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(66.) The equation

The equation at a,& + aqx + 2328=yo admits of a direct solution only when a=be. First Method. Assume

62 + a, il + QgcX2 + az x3 = (b +cx)?, and

a=2bc; then

a,' 46*a,

аз Second Method. Assume

b2 + ax + a222 + azx=(b + cx + exo)?, and

a=2bc,

ag=c? + 2be; Az

-2ce then

(E. 26, 7.) e?

46° az

If a is not a square, one solution must be obtained by trial, let this be atah + a, ha + azh'=k, then by assuming x=y+h, the equation is reducible to the above form.

(B. 184, 5; E. Pt. II, 112~27.) Solution of the equation a + ax + agw? + azx3=. [1] Suppose a=63: Assume 63 + Qqw + Qgw? + Q303 = (b +ex), then

an

362 [2] Suppose ag=c: a + a,& + 22x2 + c* x=(e +cx)),

e3 then

and x=

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assume

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and x =

3ce-a,

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[3] Suppose a=b, and ag=c?:

:
b3 + a, 6 + aqw+ c® XS = (b + cx)},

assume

then

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

a + anh + a, ho + azh=k, then by assuming w=h+y, the equation may be reduced to the form [1]

(E. 147-61; B. 1915.) Solution of the equation ax + cy' =x'. Assume x=ap3 3cpq', and y=3apřq-cq', then s=ap? +cq.

(E. 187.)
Solution of the equation 2? + axy +by'=xo.
Assume x = t - 3btu'- abu",

y = 3t u + 3atu+ (ab)u”;
then
8 =ť + atu + bu?.

(B. 196.)

FORMS OF BIQUADRATES.

(67.) Every even biquadrate *.24.n.

Every odd biquadrate # 24. n +1.
All 4th powers are of the same form to modulus a as

04, 14, 2*, &c. (La)*, when a is even,

04, 1o, 2o, įsa-1)]', when a is odd. Remainders of 4th powers from every modulus from 3 to 12.

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Impossible forms of equations of the fourth degree.

** +y*=**,
x* + a’yt=,

and generally,

ma*+may=x, 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.)

a1

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INDETERMINATE EQUATIONS OF THE FOURTH DEGREE. (68.) Solution of the equation a+a,x+Qx*+azw3 +@qw*=yo.

[1] Suppose a=52: assume b + a,x + agu? + a 3.213 + a4w* = (b + ex +fxo),

e and

f= 2b

25 then

az – 2ef

f2 — aĄ [2] Suppose an=c?: assume a + ax + aza? + azw3 + c*w*=(f+ex + cw*)',

a,

e and

2c

f – a then

a, -2ef [3] Suppose a=b', and a,=c?: assume b? + a,x + QgQ2 +2.3.23 + c*** = (b +ex + cx?)®,

a 1

e + 2bc - ag then

26'

az F2ce

аз

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2c

e=

and x=

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

a + anh + a,h? + azh + agh* =k, from which, and the given equation,

h

(K* - 8 ak + 4a). 4a? - 364

M

If the equation is a + ag X? + 04@* =yo, let one solution be found,

a + a,ho + ach'=ko; a,h + 2a,h

a + 6a,ho-p. k

2k

then if p=

and a

(qe + 3a) h - 2pq

(E. 128–46; B. 186-90.) q24 Solution of the equation xo + cy'=x4. Assume x=p* – 6cp?q? +c*q*,

y=4pq(p? — cq°); then x=po + cq°.

(E. 198.) Solution of the equation 2? + axy +by' =**. Assume x=+* 66ťuo 4abtu – (a' b)bu“,

y=4t'u + 6at u+ 4(a’ b)tu+ (a’ - 26) au; then x=t+atu + bu'.

(B. 197.)

or

(69.) Solution of the homogeneous equation
aw" + a,20-2y+Q221 < y2 + &c. + any= +e,

f(x,y)= e; in which w and y, as well as y and b, may be considered prime to each other. Assume x=cy +ex, c being such that

ac" + a_0"-1 + a,c" – 2 + &c. + ante; by substituting this value of x, the given equation is reduced to

by" + b yn – 1% + ba yu – m2 + &c. + b^x" = +1. There cannot be more than n values of x, between the limits + de and - {e, that render the integral polynomial Q(x).

(B. 88.)

p and

To determine

9,

the values of y and >, which render

f(y, %) a minimum : let aj, Ag, &c. ar +B. -1, &c. be the roots of the equation

bw" + 6,21 – 1 + b2.21 – 2 + &c. + bn = 0, then is the converging fraction nearest to one of the quantities

P

9 Q1, A2, &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 b prime to a.

If n and a-1 have a common factor C, there will be c solutions, or none.

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

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

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

Solution of the equation 2" – 1=ay.
If n is prime to a – 1, the only possible solution is x=1.
Let a – 1=en, then x= xe, x being any number prime to a.

If u is the remainder of ma, all the values of x will be found amongst the remainders of the quantities

u, u', u, &c. un-11a, unless two or more of these remainders are equal.

If 7 – 1=ay is impossible, r being a root of the equation x" – 1=ay, and m a divisor of n, then r is a primitive root.

If ni, ng, &c. are the prime factors of n, the number of primitive roots of the equation 8" – 1=ay will be expressed by

ni
1 n, 1

&c.

n.

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