FOR ANY COMPOSITE MODULUS, REAL OR COMPLEX. 
265 
i.e., if a — 1 is divisible by (1 + oi' any higher power, then a + 1 is divisible by 
(1 + if as the highest power. 
Similarly, we can show that if a higher power than (1 + 'if divides « + 1, then no 
higher power than (1 + i)^ divides a — 1. 
Therefore, if /c > 4, either 
« = 1 [mod (I -f 
or 
rt = — 1 [mod (1 + 
therefore, k being > 4, the solutions of 
are given hy 
Lemma (2).—If 
Now, if 
= 1 [mod (1 4- ■i)"] 
a = i 1 [mod (i d" 
= — 1 [mod (1 + t)*] 
(a + i) (a — i) = 0 [mod (1 + 
a = i [mod (1 + 
a i = '2i [mod (1 + if’'] 
= 0 [mod (1 + f)^]. 
Thus as above, if either of a + i, a — i is divisible by a power of (1 i) above the 
second, then (1 + if, is the highest power that divides the other. 
Therefore, if /c>4 the solutions of 
a^= — 1 [mod (1 + 0"'] 
are given by 
« = 4: [mod ( 1 + if ^]. 
is 
Lemma (3). 
a^ = 4: [a^od (1 + '^f] 
impossible for any value of k>1 . 
For if so. 
cd = 4i [mod (14- ^)^], 
or 
since 
which is impossible, for 
and 
a^ = i [mod (1 4“ '^f\ 
— i = i [mod ( I 4" if], 
1^ ^ z[mod (1 4- ty], 
d ^ i [mod (1 -f- if]. 
2 M 
MDCCCXCIII.—A. 
