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Table of the least Values of x and y in the Equation x?-ay=1, for

all Values of a from 2, to 99. (E. v. 2; p. 89.)

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9100 66 12 2 20 2574

5

59

69

4 226153980

3 19 10 7 649 15 4 33

60 61 62 63 65 66 67

17

68

2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 26 27 28 29 30 31 32 33 34 35 37 38 39 40 41 42 43 44 45 46 47

69 70 71 72 73 74 75

76

77 78

66249
485
89
15
151
19603
530

31 1766319049

63 8 129

65 48842

33 7775 251 3480

17
2281249
3699

26
57799
351
53
80
9
163
82

55
285769
10405

28 197 500001

19 1574 1151 12151 2143295

39

49 62809633

99
10

6
3
2
180
4
1
8
4
39
2
12
42
5
1
10
5
24
1820

2
273
3
4
6
1
12
6
4
3
320

2
531
30
24
3588
7
1
14
7
90

170 9 55 197 24 5 51 26 127 9801

11
1520
17
23
35
6
73
37
25
19
2049

13 3482 199 161 24335 48 7 99 50 649

79

80 82 83

1 16

8 5967

4 936 30 413

.2 267000 430

3
6630
40
6
9
1
18
9

6
30996
1122

3 21 53000

2 165 120 1260 221064

4

5 6377352

10
1

84

85 86 87 88 89 90 91 92 93 94 95

48

96 97

50 51 52

98 99

(63.) Solution of the equation me ay: = +6, in which b < Va.

If this equation is possible, b will be found amongst the denominators of the complete quotients (a,, az, az, &c. Art. 31.) of the converging fractions which express the value of Va.

Given one solution of the above equation to determine the general values of x and y. Suppose that we have found, m, n, p, q, such that

m“ – ané= +b, and aq'= +1, then

x=mpang,

y=np+ma, in which the general values of x and y obtained in the preceding Art. must be substituted for p and q respectively.

The equation på - aqi=1, or p’ - aq=-1 must be employed, according as the known solution and the given equation have the same or contrary signs.

To determine the general values of x and y in the above equation, b being > Va. [1] Suppose b to be composed of factors , , &c. each <Va: the solution of the given equation, when possible, may be deduced from those of the equations

m,' an,'= +by me - an, = + bą, &c. for (m, - an,%) (mze - ang) (mz+ang) &c. x - ay®.

Having found one integral value of x and y, the general values may be found as before by means of the equation

- aqi= +1. [2] Suppose b not to be composed of factors < Va: it will in this case be necessary to find the values of t and u in the equation

ť - au’= +bx?,

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m? - an? +b; calling this the known solution, the general values may be found as before. In this case however the values may be fractional.

(B. 181—4.) (64.) Solution of the equation a wo + bay + cy= +e.

In this equation, x and y, as well as y and e, may be considered prime to each other.

This may be reduced to a similar equation of a simpler form, by assuming

X = ,X, +by
y = 0,X, +b, y,
A=aa,? + bayag +caz,
B=2(aa,bz+cab,) +b(a, b, + a,),

C=ab,? + bb, b, + cb.
Q1, Qg, and 61, 62, being so assumed that a, b, - a,b,= +1.

If b> a, and >c, the equation may be transformed into another in which 6 $ a, or c. Suppose a < c; assume w=X1 - my,

b m being the nearest integer to i assume also

b,= am- b, and C=am' bm +c, the equation becomes a 0,2 6,819 +cıyo = +e, in which b, $ a.

If b,>c, the same process must be repeated.

In the successive transformed equations, the value of - 4ac is always the same.

If he - 4ac < 0, a and c are the least numbers contained in the transformed equation, in which 6 $a, or c.

(Leg. 53–8; B. 100_2.)

a

[1] Suppose b4ac <0, and= -f. Multiplying the equation by 4a, and assuming

x=2ax + by, we obtain

z + fy=4ae.

y<? (

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If any integral value of y< 2 which satisfies this

f equation, also renders by+x2a, a solution may be obtained; if both these conditions are not fulfilled, the equation is impossible in integers. [2] Suppose 6? —- 4ac=0.

In this case the first first side of the equation becomes a perfect square, and is therefore possible, if e is a square. [3] Suppose bo - 4ac>0, and =k?.

In this case the proposed equation may be decomposed into two simple equations

ayx + bay=eng and aq x + bey=ég. [4] Suppose b? — 4ac=4f, f not being a square.

First; let e be < Vf: expand a root of the equation

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in a continued fraction. (p. 32.) If amongst the quantities 201, 202, &c. the denominators of the complete quotients, we find the number e, the numerator and denominator of the corresponding converging fraction, if substituted for x and y respectively, will satisfy the given equation. As often as e occurs, a different solution

may

be obtained; if it does not occur at all, there is no integral solution of the proposed equation.

Secondly; suppose e > Vf: assume

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cy=ae; the equation is reduced to a,x? + 6,01. +cy=+1, which must be solved for each value of n.

The given equation is impossible in integers, unless n may be so assumed, that a, may be an integer. (Leg. 75–83.)

Solution of the equation ax + bay + cy'= +1.
Let it be transformed so as to fulfil the conditions

b$ a, or c, and a < c; multiply the equation by a, and assume x=ax + by, the equation is reduced to + eye = +a, which may be solved by preceding methods. (E. Add. 66—71.)

FORMS OF CUBES. (65.) Every cube * 4n, or 4n+1;

#7n, or 7n +1;

#9n, or 9n+1.
All cubes are of the same form to modulus a, as the cubes

03, 19, 23, &c. (a– 1)*.
as to a to modulus 6.
The equations 2013 + y3 =%,

2013 yo
+

63

202 + y =2x", and generally M 23 + naye =%, in which m is a number of an impossible form to modulus a, and n prime to a, are impossible in integers. No triangular number < 1 is a cube.

(B. 60-70; Leg. 328-32.)

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