458 Mr. Horner's new Demonstration, fyc. 



2. As this class of integers is limited, it is obvious that in 

 assigning to u the successive values 0, 1, 2, 3, to an in- 

 definite extent, the same values of R will repeatedly occur. 



N" — R N M+ * — R 

 Assume, therefore, R , = R ; then " and t 



' u+z u ■ p p 



being integers, their difference p *~ ls a ^ so i nte g ra ^ 



Therefore - is integral ; since P and N M are incommen- 

 surable. 



S. We will suppose d to represent the least value of z, which 

 satisfies this condition. Then it will also represent the least 

 interval of recurrence of the same remainder. For if the equa- 

 tion R"=R , were possible with bZd, we should have 



N" — R N" +6 — R 



2 " and !f integral ; and therefore their differ- 



P P & 



N"(N 6 — 1) N*~ 1 



ence ^-p- -; and consequently — p — integral, with 



bz.d; which is against the hypothesis. 



(B.) It follows from this that the series of remainders R l9 



R 25 R31 (R a = )l recur perpetually in the same order, and 



are all distinct from each other. 



4. If a=c 9 the proof is complete ; for by (A) it cannot ex- 

 ceed e. But if aZ.C) let q be any one of the remaining c— a 

 integers, which are less than P and prime to it ; and let 



" be integral. But * Z—l is also integral. Where- 

 fore their difference u " is integral; or ^Rm is =Q W 



rejecting the P's. 



5. Now the same reasoning as was employed in Step (1), 

 will prove that Q w is of the class (A). But it is not found in 

 the series (B). For, imagine it to be =R J; ; then qR u =zR ; 



whence * u is integral. From it deduct q times the in- 



N M — R N M (W u — o) 

 teger J, The remainder is * — p *£ integral ; 



y^V — U 



whence S is integral; or <?= &'_„ a term °*" s?r * es (B). 



