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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 uolimited problems ; being such as usually admit of an indefinite number of solutions ; although, when the question is proposed in 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 J.

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

ax-by=+c, or ax+by=c. Where a and 6 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 pumbers out of that of these two

a

a

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

-dy.
wh, or wh.

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a

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3. Take such a multiple of one of these last formulæ, 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 , and the expression

ay above used, or any multiple of it that comes near , 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

yr

=wh.=p, and y=apir, 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 x 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 always 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=C, the number of answers will always be limited ; and, in some cases, the question is impossible ; both of which circumstances may be readily discovered, from the mode of solution above given. (n)

EXAMPLES.

194 =wh.

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

14y+11

= wh., and also 19

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

=wh. 19 19

19 5y-11 20y-44

y-6
X4=

-=wh.
19
19

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

swh.

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

Also,

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y-6

с

с

m

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 asmd, 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 1002. in guineas and moidores only, the question would be impossible ; since, in the equation 213 +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 35 19 19

19

14p+5. Whence, if

р

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.

25Here

=12-y+

2 Hence, since x must be a whole number, it follows that 1-4

must also be a whole number.

Let therefore !4=wh=p;

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

And since

=12–

2 We shall have x=11+3p, and y=1-2p ; Where po 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 p must be either 0 or negative ; and consequently, from that of t, that it must be 0, -1, -2, or - 3.

Wbence, if p=0, p= -1, p=2, p=-3,
Then

Sx=11, a=8, 2=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 3:=8y- 16 to find the values of x and y in whole numbers.

=wh. ; or

2y-1

24

=wh.

Here x=
8y - 16

2y-1
=2-5+
3
3

3
Also
2y-1

4y - 2
X2=

=wh.
3
3

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

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Therefore y=3p+2,

8y– 16_8(3p+2) – 16_24p=8p.
And x=
3

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

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

5-177 Here x=

=wh. ; 21

21

2000-17y=95+

5–17y=wh.;

Or, omitting the 95,

21

214

5-17y Consequently, by addition, +

21

21

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Also,

4y + 5
21

4+20y=whi;

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21

Or, by rejecting the whole number 1,

4+20y=wh.;

21

21

21

21

217 4+20y_y_4=wh.=p ; And, by subtraction,

Whence y=21p+4, 2000—177

2000–17(21p+4) =92–17p. And x=

21

21

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