( 235 )

-1. In this example a=-1, b=1, c——1, d=i,

and r=

C

——, and consequently x=

=

ar 3 3d2 3r2d-b 91 which value of x will make x3-x2+x-1 a complete cube.

26. When both the first and last terms are cubes, as in a3+bx+cx2+d3x3, we make the formula become a cube by a different value of x from those given by the methods explained in the 24th and 25th paragraphs. We may assume a+dx for the root, this cubed and compared with the proposed formula gives a3+bx+cx2+d3 x3-a3 +3a2dx+3ad2x2+d3x3; whence bx+cx2=3a2dx+3ad2x2, and by division b+cx=3a2d+3ad2x, from which we obtain x=

3a2d-b

c-3ad2

•

Example, let 1+2x+4x2+x3 be made a cube. Here 1+x is to be assumed as the root, and we have 1+2x+4x2+x3=1+3x+3x2+x3, whence 2x+4x2=3x+3x2, or x2=x, and consequently x= which value being put for x makes 1+2x+4x2+x3 1+2+4+1=8=23.

1,

Two other values of x may be found that will render 1+2x+4x2+x3 a complete cube, by pursuing the methods taught in the 24th and 25th paragraphs.

27. When the first and second terms are both wanting, the general cubic formula becomes cx2+dx3, which is made a cube with the greatest ease. We have only to assume cx2+dx3-m3x3, which by division becomes c+dx=m3 x, from which is had x=

C

m

3

and m may be taken at pleasure.

28. Sometimes we easily obtain the general value of the unknown quantity, which will make a + bx +cx2+dx3 a complete cube. This always happens when the equation a+bx+cx2+dx3-0 has two equal roots. To exemplify this let 2-7x+8v2-3x3 be made a cube.

By putting 2-7x+8x2—3x3 0 we discover that.

there are two equal roots each unity, and therefore 2-7x+8x2-3x3 —(2—3x)×(1-x)2. Now this last is made a cube in a general manner by assuming

MB

(1-x), whence =

2n3m3 in which m and n may be taken at pleasure. 3ns-m3 If m=n=1, we find x=1, which value substituted for x will not fail to render 2-7x+8x2-3x3 a perfect cube.

29. When we have by trial or otherwise discovered one value of the unknown quantity that will render a quadratic or cubic formula a perfect cube, we may generally find as many other values as we please by a proper substitution. The method of finding these will be sufficiently understood from the fol lowing example: let 7-3 be made a cube. Here we easily find one value of x that fulfils the proposed condition, viz. x=-1, for 7—(−1)3=7+1=8=23; let us therefore put x=-1+y, and by substitution 7—x3—8—3y+3y—y3, which must be a cube; suppose its root 2-y, and we have 8—3y+3y2—y 3 = : 8 12v+62—3, whence -3y+3y2=12y+6y, or -3+3y=12+6y, and therefore y=3, and of course 3-12, which value in fact fulfils the required condition, for 7—x3—7—8——1=(~1)3. But by assuming the root differently we may find another value of y from the same equation. Let 2-ry denote the root, and we have 8-3y+3y2 — y3—8—12ry+ 62233; which by assuming 12r=3, or r=1, becomes 3-y3=6r2 y2 —r3y3, and dividing by y2

we have 3-y=6r2-r3y, and therefore y

3-6-2

=+, whence x={; and this value of ✰ makes 7---x:3

=(1)

T30. But in many instances all our endeavours are ineffectual in such researches; in fact there are innumerable formulas that never can become cubes: among these 1+x3, 1-x3, x3-1, are the most singular, or by introducing two unknown quantities x3 +y3, and 23-y3, neither of which can ever become a cube, as may be completely demonstrated.

31. Having thus briefly run through several of the most common cases of single equalities, we shall say a few words on double and triple equalities, which are of most frequent use in the Diophantine Analysis. Let us find x such that ax+b, and cx+d may be both squares.

Assume ax+b=z2, whence x=

22-b

and therefore

a square, and therefore acz2+a2d-abc must be a square, and may be treated according to the rules already given for single formulas of the second degree.

Example. To find such a value of x that 2x+3, and 4x+5 may be both squares. Assume 2x+3=z2, Z2-3 and 4x+5=2z2—1, which must

whence x=

2

be a square. By pursuing the methods taught in the preceding paragraphs for formulas of the second degree, we may find as many values of z as we please, and among others 5, 29, 169. If z=5, we have x= 72-3

2

11, and 2x+3=22+3=25-52, also 4x+-5 =44+5=49=72; if z=29, we have x=419, and if z=169, we find x=14279. The learner may further exemplify this by finding such values of x as will make 2+x, and 2-x both squares.

32. Let us find x such as will render ax+bx2, and cx+ dx2 both squares.

Dividing each of these formulas by x2, the quo

a

C

tients+b, and -+d must evidently be squares, which

x

1

by putting y become ay+b, and cy+d, formulas

already considered in the last paragraph. The reader may try as an example, to find x such that x+x2, and x-x2 may both be squares.

¶ 33. Let ax+b, and cx2+d be made squares by the same value of x.

Assume ax+b=z2, whence x=

and cx2+d

a

a2

-= a square, and therefore c(z2—b)?

+a2d=cz4—2bcz2+a2d+b2c must be a square. This is to be effected when possible by the rules given for biquadratic formulas.

2

Example. To find x such that x+1, and x2+1 may both become squares. Put x+1=z2, therefore x= z2-1, and x2+1=z4-2z2+2 must be a square, and to have x positive, z must be greater than 1. Since the formula za—2z2+2 is a square when z=1, let us put z=1+y, and by substitution z4-2z2+2=1+ 4y+4y3+4: suppose now, according to a remark in the 15th paragraph, that 1+2y2 is the root, and we have 1+4y2+4y3+j4=1+4y2+414. wherefore 4y+y44y4, or 4-3, whence y, z=1+3=3, and x=-14-140. Now this value 40 put for x will evidently make x+1, and x2+ 1 both squares, for 4+1=4(3), and x2+1=160°+1=1081. (4). The reader will no doubt observe, that the methods given in this paragraph applies equally to the formulas ax+b, and cx2+dx+e.

2

I

T34. Let us find x such that ax2+b, and cx2+d may be both squares.

We must begin with satisfying ourselves that each of the proposed formulas may become a square. Then having found by trial or otherwise a value r, which, substituted for x, will render ax2+b a square, we must put x=r+y, and by this substitution the proposed formulas are transformed into ar2+b+2ary +ay, and cr2+d+2cry+cy. And since, when x=

r, ax2+b=ar2+b= a known square e2, our first formula becomes e2+2ary+ay2; and, putting cr2+d =f, our second becomes ƒ+2cry+cy2. To make e2 +2ary+ay2 a square, imagine zy-e to be the root, and there arises the equation e2+2ary+aye1—2ezy

2artez.

+z22, whence by reduction y=

z2-a

This value

of y being substituted for it in f+2cry+cy, will produce a formula of the fourth degree in terms of %, which may, when possible, be made a square by the rules already taught for such single equalities.

We are by no means to conclude that both equations can become squares with the same value of x, merely because each may be made a square with a different value: the contrary happens frequently; the formulas x2+1. and x2-1 may both become squares, but not with the same value of x, for then (x2+1)× (x2 — 1) x4 would also be a square, which we know to be impossible.

It is easy to see that a similar method may be used to find such a value of x as will render the formulas ax2+bx+c, and dx2+ex+f complete squares.

We shall take an example. To find x such that x2+1, and x2+5 may both be squares. It is evident that each of these formulas may be made a square separately; therefore we may commence with making x2+1 a square. Suppose zx-1 to be its root, and we shall have the equation x+1=1—2zx+z2x2,

2%

5z4-6%25

22 - 1,2

and therefore x2+5

= a square, and there

fore 5z4-6z+5 must be a square. The slightest attention is sufficient to discover that when z=1, the formula 5z4-6z2+5 becomes a square; but this va

2

lue of z gives x= from which we can draw no

1

conclusion: this value of z therefore does not answer