112 SPECIAL TOPICS IN THEORETICAL ARITHMETIC

261. Th. If a, b, c are integers, a positiv or zero, b positiv,
and c not zero, if a pair of positiv or zero integers s, t can be found
such that sa — tb = c, an infinit number of such pairs of integers,

all positiv, can be found.

For if sa — tb = ¢, (s + b2)a — (¢ + az)b = c.

Let 2z be any positiv integer.

Then, since b is positiv, bz is positiv.

Therfor, whether s is positiv or zero, s + bz is positiv.

If a is positiv, since also z is positiv, az is positiv.

Then, since ¢ is positiv or zero, ¢ + az is positiv.

If a =0, ¢t is positiv. For, if £ = 0, since sa — tb = ¢,
¢ = 0, which contradicts the hypothesis.

Alsot + az = t.

Therfor in this case also ¢ + az is positiv.

Therfor, if s’ = s 4+ bz and ¢/ = ¢ 4 az, then (s, ¢) is a pair
of positiv integers such that s'a — #'b = c.

Again, if z and 2’ are two different positiv integers, since
b is positiv, bz —= bg

Hence s+ bz—=s5+ b

Thus the pairs of integers (s + bz, ¢+ az) and
(s + b2, t + a2’) are different pairs.

Therfor by choosing various positiv values for z we can find
as many different pairs (s + bz, ¢t + az), (s + b2', ¢t + a2’), -+ -
as we please, such that, if (s, ¢/) is any one of these pairs,
s'a — b = c.

262. Th. If a is a positiv or zero integer and b a positiv
integer, an infinit number of pairs of positiv integers, s and ¢,
can be found, such that

sa —th = & (a (X b)
For example, (3 + 242) X 39 — (5 4+ 392) X 24 = — 3
ypX0—1X35 -5

263. Th. If a is a positiv or zero integer and b a positiv
integer prime to a, an infinit number of pairs of positiv integers,
s and t, can be found, such that

sa —th = + 1