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INDETERMINATE EQUATIONS AND UNLIMITED
PROBLEMS.

354. When there are more unknown quantities than independent equations, the number of corresponding values which those quantities admit is indefinite (Art. 198). This number may be lessened by rejecting all the values which are not integers; it may be farther lessened by rejecting all the negative values; and still farther, by rejecting all values which are not square or cube numbers; &c.

By restrictions of this kind the number of answers may be confined within definite limits; and problems are not wanting, in which such restrictions must be made.

355. If a simple equation express the relation of two unknown quantities, and their corresponding integral values be required, divide the whole equation by the coefficient which is the less of the two, and suppose that part of the result, which is in a fractional form, equal to some whole number; thus a new simple equation is obtained, with which we may proceed as before; let the operation be repeated, till the coefficient of one of the unknown quantities is 1, and the coefficient of the other a whole number; then an integral value of the former may be obtained by substituting 0, or any whole number, for the other; and from the preceding equations integral values of the original unknown quantities may be found.

Ex. 1. Let 5x + 7y=29; to find the corresponding integral values of and y.

Dividing the whole equation by 5, the less coefficient,

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If s=0, then = 3 and y=2, the only positive whole numbers which answer the conditions of the equation; for, if x and y are positive integers, 5s cannot be greater than 2, that is, s cannot be greater

2

than and s cannot be negative, for then x would be negative.

5'

If negative values of x and y are not excluded, then an indefinite number of such solutions may be found by putting 1, 2, 3, &c. −1, −2, -3, &c. for s.

Ex. 2. To find a number which being divided by 3, 4, 5, gives the remainders 2, 3, 4, respectively.

Let a be the number,

then =p, a whole number, or x=3p+2;

2 3

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• This operation might have been abridged thus,

x=5−y+2(2−y); let 2=Y=s, or y=2-5s,

5

then x=5-y+2s=3+78. ED.

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The artifices employed in the two following examples are deserving of notice.

Ex. S. Let 11x-17y=5; to find the integral values of x and y.

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And x-2y-5p=22p+2-5p=17p+2.

If p=0, x=2, and y=1; if p=1, x=19, and y=12; &c.

Ex. 4. Let 11x-18y=63; to find the integral values of x and y.
Since 18 and 63 are divisible by 9, let x=9z, then

11×92-18y=63; and dividing by 9,

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Hence x=18p+9; and y=10p+5-3+p=11p+2;

and by giving to p the values 0, 1, 2, 3, &c. the integral values of x and y are determined.

356. If the simple equation contain more unknown quantities, their corresponding integral values may be found in the same

manner.

Ex. Let 4x+3y+10=5%; to find corresponding integral values of x, y and z.

Dividing the whole equation by 3, the least coefficient,

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y = x − x − 3 +

a whole number.

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and y = -2x + 3p+ 1 − 3 + p = 4p − x − 2 ;

and for p and substituting 0, or any whole numbers, integral values of x and y are obtained. If x=3, and p=1, then x=2 and y= −1; if x=4, and p=0, then a=7 and y=-6; &c.

357. If the number of independent equations be less by one than the number of unknown quantities, the equations may be reduced by elimination to one equation only betwixt two unknown quantities, and then the preceding method of solution may be applied.

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from which equation the values of y and z may be found by the usual method, and then the values of x may be deduced from either of the original equations.

[Exercises Zl.]

A more complete Theory of Indeterminate Equations, distinct from the preceding, is contained in the following Articles of this Section:

358. To shew that, if a is prime to b, the equation ax+by=c has one integral solution at least.

c-by

Since ax-c-by, :. x=· ;

a

...

now give to y the several successive values 0, 1, 2, 3, a-1, and since a is prime to b, the several values of c-by, divided by a, will all leave different remainders. [For, if not, let y, and y, be two of the values of y, which make c-by, divided by a, if possible, to leave the same remainder r, q, and q, being the quotients, then

c-by=aq,+r, and c-by-aq2+r,

.. by1-y2)=a (q2-91),

or by-y) is divisible by a without remainder; but b is prime to a, therefore y-y, must be divisible by a, which is impossible, since y1-y2<a].

Hence the remainders being all different, since the number of them is a, and each one necessarily less than a, it follows that one of them must be 0, that is, x is a whole number for a certain integral value of y less than a, and these values of x and y satisfy the equation ax+by=c.

COR. It is obvious, that not only is an integral solution proved possible, but also we are directed to a most simple way of obtaining it, in all cases, where the coefficient of either a or y is small.

Ex. 1. 4x+29y=150, find x and y.

150-29y

4

Here x= =a whole number for some value of y between 0 and 3 inclusive. Try 0, 1, 2, 3, successively, and we find 2 will answer; in which case

150-29y 92
4

=23; .. x=23, and y=2.

4

=

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359. To find all the integral values of x and y which satisfy an equation of the form ax+by=c.

a,

Suppose a and b prime to each other; for, if they are not, (since b, x, and y are all integers by supposition, and every common measure of a and b measures ar and by, and therefore ax±by, or c,) a, b, and c must have a common measure, and dividing by this common measure, the equation is reduced to one in which the coefficients of x and y are prime to each other. Hence it is only necessary to consider the case when a and b are prime.

Let x=a, y=ẞ be one solution of the equation ax+by=c; then

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but is in its lowest terms, since a is prime to b, .. y-ß, and a-x, are certain equimultiples of a and b, or y-ẞ-at, and a-x-bt, where t is integral and remains undetermined;

.. x=a-bt, and y=ẞ+at ;

which values of x and y are found to satisfy the original equation.

Hence, if we know one solution, x=a, y=ß, (which may often be guessed at sight, or found by Arts. 355, 358) all the required solutions are found by giving different integral values to t in the equations

x=a-bt,)
y=B+at.S

Similarly, if the proposed equation be ax-by=c, it may be shewn, (or deduced from the former by writing -b for 6,) that the general solu

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