lue of will also render the proposed expression a square.

Again, both the first and last terms of 1+2v+4v2 +8v3+4 being squares, we may assume 1+v+v2, 1+v-v2 for the root. The former of these roots being squared, we have 1+2v+4v2+8v3 +va=i+ 2v+3v2+2v3 +v4, whence 4v2 +8v3=3v2+2v3, or 4+8v=3+2v, and consequently v=-; which negative value put for v, will also make the given expression a complete square.

If 1+v-v2 be assumed for the root we shall have 1+2v+4v2 +8v3 +v4=1+2v−v2 2v3+v4, and therefore 42 +8v3 =—v2—2v3, or 4+8v——1—2v, from which is found v; which value of v equally fulfils the condition required.

Further, we may also assume 1+4v+v2, or - I +40+2 for the root. The former of these furnishes us with the equation 1+2v+40+803+v+= 1+8v+18v2+8v3 +v4; and therefore 4v2+2v=18v* +8v, or 4v+2=18v+8, whence v-. -3; a value which will not fail to render the proposed expression a complete square.

If −1 +4v+v2 be supposed the root, we have 1+ 2v+4v2+8v3 +v+=1—8v + 14v2 + 8v3 +v+; and therefore 2v+4v2= -8v+14v2, or 2+4v=-8+ 14v, consequently v=1; and this value also makes the proposed formula a complete square. We have therefore found these six values of v, viz. 4, 77, — —, —, and 1; each of which fulfils the condition required.

Let us take the formula 1+20v+4v4 for an example. Here we may suppose either 1+100+2v2, or 1+10v-2v2, or 1+2v2, or 1-2v2: in the first case we shall find v=3, in the second 2, in the third 5, and in the fourth-5.

- =

Let 12v-v4 be made a square. Assume 1-2 for the root, and therefore 1-v2+v3 — v4 = 1—2+4, consequently v3 -v4=1v4, or 1 —v=4v, whence v=4, which substituted for v makes l—v2 12 10=213=(37)2.

64 256

Let —1+v—v2+v4 be made a square. Here the root to be assumed is −2+v2; therefore squaring and equating we have-1+—v2+v+= {—v2 +04, consequently-1+, and ; a value which admits of an easy verification.

¶ 17. When a biquadratic or a cubic formula cannot be reduced to any of the cases already stated, it may be proper to inquire whether it cannot be resolved into two factors, one of which is a square. Thus, the biquadratic formula 2-4 v+v2+2v3 —v4 is not reducible to any of the forms yet explained, nevertheless it is very easy to make it a complete square; we can even assign in one general expression, all the possible values of v that can satisfy the condition required. For, on examination we find that the equation 2-4v+v2 + 2 v3 —v4 = 0 has two equal roots each unity, and therefore the proposed expression may be expressed by (1-v)2x(2-v2); it is manifest therefore that we have only to make a square of 2—v2, which is easily performed in its fullest extent by the methods already exhibited.

18. When a biquadratic formula includes only the extreme terms and the middle one, if a+bv2+ cv, it is in general very difficult to manage: there is however one case of this kind which sometimes occurs, and is very easily made a square; I mean the formula a2-b2 q2 +c2q4, which will always be -a b

α

C

a square, when we take v=z, or, or, or, four very simple values. Thus the formula 1-9v2+v4 is a square when v, or -, or 3, or 3; but the method of discovering these values is too obvious to be insisted on.

19. When we have by trial or otherwise discovered one value of the unknown quantity that makes a formula of the fourth order a complete square, we may in general find as many others as we please: an easy example will be sufficient to explain this. Let 2-4 be made a square.

Here one value of v is evidently unity; but we wish to obtain another value. For this purpose let us put =1+y, and by this substitution we have 2q← — 1 —4 y——6y2-4y—y4; and assuming the root 1-2y-5y2, we have 1-4y6y2 —433 —4— 1 — 4y -6y2+20y3+25y4, whence-4y3—4—20μ3 +25y4, therefore y=-13, and consequently v=1+y=1}

; if therefore we substitute for v, the formula 2-4 will undoubtedly become a complete square. If now we desired more values of v, we should only have to substitute for v in the formula 2-ya, and proceed as before.

20. There are yet several other particular methods for equations of the fourth order besides those already mentioned; but we have assigned a sufficient portion of our paper to such equations: we shall therefore conclude our remarks on them at present with observing that there are innumerable formulas of this order that cannot possibly become squares. 204-202+2 is such a one: whatever value we assign to v whole or fractional, this formula can never be a complete square. But the most remarkable of these impossible formulas are the following, viz. v4+ 1, 4, and 104, or which are the same in effect by introducing two unknown quantities, v4+14, and 44, neither of which can ever be a square. Also since v4y4(v2 +y2)×(v2 —y2), it follows that the formulas v2+y2, and v2—y2, cannot be both squares at the same time, and that neither of the formulas v2 + 2 v.

2

2

can be a square; these by taking v=1,

and

neither of which can ever

be a square: lastly, it is easy to demonstrate that

never can be a square.

21. These impossible formulas throw light on. some of the most abstruse subjects in mathematical

science; I mean the nature of such fluxionary expres

and y

&c.

sions as y➡y^, y No mathematician has been able to reduce such expressions into forms of which the fluents may be assignable by finite algebraic quantities, or by circular arcs and logarithms. The reason of this seems to be, that whatever substitution we make for y, or y we shall never be able to remove the surd, and obtain an expression in monomials or the form of a rational fraction, because these expressions cannot possibly become rational. Every fluxion of which the fluent is assignable in algebraic terms alone, may be transformed into one or more monomials free from surds, and having only given quantities in the denominators: and every fluxion of which the fluent is assignable by circular arcs and logarithms, may be freed from surds, and reduced to one or more rational fractions, having the variable quantity in the denominators;

ions of elliptic and hyperbolic arcs, cannot possibly become rational, and therefore cannot have their fluents assigned by circular arcs, and logarithms; in other words, elleptic and hyperbolic arcs cannot possibly be measured by the arcs of the circle and parabola.

T22. Let a+bx be made a rational cube.

Assume m for the root, a+bx=m3, from which we obtain x=

tration.

m3

6

this is too easy to require illus

23. Let a+bx+cx2 be made a rational cube. Here it is necessary that the first term should be a cube, in which circumstance our formula becomes a3+bx+cx2. Let us assume a+rx for the root, which cubed and compared with the proposed expression gives a3+bx+cx2=a3 +3a2rx+3ar2x2+

*3×3; and taking 3ar=b, which determines r, we have cx2=3ar2x2+r3x3, and by division c=3ar2+

C-3r2

3x, whence x=—

3

This case may be very

properly exemplified in the formula 1+x+x2; but this we leave to the learner as an exercise.

24. Let a + bx+cx2+dx3 be made a cube. Assuming a+rx for the root, we shall have this équation, a+bx+cx2+dx3=a3 +3a2rx+3ar2x2+ 3x3; and making 3a2r=b, which determines the value of r, we have 3ar2x2+r3x3=cx2+da3, which by division becomes 3ar2+r3x=c+dx, and therefore 3ur 2-c

the value of x=

d-r3

As an example of this method, let 1-x+x2-x3 be made a rational cube. By assuming 1-rx for the root, we have 1−x+x2—x3 = 1—3rx+3r2x2r3x3; and by making 3r=1, or r=}, we have x2— x3=3r2x2-r3x3, which divided by x2 becomes 11-3r2 x=3r2_r3x, and therefore x=

value being substituted for x, we find 1

which

25. When the last term of a cubic formula is already a cube, as in a+bx+cx2+d3x3, we may substitute for x, and the resulting formula multiplied

1

У

y

by 13 will become d3+cx+6x2+a, which coincides with that resolved in the last paragraph. But we may proceed in a more direct manner by assuming r+dx for the root, which cubed and equated with the proposed formula, gives a+bx+cx2+d3x=r3 + 3r2dx+3rd2x2+d3x3: now let us make 3rd2=c, or

and we shall have a+bxr3+5r2dr, a sim

ple equation, from which we obtain x=

a183

312db

Of

this we have an example in the formula x-x2+x