FOE, ANY COMPOSITE MODULUS, REAL OR COMPLEX. 
279 
therefore 
(688 -h 784^)2 = (344 _ q^q 784t) (mod 1000) 
= 688 + 784i (mod 1000). 
(xix. and xx.) The proofs and results are the same as in (19) and (20). 
Example .—Mod 10 = 2. 5 = — (1 -{■ if (2 i) 2i). 
To find a so that 
We shall have 
a = a-^ [mod (1 + i)^] 
= [mod (2 + i)] 
= oc;^ [mod (1 + 2{)J. 
a = + a3^3(mod 10), 
where ^3, are found thus :— 
= (2 + ^■) (1 + 2i) = 1 [mod (1 + i)^], 
and therefore 
5i Xi= 1 [mod (1 + ^)^], 
iXi = 1 [mod (1 + 0 ']’ 
= i, 
^^ = 5 (mod 10). 
£3 = (1 + if (1 + 2^) X3 = 1 (mod 2 + i), 
{2i — A) x. 2 = 1 (mod 2 + 
and therefore 
— 8X3 = 1 (mod 2 + i), 
a?3 = 3 (mod 2 + i), 
^3 = 8 + 6i (mod lO). 
^3 = (1 + if (2 + i) a-3 = 1 (mod 1 + 2i), 
(4i — 2) a?3 = 1 (mod 1 + '2i), 
— ‘^x^ = 1 (mod 1 + 2i), 
and therefore 
a‘3 = 1 (mod 1 + 2i), 
I'g = 8 + 4^ (mod 10). 
Hence 
a = 5oij^ + (8 + 6i) a3 + (8 + 4i) ot^ (mod 10), 
e.ff., if 
«! = 1 [mod (1 + ^)^]" 
% = 4 (mod 2 i) > 
063 = 3 (mod 1 + 2^) J 
