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CHAPTER X

DIOPHANTIAN EQUATIONS AND PROBLEMS*

INDETERMINATE EQUATIONS OF THE FIRST DEGREE

252. It has already been learned that, in case the number of unknown quantities is greater than the number of independent equations, there will be an unlimited number of solutions, and the equation will be indeterminate. However, it is possible to limit the number of solutions by introducing conditions which the unknown quantities must satisfy. When it is required that the unknown quantities shall be positive integers, the equations are called simple indeterminate equations.

253. In the present chapter the solution of indeterminate equations of the first degree, containing two and three unknown quantities, will be considered, in which the unknown quantities are restricted to positive integers.

Every equation in two unknown quantities can be reduced to the

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where a, b, c, are positive integers which do not have a common divisor.

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C.

by = C can

included in form (1) can not be satisfied by positive integral values of x and y; because if a, b, x, y are positive integers, ax + by must be a positive integer which can not be equal to a negative integer, Furthermore, the equations ax + by = c and ax not be solved in positive integers if a and b have a common divisor. For, if x and y are positive integers, the common divisor of a and b must also be a divisor of axby, and consequently of c; which is contrary to the hypothesis that a, b, and c have no common divisor.

* Diophanti Arithmeticorum, libri VI. Diophantus lived, according to Abulforag, about 340 A. D., in Alexandria.

254. The placing of this restriction on the variables enables one to express the solution in a very simple form.

EXAMPLE 1. Solve in positive integers 2 x + 11 y

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= 49.

where the quotient is written as a mixed expression. Transposing

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Since the values of x and y are restricted to having positive integral values, then x+5y 24 will be an integer, and, therefore,

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1- will be an integer, although written in a fractional form.

2

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Equation (2) shows that n, with respect to y, can be zero; or can have any negative value, but can not have a positive integral value.

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This equation shows that n, with respect to x, can be

1, 0, +1,

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etc., but can not have a negative integral value greater than - 1. Hence it follows from (2) and (3) that, for

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which are the only positive integral solutions of the given equation.

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above example, as will be seen in what immediately follows.

The rule is, in any case multiply the numerator of the fraction by such a number that the coefficient of the unknown quantity shall exceed some multiple of the denominator by unity.

In case this had not been done the work in the last part of the solution of Example 2 would stand as follows:

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Substitute in (1); then (6)8 x — 168 r - 63 = 33

.'. x = 21 r + 12.

The values of x and y differ in form from those found above, but the same system of values for x and y is obtained, since here for r = 0, 1, 2, 3,

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and by giving to n in system I any positive integral value, an unlimited number of values for x and y is obtained; thus

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Solution (7), or (8), is called the general integral solution of equation (1).

The student will see, in the solutions of Examples 1 and 2, that there is a further limitation to the number of solutions introduced according as the terms in x and y are connected by the plus or minus sign.

255. Should there be two equations involving three unknown quantities, they could be combined so as to eliminate one of them and have an equation containing two unknown quantities; then the process would be that of the previous examples.

EXAMPLE 3. In how many different ways can the sum of $5.10 be paid with half-dollars, quarter-dollars and dimes, so that the whole number of coins used shall be 20?

Consider that the sum $5.10 is reduced to dimes.

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and according to the last condition of the problem,

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and by substituting in (1) 10x+5y+402 (x + y) = 102

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and there are no other positive values of x, y, z which will satisfy equations (1) and (2); and therefore there are three ways in which the given sum can be paid: one half-dollar, 18 quarter-dollars, and one dime; 4 half-dollars, 10 quarter-dollars, 6 dimes; and finally, 7 half-dollars, 2 quarter-dollars, 11 dimes.

256. THEOREM I.-Given one solution of ax - by integers, to find the general solution.

Suppose that and m is one solution of ax al-bm c. By subtraction,

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0.

=c in positive

by

= c, so that

Since a is a divisor of a (x — 1), it must be a divisor of b (y — m); a must therefore be a factor of y m, since a is prime to b. mat, where t is any integer; then from (1)

Let y a (x

- 1)

=

b (y — m)

=

abt, and therefore

Hence if x =

ах by = c,

(2)

x

1 = bt. 1, y = m be one solution in integers of the equation all other solutions are given by the equations,

x lbt and y m at,

where t is a positive integer; therefore

(3)

x = 1+bt, and y = m + at.

Hence, if one solution is known, it is possible to obtain as many solutions as may be desired by assigning to t different positive integral values. It is possible also to give t such negative integral values as make bt and at less than 7 and m respectively.

EXAMPLE. The smallest positive integral numbers which satisfy 9x-5y = 1, are x = 4, and y 7; what are therefore the next five solutions?

=

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