FOR ANY COMPOSITE MODULUS, REAL OR COAIPLEX. 
277 
and therefore has exp 4. 
3 + 2? has exp 5- for mod (2 + i)^. 
3 + 2f = 2 (mod 1 + 2?), 
(3 + 2i}^ = 2l [mod (1 + 2^■)"] 
therefore 
3 + 2i has exp 4 . 5^ for mod (1 + 2^)^. 
The exponent required is the L.C.M. of 4, 5^ and 4.5^= 100. 
Therefore the exponent of 3 + 2i for mod 1000 is 100. 
(xviii.) Proof identical with (18). 
If a be not prime to m and m =^P wl)ere consists of powers of those primes 
which occur in a, and P is prime to a, then the series of residues 
a, o?, a^,. . . o’, . . . o’^'^h • • • (mod m) 
consists of periods of t terms, the first period beginning at the term : where t is the 
exponent of a for mod P, and r is the least number that makes o’" divisible by jj. 
Corollary .—When o = 1 (mod P) ^ = 1 and the period consists of one term. 
When a is divisible by p the first period starts from the first term. 
When both these conditions hold good then every power of a is = a (mod m). 
Example .—Pesidues of powers of 2 + f for mod 10. 
10= - (l+^)3(2 + ^)(l+20, 
P = 2 + b 
P = (1 + if (1 + 2i), 
therefore r 
Now 
and 
therefore 
= 1, and t is the exponent of 2 + for mod (1 + ^)^ (1 + 2i). 
2 fi- f f [mod (1 + and therefore has exp 2 
2 + f = 4 (mod 1 + 2i) 
t = 2 . 
> 
5? 5? 
The period consists of two terms and the first period begins at the first term 
2 + U 
(2 + if = 3 + ^i, 
(2 + if = 2 + 
&c. 
