3».
Euler suggested that the Fermat’s sums with F (4,4) does not equal to zero. However, in 1988, computer cal-
culations have found such a sum equal to zero [3].
(2a -e • t7on-)4 +(-eanenaa Th+(2a m • 4 naa-a )4 -—(a-oo •aa-aea7 Th4
Here, in accordance with D3 the first and the third numbers are multiple of «5», and C 4 = 2.
We have obtained the quartet identity giving F (7,4 ) = 0:
(2ab)4 + 4 -(2b3a)4 + (a4 - 2b4 )4 - (a4 + 2b4 )4 = 0. (4) In (5), in accordance with D3 there is C7 = 5, so the summands may be coprime to «5». Example. 164 + 4-(44) +144 -184 = 0.
In addition we received ternary identity, giving F (7,3) = 0:
(2ab)3 + 2p3)3 -2-(a2b)3 +
+(a3 - b3) +(a3 + b3) = 0.
Theorem D4. If p1 = p2 + p3 -1, then
(F ( P,)- F (m P 2)): Рз.
The proof based on SFT2.
m
F(m,pi)-F(m,p2)=Е(-1Г‘'op2-1(xp -xk) = p3 ■ M.
k=1
Example. p1 = 7,p2 =5,p3 = 3. (7 +37 - 77 )-
-(3+33 - 73 ) =-5460944.
Next, using the obtained theorems, we discuss FLT. We writethissumintheform F(3,n) = an + bn - cn.
If F (3,l) = a + b — c = 0 , then when n = p all F (3, p): p and these sums are negative. For example, F(3,1) = 4 + 3-0 = 0,F(3,3) = -9-28, F(3,5) =-5-310. Next, we consider the cases of F (3,1) > 0. If the last number F0 (3,1 does not is zero, then all sums F(3,1 + 4k) Ф 0 , i. e. FLT is valid for them. The numbers included in the sum of Fermat F (3,1) > 0 is not entirely arbitrary. Take a - even, b and c odd, then (a,0) = 1,c > a ,c > b ,(a + 0) > c ,b > a .With the growth of n, F (3, n) changes from a positive value to a negative sometimes monotonically decreasing, sometimes non-monotonically, going through a maximum. F (3,n) < 0 where c>c = b-a(a,b,n}. Here we introduced c - an exact solution of the equation an + bn -cn = 0 by the
function a (a ,b, p) = (l + (a / b ) .In these terms we can formulate the PTF as follows: when n > 3 c is irrational. Hence there is the more general inequality F(3,p) < 0,where c > bV2.
F(3,1 = a2 + b2 - c2 = 0, if both a and b are not multiples of «3», or a:3, or b :3, but there are no a:3 and b: 3 together. c is find by a quadratic identity. If F(3,2) = 0, all F(3,2 + 4fc) end at «0». At the same time we do not claim that F (3,2 + 4k) Ф 0.
19
Section 3. Mathematics
Fact that F (3,3) Ф 0 proved by Euler. This inequality indicates that a(a,b,p)e I where p > 3 i. e. This number is an irrational. Indeed, the equation a3 + b3 — c3 = 0 gives c = b -a (a ,b,3) — in this case c is not an integer. If a(a,b, p) = (s /1), where (5 ,t) = 1 a rational number, then c is a rational number c = b ■ s /1, and the original equation a3 + b3 - c3 = 0 converted into (t-a) + (t-b) -(5-b) = 0 in integers, but t:s. This is contrary to the initial condition for the Fermat’s sums. Since a (a,b,p) e I, if the last digit F0 (3,p) Ф 0, then the sum itself F (3,3 + 4k) Ф 0 . This means that Table. № 1 and inequality F(3,3) Ф 0 provide 95% of the Fermat’s sums.
Note that a(a,b,2) is a rational number, where a < b and a = 4xy,b = |x2 -y2| or a = |x2 -y2|, b = 4xy.
As for a (a,b,2) ,this result was obtained by Fermat. We show how this happens when (a,5) = (b,5) = (c ,5) = 1. Sum F(3,4) = a4 + b4 -c4 = a5-1 + b5-1 -c5-1 =[МТФ2] = = 1 + 5 -М Ф 0.It follows that F (3,2n) = a2n + b2n — c2n Ф 0 when n > 1, since this sum can always be reduced to the form F (3, p).
There remain around 5% of the Fermat’s sums on which the Table. № 1 do es not work. This sums F (3 p )* 0, in which the last number F0 (3,p) = 0. At the next related Fermat’s sum F0 (3,p + 4k) = 0, but it does not guarantee that the sum F (3,p + 4k) Ф 0. Analysis penultimate digits F (3, p) = 100 •F1 +10 • F1 ,namely F1 where F0 = 0 shows that for odd F1 sums F (3, p + 4k) have №{Ф 0 . This indicates that F (3, p + 4k) 0 . For example, F (3,1) = 12 +11 -13 = 10 — the smallest of
these Fermat’s sums. F(3,3) = 125 +115 -135 = 38590. Here Fj = 9. FLT is performed. Next related sums are F(3,9) = 129 +119 -139 =-30 8 6 77 1 330.
Here the last number is saved, the penultimate F1 = 3 is not zero. In addition, the this sum is negative. Next check is not necessary, since all such sums are negative and are applicable to them FLT. Two factors: a non-zero second number Ft and entering a negative value includes an additional 4% of the sums in the FLT. But still a small part of our approach Fermat’s sums is not checked. The yield is that the function a(a,b,p) e I where p > 3.
For the Fermat’s sums with a large number of summands the function a (a,b,p) is replaced by a function
a
( x 2,..., p ))l +
Jl+ p X2
1 V xi j
a(*i, x2,.
f Y Хз
V xi J
+ ...+
f V I p
Xm-L 1
V Xi J
„_!, p) ,and it may be a rational number equal to (-xm / xj . This gives F (m, p) = 0, as there are more degrees of freedom.
We show this by two numerical examples.
F(5,e)o eke FFa F-s o 6; F(5,3)o e3 Fe3 F 53 k 63 -- s3 o 0; F (1,1 o e F n k n - 3 k 4 - S - S k 1 o 0;
F (1,5) o e5 k n5 k n5 - 35 k 45 - S5 - S5 k 15 o 0.
Thus, we believe that our approach makes the analysis of FLT more transparent, compared with the theory of Wiles. In fact, the small Fermat’s theorem proving great Fermat’s theorem.
The author thanks Prof. N. S. Shevyahova, A. A. Lazarev and the participants of the department for a discussion of this work and for useful comments.
References:
1. Wiles A. Annals of Mathematics 141 (3): 443-551, 1995.
2. Alvares L. F. A. Fermat’s Last Theorem. M. De Agostini, 2015.
3. Druzhinin V. V. NTVP, № 1, 19-23, 2013.
20
