INDETERMINATE EQUATIONS OF FIRST DEGREE 163

For then ax + by = ¢
and ax; + by = ¢
Hence a(x —x1) +0(y —v) =0
and alx —x1) = — by — 1)
Therfor a(lx —x1) >0
But alld
Therfor x—x 30 § 267.

Hence x = x; (mod b) and an integer n exists such that
x — x1 = nb, whence x = x; + nb.

Also by —v1) = —alx — x1) = — anb
Hence y—y1 = —an
Y — Y
y = y; (mod a)
and Yy =9y — na

407. Th. If (x1, 1) is a solution, then every solution (x, vy)
can be put in the form (x; + nb, y1 — na), where n is some
integer, and conversely, for every integral walue of n
(%1 + nb, y1 — ma) is a solution.

The first part of the theorem follows immediately from the
preceding theorem.

For the converse, a(x; + nb) + b(y1 — na) = ax1 + by, = ¢

Therfor, whatever integer #n is, (x; + nb, y1 — na) is a
solution.

Otherwise stated: If (x1, y1) is a solution, the expression
(x1 + nb, y1 — na) represents every solution and nothing but
solutions. i

Def. On this account (x; + #nb, y1 — na) is called a com-
plete solution.

408. Th. The complete solution of the equation can be written
in the form (lc + nb, mc — na), where al + bm = 1.

This follows from §§ 405, 407.

Example. A complete solution of 49x + 19y = 900 is
(6300 + 197, — 16200 — 49x).

The numbers lc and mc¢ may be numerically large, as in this
example. Shortly (§§ 410, 411) we will show how to get a
simpler solution.