408 BALFOUR STEWART, ESQ., ON A PROPOSITION 
1 to m—1 inclusive, though not in order of magnitude. We have exhibited this 
in the margin for m=7, the multipliers being 3 and 4. 
IV. If m be not a prime, but composed of the prime factors axbxex 
dxexfx ...., and if a or one of its prime factors (7) be also one of the prime 
factors of m, then in some case will a@=ym, where @ is one of the numbers 
1,2,3,.... m—1. 
For 7 and © are whole numbers <m; and a.” =“. m, which satisfies the 
f 
condition. 
V. If neither a nor any of its prime factors is also a prime factor of m, then 
there will be a remainder 6 <m whatever be /. 
For if not, let aG=ym; then since m>8..y<a. Therefore y is either 
prime to a, or there is a factor in a which is not in y: but this factor is, by hy- 
pothesis, not in m: it is, consequently, not in ym, which is absurd. 
VI. The remainders will be different for every different value of 6. For 
if possible, let a B=ym+6 
a (64+ B)=7,m+0 
=(%1—7) m+aB 
therefore, a 6B, =(y,—‘y) m which, as in the last Prop., is impossible. 
VII. If, then, we arrange the numbers as in Prop. UL., 
we shall have 0 amongst the remainders, for all values 
of a which either divide m, or have a prime factor in com- 
mon with it; whilst, for all other values of @, we shall 
have the same results as in Prop. III. This is shown for 
m=8 in the margin, for the multipliers 3, 4, and 6. 
or 
~I 
bo 
on 
1 
2 
3 1 
40 4 
5 
6 
vi 
Problem 1. If m be a prime number, the roots of the equation #"—1=0 are 
1, Py Py, - - - pm—1 where p, = cos a /—1 sin stil 
For, from the theory of equations, 
#™—1=(@—1) (w—p,) (a—p,2) . « . (e—p,"-) 
or am—1ligm—-2 +1=(@—p,) (w—p,?) oh af 
putting 1 for 2, m= (1—p,) (l—p,2) . . . (l—p,"—. 
If p=p,* since p= ae PY m—]; 
and if a B=ym=6 pif=p, 
