FOR ANY COMPOSITE MODULUS, REAL OR COMPLEX. 
257 
The ^(48) =16 primitive roots are 
2 “b ^ 
4 + tU' 
2 -b 6z 
3 -b 6 1 
6 “b 2 i 
1 -b 4f 
1 + 
1 “b 3i 
5 -b 6f 
3 +^ 
5 -b ^ 
4+7 
1 + 
6 -f 3i 
6 “b 
6 + 4i 
(xii.) The exponents to which a number C6 belongs for successive powers of a prime 
as moduli. 
If we make one slight alteration the proof of (12) holds good throughout in the 
case of complex numbers for powers of a pu)‘e prime p as moduli. [In place of “ x < P” 
read “x = one of the^® — 1 residues, mod 
The exponent of a for mod jP is 
t if X < s, 
tjp~^ if X > 6‘, 
where t is the exponent of a for mod p, and is the greatest power of p that divides 
d — 1 . 
Corollary .—The greatest value of t is p^ — 1. (Proposition xi.) 
The greatest value that can have is got by making 6- = 1, i.e., by taking a so 
that a/~^ — 1 (though necessarily divisible by ^ 3 ) is not divisible by p”. 
Hence the greatest exponent that a number can have for mod p^ is 
(/>2 _ 
Now <1) (^/) = ^7“^ — = (yj-— which is p^~^ times the highest 
exponent. 
Hence for -o. power of a pure prime as modulus primitive roots do not exist. 
Consider next the case of a mixed prime (not 1 + i). 
Suppose a is the prime. 
Then, if (a + = A + Bt, A and B must be co-prime, for otherwise A + Bi 
would be divisible by some number other than a + /3 l 
Hence the N (a 4 - = (a~ d- (^^Y residues can be taken to be the pure numbers 
U, L, 2, . . . (a-^ /3~Y — !• 
Suppose that a is any one of these numbers. 
Then, if we have any congruence of the form 
cY = h [mod (a + /3f)^], 
a’’ — 6 = 0 [mod (a + 
2 L 
MDCCCXGJIJ.—A. 
