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OF THE

INDETERMINATE ANALYSIS.

In the common rules of Algebra, such questions are usually proposed as require some certain or definite answer ; in which case, it is necessary that there should be as many independent equations, expressing their conditions, as there are unknown quantities to be determined ; or otherwise the problem would not be limited.

But in other branches of the science, questions frequently arise that involve a greater number of unknown quantities than there are equations to express them ; in which instances they are called indeterminate or unlimited problems ; being such as usually admit of an indefinite number of solutions ; although, when the question is proposed io integers, and the answers are required only in whole positive numbers, they are, in some cases, confined within certain limits, and in others, the problem may become impossible.

PROBLEM I.

To find the integral values of the unknown quantities x and y in the equation.

ax-by=+c, or ax+by=c. Where a and b are supposed to be given whole numbers, which admit of no common divisor, except when it is also a divisor of c.

RULE.

1. Let wh denote a whole, or integral number ; and reduce the equation to the form by #C

c-by -wh, or x= wh. 2. Throw all whole numbers out of that of these two

a

a

e

expressions, to which the question belongs, so that the numbers d and e in the remaining parts, may be each less than a; then

dyte

=wh, or

e-dy=wh

a

a

3. Take such a multiple of one of these last formule, corresponding with that above mentioned, as will make the coeficient of y nearly equal to a, and throw the whole numbers out of it as before. Or find the sum or difference of all, and the expression

ay above used, or any multiple of it that comes near ag, and the result, in either of these cases, will still be =wh, a whole number.

4. Proceed. in the same manner with this last result ; and so on, till the coefficient of y becomes = 1, and the remainder = some number y; then

ytr

=wh.=p, and yap Fr, Where p may be o, or any integral number whatever, that makes y positive ; and as the value of y is bow known, that of 2 may be found from the given equation, when the question is possible (m). Note. Any indeterminate equation of the form

ax-by=+c, in which a and b are prime to each other, is alwaye possible, and will admit of an infinite number of answers in whole numbers.

a

(m) This rule is founded on the obvious principle, that the sum, difference, or product of any two whole numbers, is a whole number ; and that, if a number divides the whole of any other number and a part of it, it will also diyide the remaining part,

But if the proposed equation be of the forma

ax+by=sc, the number of answers will always be limited ; and, in some cases, the question is impossible ; both of wbich circumstances may be readily discovered, from the mode of solution above given. (n)

EXAMPLES.

19y wh.

[ocr errors]

1. Given 19x – 14y=11, to find x and y in whole numbers. Here x =

14y+11

=wh., and also 19

19 Whence, by subtraction, 19y_14y+11_5y - 11

=wh. 19 19

19 5-11

20-44
X4=

=wh.
19
19

19
And by rejecting y-2, which is a whole number,

2. =P
19
Whence we have y=19p+6.

Also,

y-6

[ocr errors]

-=Y-2+

y-6 wh.

с

C

[ocr errors]

m

(n) That the coefficients a and b, when these two formulæ are possible, should have no common divisor, which is not, at the same time,a divisor of c, is evident; for if a=md, and b=me, we shall have ax+by=mdx + mey=c; and consequently dx +eyBut d, e, x, y, being supposed to be whole numbers

must also be a whole number, which it cannot be, except when m is a divisor of c.

Hence, if it were required to pay 100L. in guineas and moidores only, the question would be impossible ; since, in the equation 210 +-27y=2000 which represents the conditions of the problem, the coefficients, 21 and 27, are each divisible by 3, whilst the ab. solute term 2000 is not divisible by it. See my Treatise of Algebra, for the method of resolving questions of this kind, by means of Continued Fractions.

14y+11 14(19p+-6)+11 266p+95 And %= 19 19

19

14p+5. Whence, if p be taken =0 we shall have x=5 and y=6, for their least values ; the number of solutions being obviously indefinite.

2. Given 2x+3y=25, to determine x and y in whole positive numbers. Here =

1-Y 2

2 Hence, since x must be a whole number, it follows that -- 4must also be a whole number. 2 Let therefore =w=p;

2
Then 1-y=2p, or y=1-2p.

And since
=12–y+'}=12–(1–2p)+p=12+3p=1,
We shall have x=11+3p, and y=1-2p ;

= Where

p may be any whole number whatever, that will render the values of x and y in these two equations positive. But it is evident, from the value of y,

that

р

must be either 0 or negative ; and consequently, from that of 2, that it must be 0, -1, -2, or -3.

W bence, if p=0, p= -1, p=-2, p=-3,
Then

x=11, x=8, x=5, x=2

y=1, y=8, y=5, y=7 Which are all the answers in whole positive numbers that the question admits of.

3. Given 3x=8y- 16 to find the values of x and y in whole numbers.

[graphic]
[ocr errors]

=wh. ; or 245?

=

=wh.

Here x=

8y - 16
=y-5+
2y - 1

2y1
3
3

3
2y-1

4y - 2 X2=

=wh.

3 Or, by rejecting y, which is a whole number, there will remain

Also 245?

u

[graphic]

Therefore y=3p+2,

8y- 16_8(3p+2) – 16_24p=8p.
And =
3

3

3 Where, if p be put =1, we shall have x=8 and y=5, for their least values; the number of answers being, as in the first question, indefinite.

4. Given 21% +17y=2000, to find all the possible values of x and y in whole numbers.

y

4y+5

=wh. 21

4+20y=wh.;

4+20y=wh.;

217_4+20y=y=+=wh.=p;

-4 And, by subtraction,

21 21 21

Whence y=21p+4, 2000—177

2000–17(21p+4) 21

21

=92-17p

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