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And consequently, 1 – 2x + 3x2 - 4x + 5x4 =1-1+2

5 9 -1+

which is a square number, as was required. 16

16' 2. It is required to find such a value of x as will make 4224 2:43 22 + 30

2 a square. Here, the first term being a square, let 4x4 2x3 22 + 5

5 25 3х - 2 = (2x2 – Ža )2= 4x4 - 2x - 2x2 +

X + 16

16 256 according to the method in Case 2.

5 25

5 Then we shall have 3x - 2 = ac +

2 16 256

16 25 +

Whence 768x 80x = 512 +25; and conse

or 3x

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256*

512 + 25 537 quently, a =

768 - 80 688

1 Or, if we put x =

the formula in that case will become

and x =

4 2 1 3

+ - 2. y уу?

Y And, therefore, multiplying this by y', which is a square, it will be 4 - 2y

- 2y - y? + 3y - 2y4. Where the first term being now a square, if the expression, so transformed, be

688

1 resolved by Case 1, we shall have y

537'

Y 537

as before. 688'

3. It is required to find such values of x as will make 1 + 3x + 7x2

2003 + 4x4 a square. Here, both the first and last terms being squares, let 1+

3

25 3x + 722

x 2x22 = + 2 + 2.

4 6x3 + 4x*, according to the method in Case 3.

25 Then, we shall have 6x03 + 30% = 7X2

4

233; or 6x + 200 25

3 =

; and, by reduction, x = 4

32 And if we put the same formula, 1 + 3x + 7802 - 223 + 4x4 (1+30 - 2x2)2 =1+ 3x

6x3 + 4x4, we shall have, by cancelling, 722 - 2x = - 2002 633 ; whence 6x

35

35 7.

; or x = 4

16

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1 ty

And, in a similar manner, other values of x may be found, by employing the method of substitution pointed out in the latter part of Case 3.

4. It is required to find such values of x as will make 2x4

- 1 a square. Here, 1 being an obvious value of x, let, according to Case 4, X

Then 2x* - 1=2(1 + y) - 1=2(1 + 4y + 6y2 + 4y? + y) - 1=1+ 8y + 12y2 + 8y + 2y*. And since the first term of this last expression is now a square, we shall have, by Case 1, 1 + 8y + 12y2 + 8y + 2y = (1 + 4y -272)2 = 1 + 8y + 12y2 - 1673 + 4y4.

Whence, as the three first terms of the two members of this equation destroy each other, there will remain 444 16y3 = 244 + 8yo; or y == 12; and consequently x = 1 + y

13; which value being substituted for x, makes 234 1

57121 = (239), as required. And if 13 be now taken as the known value of x, and the operation be repeated as before, we shall obtain, for another value of x, the complicated

10607469769 fraction

2447192159 PROBLEM 4.-To find such values of x as will make 3 (ax3 + bx2 + cx + d) rational, or ax3 + bx2 + cx + d a cube. This formula, like the two latter of those relating to squares, cannot be resolved by any direct method, except in the cases where the first or last terms of the expression are cubes; it being necessary, in all the rest, that some simple number answering the conditions of the question, should be first found by trial before we can hope to obtain others, but when this can be done, the problem, in each of the cases here mentioned, may be resolved as follows:

RULE 1.- When the last term d of the given formula is a cube, put it co?, and make e3 + cx + 6x2 + ax3

c3 ( c 3 1 73

x2 +

23.

. 3e

27 28 Then, by expunging the two first terms on each side of the equation, which are common, there will remain ax: +

C3 bx2

az ; whence, by division and reduction, we shall have 27ae® x + 27be6 = cx + 9c2e", and consequently 9e(3be — c2)

; which form fails when the coefficients 6 03 27 aer and c, or a and c, are each equal 0.

C2

C2

x3 +

278

3e3

x + 276

27f° ? i

2. When the coefficients a of the first term is a cube, put it

b =fo, and make f*x3 + bx2 + ca + d == (fæ+372)* = f*x* 62

63 + bx2 +

3f** * " Then, by expunging the two first terms on each side of

32 the equation, as before, there will remain cx + d

to

3f3 33

; whence, by multiplying by 27f6 we shall have 27f6cx + 270f -= 9b2f3x + b3, and consequently, x =13 - 27df

; which form likewise fails when b and c, or 9f3 (3cf3 52) b and d, are each = 0.

3. When the first and last terms are both cubes, put a = fy and d=e, and make e + cx+622 +f323 = (e +f«)' = 63 + 3fe*x +3f4ex2 +f***.

Then cx + = 3fe-x+3f2ex"; Whence we shall have ba 3f4ex = 3f02 — C, and conse

3fea --C quently, x=

which formula may also be resolved

5 3fe by either of the two first cases.

4. When neither the first nor the last terms are cubes, let P be a value of x, found by inspection or by trials, and make ap + bpa + cp + d= q*. Then, by putting x= y +p, we shall have ap3 + bpa + cp

= a(y + p)3 + b (y + p) toly +p+d=ays + (3ap +7) ya + (3apa + 2bp + c) y + ap3 + bp2 + cp + d, or ax? + bx2 + cx +d=ays + (3ap+b) y + (3apa + 2bp + c)y + q.

From which latter form, the value of y, and consequently that of x, may be found as in Case 1.

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EXAMPLES.

1. It is required to find such a value of as will make 202 + x + 1 a cube.

Here, the last term being a cube, let the root of the cube sought = 1 + fx according to Case 1.

Then, by cubing, we shall have 1 +-x + x2 = 1 + a + 3x2 +27 22;

And since the two first terms on each side of this equation destroy each other, there will remain 22 = 22 + y oci?.

Whence, dividing by xa, we shall have 2 + }=1, or x +9= 27; and consequently, x=27 – 9=18; which num

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ber, by substitution, makes 1 + x +22=1+18+ 324= 343

173 a cube number, as was required.

And if we now take this value of x, and proceed according to the method employed in Case 4, we shall obtain x = 137826

which last number will also lead, in like manner, to 50653 other new values.

2. It is required to find such a value of x as will make 2013 + 3x2 + 133 a cube.

Here, the first term being a cube, let its root = 1 + x, according to Case 2.

Then, by cubing, we shall have 133 + 3x2 + x3 = (1+x)3 =13x 3x2 + 23.

And since the two last terms of this equation destroy each other, there will remain 1 + 3x 133, or 3x = 133 - 1

132 132; whence x = 44, and 23 + 302 + 133 == 91125

3 = (45)], a cube number, as was required.

And if 45 be now taken as a known value of x, other values of it may be found, as in the last example.

3. It is required to find such a value of x as will make 8 +- 28x + 89x? 125203 a cube.

Here, let the root sought = 2 -- 5x, according to Case 3. . Then, by cubing, we shall have 8 +- 28x + 89.22 — 125803 = (2 -- 5x)3 = 8 60x --- 150X? 125x3.

Ind since the first and last terms of this equation destroy ea h other, there will remain 28x + 89x 60x + 150x2.

Whence, by dividing by x, and transposing the terms, we shall have 150x --- 89x = 28 + 60, or 61. x = 88; and conse

88 quently x =

61 And as this formula can also be resolved either by the first or second Case, other values of x may be obtainerl, that will equally answer the conditions of the question.

4. It is required to find such a value of x as will make 273 3x + 7 a cube.

Here, - 1 being a value of x that is readily found, by inspection, let a

y 1, agreeably to Case 4. Then, by substitution, we shall have 223 3x +7= 2(y - 1) — 3(y 1)+7= 2y3 - 6y2 + 3y + 8. And as the last term of this expression is a cube, let

1

3

1 8 + 3y - 6y2 + 2yo = (2+ày)?= 8 + 3y + y

64.30

643?

according to Case 1. Then, by expunging the equal terms on

1 each side, there will remain 243 – 6y2 = y +

Whence, dividing by yé, and reducing the terms, we shall have 1284 — 384 = y + 24, or 127y 408; and conse408

408

281 quently, y =

1
127
127

127 Which number, by substitution, makes 2x3 - 3x +7 2 x (281)

281

45118016 356 +7

as required. (127)

2048383 127) And, by taking this last as a new value of X, others may be deternined by the same method.

and x =

-3 127

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3

PROBLEM 5.- Of the resolutions of double and triple equalities

When a single formula, containing one or more unknown quantities, is to be transformed into a perfect power, such as a square or a cube, this is called, in the Diophantine Analysis, a simple equality; and when two formula, containing the same unknown quantity or quantities, are to be each transformed to some perfect power, it is then called a double equality, and so on; the methods of resolving which, in such cases as admit of any direct rule, are as follows:

RULE 1.--In the case where the unknown quantity does not exceed the first degree, as in the double equality,

ax +b= 0, and cx + d= ], let the first of these formulæ ax + b =%, and the second cx + d = wa.

Then, by equating the two values of X, as found from these equations, we shall have cza + ad · 5c awa, or acza ta(ad bc) = awa.

And since the quantity on the righthand side of this equation is now a square, it only remains to find such a value of z as will make, when the question is resolvable, acza + a (ad bc) = 0); which being done, according to the method

22 pointed out in Problem 1, we shall have x =

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a

2. When the unknown quantity does not exceed the second degree, and is found in each of the terms of the two formulæ ; as in the double equality

axa + bx = [], and cxa + dx Let x=; then, by substitution, and multiplying each of the resulting expressions by y?, we shall have

a +by = 0, and c + dy + 0,

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