166 SPECIAL TOPICS IN THEORETICAL ARITHMETIC

But alld

Hence x>h

Then there exists an integer 7, such that x = #nb.
Hence, since by = — ax = — anb, y = — na.

So that (x, ¥) = (nb, — na)

For the converse it is easily seen by substitution that for
every value of n (#nb, — na) is a solution.

Example. The solutions of 3x — 7y = 0 are of the form
(= 7n, — 3n) or of the form (7m, 3m).

416. It may be worth our while to solve one or two examples
without using the general principles that we have derived.

Example 1. 17x + 43y = 60

Of the two coefficients 17 and 43 we choose the numerically
smallest as a modulus.

We have, if (x, y) is a solution, 60 — 43y > 17

Hence 60 = 43y (mod 17)

But 60 = — 8 (mod 17) and 43 = — 8 (mod 17)

Hence — 8 = — 8y (mod 17)

But 8 ]| 17. Therfor y =1 (mod 17) § 281.

Therfory — 1 =17nand y = 17n + 1

Hence 17x =60 —43y=60—43X17n —43=17—-43 X 17n
Therfor x=1—43n

Thus our complete solution is (1 — 43n, 1 4+ 17#)

For 1 — 43n to be positiv, # must be less than 1/43

For 1 4+ 177 to be positiv, » must be greater than — 1/17
For both x and y to be positiv, # must be 0.

The only positiv solution is (1, 1)

Example 2. 15 — 22y = 41
Choosing the numerically smallest coefficient 15 as a
modulus,

22y 4+ 41 > 15
22y = — 41 (mod 15)
22 = 7 (mod 15)
— 41 = 4 (mod 15)

7y = 4 (mod 15)
7y — 4 = 15m