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9MATHEMATICAL SCIENCES
CLASSES OF THE DIOPHATINE EUQUATIONS
Druzhinin V.
Doctor of phys-mathematical Sciences, Professor the head of the Department of higher mathematics SarFTI NRNU MEPhI, Sarov.
Korableva E.
schoolgirl Lyceum 15 name Yu. Khariton,
Sarov
Abstract
Classification of the of Diophantine equations proposed and it allows you to quickly find solutions for them. Determinant of divisibility method is used to solve linear and nonlinear equations.
Keywords: Diophantine equation, determinant of divisibility.
Linear of the Diophantine equation (DE) with two unknowns of the form AxL + ByL = C have countless solutions xt = x + Bl, yt=y- Al, if the natural coefficients A and B are mutually simple (A 1 B) and I is an arbitrary integer. To find the initial integers of these arithmetic progressions «x» and «y» we use Euclid's algorithm, continued fractions and Euler's algorithm [1,2]. This procedure is rather cumbersome and we will offer a more General and simple solution method. Consider the DE of the form Ax + By = 1. Let us always: A> B; A-B = K; A = Kt + m; K 1m; B = Kt — n; n = K-m; 1<m<K— 1; 1 < n < K — 1; K = m + n. It turns out that the solution x = at + ft at lal < IP I defines the structure of the second solution y = —at + a and the General form of the coefficients A and B. That to do this, we first make an auxiliary DE in the form
ftK + an = 1. (1)
In this DE we know a and ft and we have the unknowns are K and n. Finding them we obtain m = K — n and coefficients of the original DE A = Kt + m and B = Kt — n. Moreover, we find an unknown parameter
o = , (2)
k ' v '
defining the roots of the equation DE y.
As an example, consider the case x = 2t -1. DE (1) has the form -K + 2n = 1. As solutions it is possible to take K = 7, n = 4, m = 3. Next we find a = (-2 • 3 — 1)/7 = —1 and we have the second root y = —2t — 1. As a result, we get DE: every type (7t + 3)x + (7t -4)y = 1 has the roots x = 2t-1 and y = —2t -1. At t = 2 DE 17x + 10y = 1 you
has x = 3, y = -5. So we have the first sub-class DE: all DE with K = 7 and m = 3 form a sub-class DE with roots x = 2t - 1 and y = -2t - 1. The General class DE with x = 2t - 1 is obtained by multiplying the primary roots K = 7 and n = 4 of the auxiliary equation (-K + 2n = 1) in the standard way: K =7 -21, n = 4-1, m = 3-1. Thus, we obtain a class DE with two
parameters { t;l }
[(7 - 2l)t + (3- l)]x + [(7 - 2l)(t - 1) + (3 - l)]y = 1 (3) whose roots are x = 2t - 1 and y = -2t - 1. If you fix in the DE parameters {K; m}, the class (3) forms a sequence of pairs [{5; 2}, {7; 3}, {9; 4}, {11; 5},... ]. If you go to another roots x = -2t + 1 and y = 2t + 1, they can be combined with the previous result and write the class DE
A = (2m ±1)t + m; x = ±(2t - 1); y = ±(-2t-1). (4)
For example, for m = 5,t = 3 and the minus sign we get up to 32x + 23y = 1 with roots: x = -5,y = 7.
We give some General classes of solutions.
1. If m = 1, then x = -t + 1; y = t.
2. If n = 1, then x = t;y = -t-1. (5)
3. If K = 2n + 1; m = 2, then x = ^t- (p.-1);y = -¡it - 1.
4. If K = 2n + 1; n = 2, then x = -pt+ 1;y = lit + (ji- 1).
For other values {K; m} the control solutions are shown in the table 1. In table 1 class (3) is highlighted by some dimming. Table 1 on the structure of the integers A and B allows you to find the {K; m} and find the roots. The table 1 can be expanded to any size and with its help instantly find the roots of DE.
Ta6n. 1.
Solutions DE Ax + By=1 if A = Kt + m, B = A- K_
K; m x y K;m x y K;m x y
7; 3 2t — 1 —2t — 1 11; 3 —4t + 3 4t + 1 13; 3 4t — 3 —4t — 1
7; 4 —2t + 1 2t + 1 11; 4 —3t + 2 3t + 1 13; 4 3t — 2 —3t — 1
8; 3 —3t + 2 3t + 1 11; 5 2t — 1 —2t — 1 13; 5 5t — 3 —5t — 2
8; 5 3t — 1 —3t — 2 11; 6 —2t + 1 2t + 1 13; 6 2t — 1 —2t — 1
9; 4 2t — 1 —2t — 1 11; 7 3t — 1 —3t — 2 13; 7 —2t + 1 2t + 1
9; 5 —2t + 1 2t + 1 11; 8 4t — 1 —4t — 3 13; 8 —5t + 2 5t+3
10; 3 3t — 2 —3t — 1 12; 5 —5t + 3 5t + 2 13; 9 —3t + 1 3t + 2
10; 7 —3t + 1 3t + 2 12; 7 5t — 2 —5t — 3 13; 10 —4t + 1 4t+3
Here are some classes of control included in the table 1:
A = (3m ±1)t + m; x = ±(3t — 2); y = ±(—3t—1); A = (3l + 1.5 ± 0.5)t + (21 + 1); x =
±(3t — 1); y = ±(-3t-2).
(6)
Next, we consider some new approaches to direct solution of the DE but somewhat different from the Euclid-Euler method. When constructing a continued fraction, it is sometimes useful to use not the remainder of the division, the smallest modulo the deduction. For example, DE 163x + 42y = 1 on a classical set gives such a set of equals: 163 = 3^ 42 + 37; 42 = 1 • 37 + 5; 37 = 7 • 5 + 2; 5 = 2^2 + 1. Next, the Euler
algorithm from the set {2; 7; 1; 3} yields modules of roots {17; 66}. If we choose a deduction in the first equation, and write 163 = 4 • 42 — 5, then we can proceed to DE —5x + 42z = 1. This DE is easily solved
by "eyes": x = —17, z = —2, i.e. the first root we found quickly.
The second approach is based on the determinant of divisibility method (DDM) proposed earlier by Dru-zhinin [3]. The meaning of this metods is as follows. Let the dividend F = F^^b1 + f0 and the divisor S = S-tb + s0. Then necessary and sufficient condition of divisibility F on S (SlF; F = 0(mod S)) is the divisibility on S determinant D(F,S), i.e. SID(F,S) or D(F,S) = 0(mod S), where D(F,S)
D(F,S) =
F
(—1)t+1Sl
fu
= FisO + (—íyfoSl
(7)
In many cases D (F, S) « F, therefore DDM is convenient in a number of operations. Let's see how the DDM participates in the decision of the do. We have DE 121x + 34y = 1. The chain of Euclid: 121 = 3^34 + 19; 34=1^19 + 15; 19 = 1 • 15 + 4; 15 = 3-4 + 3; 4 = 1^3 + 1, gives multipliers {1; 3; 1; 1; 3} and modules roots {9; 32}. There are 17 arithmetic operations. If we rewrite the original DE in the form (112x — 1) = y(3 • 11 + 1) then use (7) we obtain the equation
—1 9 1 I
= (x — 9) = 0.
(8)
DDM
—1
= (8x + 1) = k9.
(9)
From (8) it follows that x = 9. There are only three arithmetic operations. Here is another example for the use of DDM. We have DE 343x + 9y = 1. We write it in the form 73x — 1 = (7 + 2)y and make a
From (9) we immediately find x = 1, k = 1 which gives y = —38.
DDM allows to solve non-linear DE. For example, 7x3 + 3y2 = 83 is overwritten as (7x3 — 83) = —(7 — 4)y2. For both brackets we make a determinant (7) and get the equation (—4x3 + 83) = 3k. There is a solution for x = 2. Because 7x3 + 3y2 = 83 there is x = 2 and y = 3.
REFERENCES:
1. Graham, Z., Knuth, D., Patashnik O. Concrete mathematics, Moscow, MIR, p. 322, 1998 .
2. Buhshtab V. V. The theory of numbers. M. S.-P. Lan, 2015.
3. Druzhinin V.V., Lobov L. A., Sirotkina A.G. // NTVP, No. 4, p. 17-19, 2012.
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