168 SPECIAL TOPICS IN THEORETICAL ARITHMETIC

Example 3. 23x — 17y = 63
The complete solution is (177 + 2, 237z — 1).
x and y are both positiv when, and only when, # is positiv.

417. The process used in working the examples of the last
article is that which would be suggested by the following
theorem:

Th. If ay, by, c1, %1, x2 are integers, a;, —= 0, and

a1x1 + bixe = C1,

if also by = by (mod ay) and ¢ = ¢y (mod ay), then there exists
an integer x3, such that

ai1X3 + b29€2 = (9 (1)

For by hypothesis bixs — ¢; = — a1 >> o4

Hence bixe = ¢; (mod a;)

Also byxs = bixe = ¢1 = ¢3 (mod a,)

Hence baxs = ¢3 (mod a,)

Hence boxs — €2 > ay

Therfor there exists an integer — x3 such that

boxs — ¢2 = a;(— x3) and hence
a1xs + bexe = co.

418. In applying this theorem to the solution of the equa-
tion aix; + bixs = c¢1, assuming that neither @; nor b, is zero
and that a; || b, we form a series of equations like the
following :

a1xy + b = ¢
a1X3 + b2x2 =02
a2x3 + boxy = c3
asXs5 + bsxs = €4

in which the a’s and b’s gradually decrease in value numer-
ically. When one of them is equal to 1 numerically, our series
is finisht. It is easy then to get the general solution of the
original equation.

() This theorem discoverd Feb. 14, 1931.