eliminating f, g, and putting y for ei, we obtain

This equation may be solved by either of the preceding methods, if two of its roots are impossible, and the possible value of y, or e2 obtained; the values of a are

The same reducing cubic equation may be obtained by substituting y + for a, and in the resulting equation putting

4xy3+(4x+2a2+a1)y=0.

[2] Waring's Method. Let the equation be

x2 + αz x3 + α2x2 +α ̧x+a=0,

(F. 551—2.)

y.x+ to both sides of this equation, 4

the second side becomes a perfect square,

if

y3 — α ̧y2 + (α1 · α, — 4a) y — a (a3 — 4a,) — a} = 0.

By substituting the possible value of y obtained from this equation, the preceding is reduced to two quadratics, whence the roots may be found. (W. 334, 5.)

This and the preceding method are applicable only when the proposed equation has two possible, and two impossible,

roots.

[3] Euler's Method. Let the equation be reduced to the

Assume = VB1 + √ß2 + √ß ̧‚

(1)

B1, B2, B3, being the roots of the equation y3 +b2y2+b1y+b=0.

Raising (1) to the 4th power, we have

'+2b2x2 — 8 (− b)*x + b2 − b1 = 0 ;

Solution of a biquadratic equation by the method of divisors.

Solution by the theory of symmetrical functions.

(E. 759.)

(Bour. 416, 7; G. A. 53, 5.)

THE EQUATION y" p=0.

(48.) Properties of the equation yn+p=0. This equation

may be reduced to "+1=0, by assuming y=p" x.

x2m-1—1=0 has only one possible root, 1.

m

x2-10 has only two possible roots, 1 and - 1.

x2m−1+1=0 has only one possible root, -1.

x2+1=0 has no possible root.

"+1 has no equal roots.

If a is a root of the equation a"-1=0, a" is also a root, being any integer. And a" =a”-n.

If a is a root of the equation "+1=0, a2r-1 is also

a root.

S+m=0, unless m÷n, in which case S+ ±m=n.

P3

'...&c.

If nn11.n.n.P3... &c. n1, n2, &c. being prime numbers, the solution of the equation "-1=0 may be made to depend on the solution of the equations "-1=0, x-1=0; &c.

(Bour. 393-401; L. C. 15—8; G. A. 46, 7.)

(49.) Theorem of Fermat. If p is a prime number, and a prime to p, then

a2-1-1p.

-1

If a is such a number that no power of a <a2-1p, and a, a2, a3,... a-1 be respectively divided by p, the remainders will all be unequal.

The values of a for all values of p from 3 to 37 inclusive will be found in the annexed Table.

(G. A. 48, 49; Mem. Acad. Berlin, Ann. 1771.)

(50.) Algebraical solution of the equation xo-1=0, p being a prime number.

The roots of this equation,

a, a2, a3, a1, &c. q-1

may be more conveniently represented by

α, aa, aa2, àa3,
aa3, &c. qa1-2

in which a may have any value corresponding to the value of p in the preceding Table.

If aa be substituted for a, the roots become

aa, aa2, aa3, aa1, &c. qa2-2,

In this case the roots are evidently the same as before, but arranged in a different order.

1

Let B, B2, B3, BP-2, &c. be the roots of the equation,
-1=0;
; assume

-2

e=a+Ba" + ẞ2 aa2 + ... + B2 - 2 au2-2;

then 'c”-1= (a + Baa + ß2 aa2 +

observing that all indices of a > a2-1, and of ß >p-1 may be respectively divided by those quantities, and the quotients neglected.

Each of the quantities A, 41, 42, &c. may be reduced to the form

in which в and c are known quantities independent of a; hence the values of 4, 41, &c. are not altered by substituting successively a", &c. in the place of a.

-1

Let cp-1, c-1, &c. cp-2-1, be the values of c-1, when ß3, ß3, &c. ßo-1, are substituted for ß, then