FOR ANY COMPOSITE MODULUS, REAL OR COMPLEX. 
237 
Therefore 
(^a+l.K + lj • • • (^a +1.«+ 1; • • • (^6 + 1.6 + I> • ■ • (oiOcl 
and, since 
{h+i.a + i, • • • is prime to 2^, 
therefore 
(4 + i.« + n • • ■ hb) and (4+ 1.6 + 1 , . . . O are both prime to^5, 
and so on. 
Hence, as stated above, the single condition (iji, . . . prime to j?, includes all the 
other conditions found necessary in Proposition (29), and is therefore a necessary and 
sufficient condition that the numbers U (having the proper set of exponents) may com¬ 
pletely generate the p-power-exponent numbers. 
Also, tbis single condition may be replaced (with much advantage practically) by 
the equivalent set of conditions— 
(^11 3 • • • ^(la) I 
(h.+ l3 hj+l3 • • • ^bb) 1 n 
y all prime to p. 
(4+13 ^6 + 13 • • • 4c) j 
&C. ! 
(31.) We have now a comjilete series of methods for ascertaining whether a given 
set of numbers G can act as generators of the complete system of ^ (m) numbers 
(modulus m). 
(1.) Find the exponent to which each belongs. (Proposition 17.) 
(2.) The principal factors of these exponents must be the numbers found in 
Proposition (26). 
(3.) Express each number G as a product of numbers with principal factors 
of exponent of G as exponents. (Proposition 10.) 
(4.) Express each of these last as a product of powers of unitary generators, 
and thus get for each a set of indices. 
(5.) Apply the series of conditions of Proposition (30). 
If (2) and (5) are both satisfied then the numbers G are complete generators. 
Conversely to form any set of generators— 
(1.) For each prime p in ^ [m) form a set of unitary generators. 
(2.) Form from these any set of jji-pow'er-exponent generators satisfying the 
conditions of Proposition 30. 
(3.) Multiply together one generator from each of the sets just formed. 
Examvles. 
u. 
Let us form a set of generators for mod 308 = 2^. 7. 11. 
m = 7. 11. 
(/) (7) = 2. 3. 
(f> (11) = 2. 5. 
