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XXVIII. — On a Proposition in the Theory of Numbers. By Balfour Stewart, 



Esq., of the Kew Observatory. 



(Read 21st April 1856.) 



Problem. Yip be one of the roots of the equation ^™-l = 0, (not 1,) then 

 (1— p){l~p 2 ) .... (1— p m ~ 1 )=m, provided m is a prime number. 



If m be not a prime number, and if » 1 = cos — + \/^l sin — , the same will 



1 m m 



hold for all roots p=Pi K , where a is a number < m, and prime to m. But for all 

 roots p =p 1 K , where a, or one of its prime factors, is also a prime factor of m, the 

 product (1-p) (1-p 2 ) .... 0--p m ~ l ) will be equal to 0. 



Preliminary Propositions. 



I. If m be a prime number, and a, (3 two numbers, each less than m ; and if 

 a^=ym + d, where 8 is less than m ; then 8 is neither equal to a nor to (3. 



For if 8= a, we shall have a (J3-l) = ym, where a and j3-l 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. 



II. Again, if a (/3 + /3 1 )=7 1 m + 8 1 (where /3 + /3 t < m) ; then 8 X is not equal to 8. 

 For if £ 1 = 3, then a (/3 + ft) = y x m + 5 



= (7 1 -7)m + 7m + S 

 — (fy _ ry) m + a /3 



therefore a /3 t = (7 X - 7) 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-l inclusive, and the same in a vertical row downwards, 

 so that the two columns shall form the adjacent sides 

 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 

 by m, and write down the remainders : then none of these 

 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 



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