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Determinant sign of divisibility for polynomials
Section 6. Mathematics
Druzhinin Victor Vladimirinich, National research nuclear university “MIFI", Sarov physics and technology institute, department of the higher mathematics E-mail: Sarov, [email protected]
Determinant sign of divisibility for polynomials
Abstract: Anntation: We have proposed a new method of working with polynomials — a new sign of divisibility of each other. The method significantly reduces the number of operations when working with polynomials. We have given some examples, and have formulated a generalization of Fermat’s last theorem.
Keywords: Determinant sign of divisibility, Fermat’s last theorem.
We have proposed a new method of working with polynomials — a new sign of divisibility of each other. The method significantly reduces the number of operations when working with polynomials. We have given some examples,and have formulated a generalization of Fermat’s last theorem.
Determinant sign of divisibility (DSD) for the numbers was formulated by the author in [1-3] and is expressed by the following theorem.
Dividend A = aÄ"A A0 is multiple of the divisor B = ßAxB B0 if and only if D (A;B ) = k ■ B, where k is an integer, and D (A;B) is the determinant
D (A; B ) =
a
\ n+1
(-1) ßn вп
When working with integers, in symbo looks so
(aA1 + A ):(ßAi + A)
A
. (1) ics this theorem
a
= k ■ В.
(2)
(-i) ß m
As follows from (1-2), by using DSD A1 is removed from the calculations, that often greatly simplifies the definition of divisibility.
Example. Let’s check the divisibility of A = 100006 by B = 31. Here
a = 1, A = 10,n = 5,ß =3, A0 =6, B0 =1,
D (A;B) = 1 - 35 ■ 6 = -1457 = -47 ■ 31. is multiple of B: 100006 = 3226 ■ 31.
Since a polynomial of n-th order Pn(x) = ^ak ■ xk
k=0
determines the number, then the statement (2) is automatically transferred to the polynomials and is written as
{ ={aPm + PNa A =AP + PLA A
Indeed, A
a
\ m+1
P
(-1) ßm Pm
= k ■ Pl•
(3)
First we give a simple example proving the theorem. Let PN =A Ay2>A = x + y , then Pn = e = x,a=ß=1 =
1 - V2
= -y 2,a = 2, Plo a a, D (PN; PL ) = ' = 0. There is
-1 V
divisibility, and we have a school rule: x2 -y2 =(x + y)(x-y).
Next, we consider PN (x,y ) = x3 + 2x2y + 2xy3 + y3 and check its divisibility by PL = x + y. So Pn = P3 = x 3,a = ß = 1, PN 0 = 2x2 y + 2xy3 + y 3,a = B,
PL 0 = У , D ( PN ; PL ) =
2x2 y + 2xy3 + y3
y3
= -2xy (x + y).
Since D(PN;PL ):( + У), then PN (x,y ):(x + y). Indeed,
x3 + 2x2y + 2xy3 + y3 =(x + y)(x2 + xy + y2). In the general case we obtain
(4)
(x + y)^£xn kyk = xn + 2^x" kyk ■
■y
Check divisibility P3 (x, y, z ) = x3 + y3 + z3 — 3xyz by
Pj (x, y, z ) = (x + y + z 1 y3 + z3 - 3xyz
have
1
(y+z )
. We work with variable « x ». We = 3yz(x + y + z), that indicating at
divisibility. Indeed, о3 + y3 + z3 — 3oyi = ( + y + z )x x^2 — xy + y2 — oz + z2 — xy).
The next example is related to the famous Euler chain of primes (x2 + x + 4l). For all integer numbers 0 < x < 39, it gives 40 consecutive primes. When x = 40 we have a composite number 412. The end of this chain can be found by using theorem (3). We write the divider of Euler’s equation in the form PL = x + y, where Y — an unknown number. By taking Pn = P2 = x2, we make a quadratic equation
1 x + 41 2 \
^ 2 =b + x + 41 = A (x + Y).
The discriminant { — 4x(l — k) —164} = 0, where x = 40,k = 2, gives the divider of bracket (x2 + x + 4l) equal to «41».
п
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Section 6. Mathematics
With DSD for polynomials, we can simplify the decision-governmental nonlinear Diophantine equations.. For example, there is x2 — 8x + 2y2 = 1 . We write 1 = x + a and make D =a2 — 8x + 2y2 = 0 . Its solution a = 0, x = 1,у = 2 satisfies the initial equation.
Considering the solution ofthe equation x2 + y3 — z4 = 0, we introduce the divisor {(x + z2) + a } andmake an equation
D =
x - z
1 a
= a(x - z2)-y3 = 0.
By introducing a new variable a, we have reduced the degree of the polynomial. His solution: x x —27; x x 28; x = 6; x = 8 satisfies the original equation.
With using DSD you can solve the algebraic equation of high degree. For example, the equation x5 — 9x3 — 8x2 + 72 = 0 can have the following roots ( ±) -(l; 2; 3; 4; 9; 12; 18; 24; 36; 72} by theorem Vietta. We check the root x = 3 by checking the divisibility of the equation to (x — 3).
D =
1
1
-9x3 - 8x2 + 72
-243
= 9x3 + 8x2 - 315 = 0.
When x = 3, equality occurs. We have two other valid roots {2; — 3}.
Applying DSD to Fermat’s Last Theorem xn + yn Ф zn, when {x; y; z } natural number, and n an integer bigger than «2». Formulated 350 years ago, it was proved in 1994 by English mathematician Wiles [4,5]. We will consider odd n = 2m +1 and write the left part of the equality in the form
2m
x2m+1 + 42m+1 =(x + yЩ(-1) x2m-k ■ yk =(x + y)S2m(x,y).
We decompose the polynomial S2m — incomplete degree « 2m » the difference of two numbers in Taylor series in powers of (x + y ).
2m
П(-1)‘х2m-k ■ yk =
k=0
=Z(-i)
k
t=0
(2m +1) (t +1)
C2m ( + У ) У
2m-t
(5)
In (5), C2m = {m )!/1! (2m -1)!}- binomial
coefficient. For example,
S2 = x2 - xy + y2 = (6)
= (x + y) -3y(x + y) + 3y2 =(x + y) -3xy.
We wondered whether xn + yn = z2 ? We take n = 3 and get
x3 + y3 =(x + y) -3y(x + y) + 3y2(x + y). (7)
Assuming that x = 3y2 — y, we substitute x in (7) and get
x3 + у3 = 9y4 (y2 -3y +1).
At y = 1, x = 2 get 23 +13 = 32. There are other sets of numbers, giving x3 + y3 = z2:
1843 + 83 = 24 962;1543 + ( -7 )3 = 19112.
As for equations x5 + y5 = z2; x7 + y1 = z2 and so on, our computer search found no such triples of integers, except in the trivial solution x = y = z = 0 and x = z = l,y = 0. We consider x5 + y5 = (x + y ) S4 = z2. S4 can be written, except (5), as S4 =(x + y) -5xy(x + y) + 5x2y2 . Let
5x2 y2 =ß(x + y ) .If y2 = ß (x + y ),then ßx = y (y - ß)
, that is forbidden because x and y are coprimality. If x2 y2 =ß(x + y), then y2 =(ß/x ) + (ßy / x2), y is not an integer. We consider (x + y) = 5(x + y) = 5 . Then x5 + y5 =(x + y) ^125 — 25xy + x2y2). There are two sets of (x;y):x = 1,y = 4 andx = 2,y = 3 . The second parenthesis in the last equality is equal to «41» or — «11», i. e., x5 + y5 Ф z2. Qualitative and computer analysis of this equation for different values of (x + y ) also gives x5 + y5 Ф zk for k = 2;3;4 . For other powers n , this approach also provides such inequality. At the moment, we do not claim that our proof is absolutely correct.
Therefore, we put forward the hypothesis: xn + yn Ф z2 when 2 < k <(n — l), z is natural odd, n — 5 , x and y are different parity and coprime. If the hypothesis is correct, then we obtain a generalization of Fermat’s last theorem in the form: xn + yn Ф z1+m when meN, which includes the Fermat’s theorem.
The author thanks professor, the doctor of physical and mathematical sciences Yu. N. Deryugin and N. S. Shevyahova for discussion and support.
References:
1. Druzhinin V V Determinant sign of divisibility, Sarov, Alfa, 2012.
2. Druzhinin VV. European science review. № 1-2. P. 24-26, 2015.
3. Druzhinin V V Lobov L. A., Sirotkina A. G. NTVP, 2012, № 5, P. 17.
4. Sizyi S. V. Lectures on the theory of numbers, M. Fizmatgiz 2007.
5. De Agostini. Fermat’s Last Theorem, Moscow, 2015.
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