FOR ANY COMPOSITE MODULUS, REAL OR COMPLEX. 
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If \ = 2/x, we can similarly see that we must take three generators 
one generator with exp 2^" “ ^ 
„ „ 2'^-^ > . 
„ 23 
•/ 
We have now to show how these may he selected, so that all the numbers generated 
may be incongruent. 
Any number, with exponent > 2 modulus (1 + iY, has one of its powers, and 
one only, which has exponent 2. 
Suppose all the numbers modulus (1 + iY, arranged in 7 groups, so that all the 
numbers in a group may have the same number, with exponent 2, as a power. 
We shall now prove four Lemmas with regard to these groups. 
Lemma (i.). A power of a number belonging to any group belongs to the same 
group. 
Let a have exponent 2h and belong to the group a, i.e., 
^ = a [mod (1 + i)^]. 
Take any power of », say where i is odd. 
Its exponent is (Proposition iv.) 
Therefore 
^.e., 
a- 
.,s-l 
^ has exp 
and it is = a\ and, therefore, = a, because i is odd. 
Therefore, any power of a belongs to the same group. 
Lemma (ii.). If a and a have exponent '2% and belong to the same group a, then 
For 
Therefore 
Lemma (hi). If 
and 
{aa'Y' ^ = 1 [mod (1 + ^y]. 
^ 1 
or = a. 
a cc. 
{aa'Y' * = = 1 [mod (1 + i)^]. 
a has exp 2'’ and belongs to group a, 
b 2^ ,, 13, 
os 
,, 5 > 
then 
ah 
