CONGRUENCE 129

if k is an even multiple of x, a* = 1 (modm),

 

if k is an odd multiple of x, a* = — 1 (mod m).

First a?* = 1 (mod m) § 219.

Then, if & is an even multiple of x, & > 2x.

And a* =1 (mod m) § 303.

If % is an odd multiple of x, 2 = x (mod 2x) § 307.

Therfor, since a?* = 1 (mod m), a* = a* (mod m) § 304.

But a* = — 1 (mod m)

Therfor a* = — 1 (mod m)

Ex.1. 2= — 1 (mod 3) Ex.2. 33= — 1 (mod 4)

20 = 1 i 36 = ¥ -
260= 1 5 P=-—1 o
29=—1 &

Ex.3. 10! = — 1 (mod 11)

102 = 1 o
13= — 1 &
Ex. 4. 103= — 1 (mod 7 or 11 or 13)
106 = 1 -
100= — 1 "

300. Th. If a, m, and x are positiv 1integers,
a* = — 1 (mod m), and k and | are positiv or zero integers; then,
if k and | differ by an even multiple of x, a* = a' (mod m),
if k and 1 differ by an odd multiple of x, a* = — a’ (mod m).

Suppose k = 1. Then k — [ is positiv or zero.
Now, if 2 — [ is an even multiple of x,

 

 

since a* = — 1 (mod m),
a*'= 1 (mod m) § 308.
Hence atl.g' = a' (mod m)
Or a* = a' (mod m)
The case when 2 — I is an odd multiple of x is proved
similarly.
For the case when k& < [, compare the proof in § 304.
For example, since 103 = — 1 (mod 7),
100 = 10t = — 107 (mod 7)