2G4 
MR. G. T. BEXNETT OX THE RESIDUES OF POWERS OF XOiBERS 
Mod (1 + <4>(1 i)'‘ = 64. The 64 numbers, with their exponents, are 
arranged below. 
1. Exp 2. Exp 4. 
7 
4+7 
12+7 
3i 
8 + 3 i 
57 
8 + 57 
4 + / 7 
12 + 7i 
9 
3 
5 
11 
13 
1 + 47 
7 + 47 
9 + 4i 
15 + 4t‘ 
15 
7 
8+7 
4 + 3i 
12 + 37 
4+57 
12 + 57 
77 
8 + / i 
3 + 4i 
2+7 
10+7 
2 + 37 
10 + 37 
6 + 57 
14 + 57 
6+77 
14 + 7i 
5 + 47 
5 + 27 
7 + 2i 
13 + 27 
15 + 27 
1 + 67 
3 + 6i 
9 + 67 
ll + 6i 
11 + 47 
6 + 'i 
14+i 
6 + 37 
14 + 37 
2 + 57 
10 + 57 
2 + 77 
10+77 
13 + Ai 
1 + 27 
3 + 27 
9 + 27 
11 + 27 
5 + 67 
7 + 6i 
13 + 67 
15 + 6j 
The 56 numbers with exponent 4 are arranged in rows collineaiTy with the 
numbers, with exponent 2, to which their squares are congruent. 
The 7 numbers with exponent 2 can be arranged in 7 sets of 3, so that of each 3 
the product of any two is congruent to the third (or the product of the three 
congruent to unity). Thus : 
7. 9. 15 = O 
7 (13 + 4i) (3 + ii) = 1 
7 (5 + 4{) (11 + Ai) = 1 
9 (3 + Ti) (11 + E) = 1 (mod 1 + if. 
9 (5 + 4i) (13 + Ai) = 1 
15 (3 + Ai) (5 + di) = I 
15 (11 + (13 + 4i) = 1 J 
To generate the 64 numbers, we must take three numbers each with exponent 4. 
The product of powers of three such numbers can only be congruent to unity without 
each being separately congruent to unity, if the product of their squares be so. 
Hence, the three generators must be chosen, one each from three of the above 
rows, and the numbers, with exponent 2 found in these rows, must not have their 
product congruent to unity. 
(xiv.) The nutnbers which have as exponents 2, 4, and 8 for mod (1 + if. 
We shall use the results of the three following: Lemmas : 
Lemma (1).—If 
a^ = 1 [mod (1 + iY~\ 
{a — 1) {a 1) = 0 [mod (1 + 
a=l [mod (1 + i)®] 
« +1=2 [mo(.l (1 + t)®] 
= 0 [mod (1 + 0']> 
Now, if 
