292 
MR. G. T. BENNETT ON THE RESIDUES OF POWERS OF NUMBERS 
Examiple 2. 
From the tables 
(2 + ^) X® = 3 + 2i (mod 9). 
(24) (3) 
3 + 2i = (22. 0) 
therefore 
2 + ^• = ( 1. 0), 
therefore, if 
x3 = (21. 0), 
X = («. h). 
therefore 
3a = 21 (mod 24), 
a=7, 15, 23 (mod 24) 
therefore 
36=0 (mod 3), 
h = 0, 1,2 (mod 3). 
Tliere are thus nine 
solutions, viz. : 
(7. 0)=l+7i (15. 0) = 7 + 7^ (23. 0) = 4+7i 
(7. i)=l + 4i (15. 1) = 7 + 4j (23. l) = 4+4i 
(7. 2) = l+t (15. 2)=7^-^ (23. 2) = 4 + i 
Exam2')le 3. 
From tlie tables 
therefore 
and therefore 
Example 4. 
and so 
Let 
ox^ = 3 + 2^ (mod 10). 
(4) (4) (2) 
3 H- 2i = {l. 2. 0) 
3 =(3.3.0), 
x^ = {2. 3. 0), 
X = (2. 1. 0) 
X = 3 -f- 8i. 
3x® — 3 -j~ ' 2 '^ (mod 30 -|- 20i), 
:c = (3+2i)f. 
Then 
3 (3 + 2i)~ P= 1 (mod 10) 
