(66.) The equation a+a1x+α2x2 +α ̧x3=y3

admits of a direct solution only when a = b2.

and

then

First Method. Assume

b2 + a1x +a¿x2 + α ̧x3 = (b+cx)3,

If a is not a square, one solution must be obtained by trial, let this be a + a1h + a ̧ḥ2+a ̧h3=k2,

then by assuming x-y+h, the equation is reducible to the above form. (B. 184, 5; E. Pt. 11, 112–27.)

Solution of the equation a+a1x+α ̧x2 +α ̧x3=y3.

[1] Suppose a=b3:

Assume b3 + а ̧x+α ̧x2 + a ̧x3 = (b+ex)3,

If neither of these conditions is fulfilled, one solution must be obtained; let this be

a + a1h+a2h2 + a ̧h3=k3,

then by assuming a=h+y, the equation may be reduced to the form [1]. (E. 147-61; B. 191-5.)

Solution of the equation ax2+cy2=x3.

Assume x=ap3-3cpq2, and y=3ap3q-cq3, then

x= ap2+cq2.

x=t3-3btu-abu3,

Solution of the equation x2+axy + by3 =≈3.

Assume

y =3ť3u+3atu2 + (a2 — b)u2;

(E. 187.)

All 4th

powers are of the same form to modulus a as

0a, 1*, 2*, &c. (±a)*, when a is even,

0a, 1a, 2a, 1⁄2 (a — 1)]†, when a is odd.

Remainders of 4th powers from every modulus from 3 to 12.

m being a number of an impossible form to modulus a, and n prime to a.

No triangular number >1 is a biquadrate.

(Leg. 324-7; B. 71—6.)

INDETERMINATE EQUATIONS OF THE FOURTH DEGREE. (68.) Solution of the equation a+¤ ̧x+α ̧x2+¤ ̧Ã3+A4x*=y2.

[1] Suppose a=b2:

3

assume b2+a1∞ +a1⁄2œ2 +α ̧Ã3 +α ̧x* = (b + ex+ƒx2)3,

assume a + a1x+α ̧‹x2 +a ̧x3 + c2x* = (f+ex + cx2),

[3] Suppose a = b2, and a1=c2:

assume b2 + a1x +α2x2 + α ̧x3 + c2x2 = (b + ex+cx2)o,

If neither of these conditions exists, let one solution be found,

a + a1h+a ̧h2 + a ̧h3 + a ̧h1 = k2,

from which, and the given equation,

then

x=

(q2+3a4)h-2pq

2

2k

2

(E. 128-46; B. 186-90.)

Solution of the equation x2+cy2 =≈1.

Assume xp — 6cp2 q2 + c2 q*,

=

y=4pq(p2 — cq2);

x = p2 + cq2.

Solution of the equation x2+axy + by2 = x2.

Assume xt — 6bť2 u2 - 4abtu3 — (a2 — b)bu*,

=

(E. 198.)

y=4t3u+6at2 u2 + 4 (a2 —b) tu3 + (a2 — 2b) au1; =t2+atu+bu2.

(69.) Solution of the homogeneous equation

or

2-1

(B. 197.)

а x2 +α1x1· 'y + α ̧x2 − 2y2 + &c. + any" = ±e,

f(x,y) = ±e;

in which x and y, as well as y and b, may be considered prime to each other.

Assume x=cy+ex, c being such that

by substituting this value of a, the given equation is reduced to

-1

by" + b1y"−1×+ b2yn−2x2 + &c. + b2x2= +1.

There cannot be more than n values of x, between the limitse and -e, that render the integral polynomial

To determine

p and q, the values of y and x, which render f(y, x) a minimum :

let a1, a2, &c. a, + B-1, &c. be the roots of the equation

is the converging fraction nearest to one of the quantities

a1, α, &c. a,,' &c. which must be determined by experiment.

(Leg. 120-8. E. Add. 28.)

(70.) Solution of the equation x"-b=ay,

a being a prime number, and 6 prime to a.

1 have a common factor c, there will be

c solutions, or none.

4-1

If the given equation is possible, b -1a.

If one solution, xe, has been obtained, the others may be found by multiplying e by the several roots of the equation

xn 1 =ay.

The proposed equation is always possible, if n is prime to a-1.

Solution of the equation "-1=ay.

If n is prime to a-1, the only possible solution is x=1.

Let a

1=en, then x=*, ≈ being any number prime to a.

If u is the remainder of÷a, all the values of a will be found amongst the remainders of the quantities

-1

u, u2, u3, &c. un−1÷a,

unless two or more of these remainders are equal.

If TM — 1=ay is impossible, r being a root of the equation x2-1=ay, and m a divisor of n, then r is a primitive root.

If n1, n2, &c. are the prime factors of n, the number of primitive roots of the equation "-1=ay will be expressed by