4x4 2x3 x2 + 3x Here, the first term

+

And consequently, 1-2x + 3x2 - 4x3 5x = 1−1+3

5

9

+

16

16'

which is a square number, as was required.

2. It is required to find such a value of x as will make

3x 2 (2x2 - 1 x

= 4x4 - 2x3- 2x2 +

x+

16

16

256'

+ y2 y

2.

ม

And, therefore, multiplying this by y1, which is a square, it will be 4-2y — y2 + 3y3 — 2y1. Where the first term being now a square, if the expression, so transformed, be

resolved by Case 1, we shall have y

3. It is required to find such values of x as will make 1+ 3x + 7x2 2x2 + 4x4 a square.

Here, both the first and last terms being squares, let 1 +

And if we put the same formula, 1 + 3x + 7x2

2x3 + 4x4 6x3 + 4x4, we shall

6x3; whence 6x

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 = 1+y.

Then 2x-1= 2 (1 + y)* − 1 = 2 (1 + 4y + 6y2 + 4y3 + y1) − 1 = 1 + 8y + 12y2 + 8y3 + 2y1. And since the first term of this last expression is now a square, we shall have, by Case 1, 18y + 12y2 + 8y3 + 2y1 = (1 + 4y — Qy2)2 = 1 + 8y + 12y2 — 16y3 + 4ya.

. 1

Whence, as the three first terms of the two members of this equation destroy each other, there will remain 4y1 16y3 — 2y1 + 8y3; or y 12; and consequently x = 1+ y 13; which value being substituted for x, makes 2xa 57121 = (239)2, 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 = e3, and make e3 + cx + bx2 + ax3

Then, by expunging the two first terms on each side of the equation, which are common, there will remain ax3 +

bx2

C2

3e3

278+2; whence, by division and reduction, we shall have 27ae6x+27becx+9c2e3, and consequently 9e3 (3be — c2)

; which form fails when the coefficients b

and c, or a and c, are each equal 0.

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

b

= ƒ3, and make ƒ3æ3 + bx2 + cx + d = ( ƒ x + 3ƒ2)3

b2

b3

73.3

Then, by expunging the two first terms on each side of

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

27fe;

b2

3f3

x +

whence, by multiplying by 27f we shall have 27ƒ°cx + 27df9b2f3x + b3, and consequently, x= b3 - 27dfo

9f3 (3cf3b2)

6

; which form likewise fails when b and c, or

b and d, are each = 0.

3. When the first and last terms are both cubes, put a = ƒ3 and d = e3, and make e2+ cx + bx2+ƒ3x3 = (e +ƒœ)3 = e3+3fe2x+3ƒaex2 +ƒ3x3.

Then cx+bx

Whence we shall have bx

=

3ƒe2x +3ƒ3ex2; 3ƒ3ex=3fe2

C, and conse

which formula may also be resolved

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 ap3 + bp2 + cp+d=q3.

Then, by putting ≈ = y+p, we shall have ap3 + bp2 + cp + d a (y + p)3 + b (y + p)2 + c (y + p) + d = ay3 + (3ap + b) y2 + (3ap2 + 2bp + c) y + ap3 + bp2 + cp + d, or ax3 + bx2 + cx + d = ay3 + (3ap + b) y2 + (3ap2 + 2bp + c)y + q3. From which latter form, the value of y, and consequently that of x, may be found as in Case 1.

EXAMPLES.

1. It is required to find such a value of x as will make x2 + x + 1 a cube.

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

Then, by cubing, we shall have 1 + x + x2 = 1 + x + x2 + 27 203;

And since the two first terms on each side of this equation destroy each other, there will remain a22+2. Whence, dividing by a2, we shall have += 1, or x+9= 27; and consequently, x=27—9—18; which num

3

ber, by substitution, makes 1 + x + x2

1+18324343

73 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 =

which last number will also lead, in like manner, to

other new values.

2. It is required to find such a value of x as will make x3-3x2 + 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 + x23 = (1+x)3 = 1 + 3x + 3x2 + x3.

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 3 + 3x2 + 133 = 91125

3

(45)3, 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 828x+89x2 125x3 a cube.

Here, let the root sought 2 – 5x, according to Case 3. Then, by cubing, we shall have 8+ 28x + 89x2 — 125x3 (2 – 5)3 = 8 60x150x2 125x3.

And since the first and last terms of this equation destroy ea h other, there will remain 28x+89x2 60x + 150x2. Whence, by dividing by x, and transposing the terms, we shall have 150x 89x28+ 60, or 61x 88; and conse

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

4. It is required to find such a value of x as will make

Here, 1 being a value of x that is readily found, by inspection, let x y-1, agreeably to Case 4.

Then, by substitution, we shall have 2x3 2(y-1)-3(y-1) + 7 = 2y3 — 6ya +3y+8.

3

3x+7

And as the last term of this expression is a cube, let

according to Case 1.

Then, by expunging the equal terms on

each side, there will remain 2y3 — 6y2 — 3 y2 +

1

Whence, dividing by y2, and reducing the terms, we shall 384 = y + 24, or 127y 408; and conse

have 128y

And, by taking this last as a new value of x, others may be determined by the same method.

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 formulæ, 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=□, and cx + d = 0,

let the first of these formulæ ax + b = 22, and the second cx + d = w2.

Then, by equating the two values of x, as found from these equations, we shall have cz2 + ad bc aw2, or acz2 + a (ad — bc) = a2w2.

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, acz2 + a (ad — be) = □ ; which being done, according to the method

pointed out in Problem 1, we shall have x =

22

α

b

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

1

ax2 + bx , and cx2 + dx 0.

Let x; then, by substitution, and multiplying each of the resulting expressions by y2, we shall have

a+by, and c + dy + □,