__ by m, and write down the remainders: then none of these 
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XXVIII.—On a Proposition in the Theory of Numbers. By Batrour STEWART, 
Esq., of the Kew Observatory. 
(Read 21st April 1856.) 
Problem. Tf p be one of the roots of the equation «—1=0, (not 1,) then 
(1—p)(1—p”). . . . (L—p™-1)=m, provided m is a prime number. 
If m be not a prime number, and if p,=cos 284 4/ —1 sin = the same will 
hold for all roots p=p,*, where a is a number <m and prime to m. But for all 
roots p=p,*, where a, or one of its prime factors, is also a prime factor of m, the 
product (1—p) (l—p?) ... . @—p™~1) will be equal to 0. 
Preliminary Propositions. 
I. If m be a prime number, and a, 2 two numbers, each less than m; and if 
aS=ym+0, where 0 is less than m; then 0 is neither equal to a nor to . 
For if d=a, we shall have a (@—1)=ym, where a and 6—1 are both less than 
m, which clearly violates the well-known theorem that a number cannot be made 
up in two ways of prime factors. 
Il. Again, ifa (8+8,)=y, m+, (where 8+8,<m); then 6, is not equal to 6. 
For if 6,=0, then a (@+8,) = y,m+60 
= (1-7) m+yms+o6 
= (m7) m+a8 
therefore a6,=(y,—Y)m 
_ or a number is made up in two ways of prime factors, which is impossible. 
III. If, therefore, we arrange in a horizontal row all the numbers in order of 
magnitude from 1 to m—1 inclusive, and the same in a vertical row downwards, 
so that the two columns shall form the adjacent sides T_ . , , , 
_ of a square, and if we multiply each successive number 
in the top horizontal column by the number in the ver- 
_ tical (as in the multiplication table), divide the product 
E remainders will be the same, for they will not be the 
multiplier itself (Prop. I.), nor, 2°, will any two be alike 
(Prop. II.). The multiplier and remainders will contain all the numbers from 
VOL. XXI. PART III. De 
