82 Mr. Horner on a Theorem of Fermat. [Feb. 



Demonstration. — 1. If m is not a multiple of/), neither is n P. 

 For, conceive the three to be resolved into their prime factors ; 

 then P, p, have no factors in common ; and p contains at least 

 one factor which is not in xi, and /. not in u P. 



II p . 



2. The remainder of — is commensurate \vithj9, or not, ac- 

 cording as M is or is not commensurate with p. For, let the 

 quotient be Q, and the remainder r ; then ?< P — r = p Q ; 

 and if ?• and u P be severally divided by any number that mea- 

 sures p, the two remainders will be identical. Let the common 

 remainder be O ; then ;• and u P, and .-. /• and m, since P is prime 

 to^, are each of them commensurate with j9. Let the remainder 

 be > O ; then each of these is incumraensurate with p. 



3. If tt — 1) is not a multiple of p, -— and — leave different 



remainders. For if both leave the same remainder r, we shall 

 have u P — r, and vV — r, both divisible hy p\ and .-. their 

 difference (« — ' v) P will be divisible by^, which is impossible 



4. If each term of the series P, 2 P, 3 P, (p — 1) P, be 



divided by p, all the remainders will be distinct (3). And they 

 are p — 1 in number. Consequently they comprise every 

 integer < p \ or, the series of remainders, when progressively 

 arranged, is identical with the natural series of multipliers. 



5. ^separate this series into two classes, distinguished by 

 having the multipliers prime to p, or commensurate with it. 

 The entire set of remainders in each class, if progressively 

 arranged, will prove identical with the series of multipliers in the 

 same class. For each of these two series is selected from an 

 identical set of numbers (4), and by an identical mode of 

 choice (2). 



6. Let the multiphers prime top be 1, r, r,, r,,, &c. and let 

 M p be understood to mean some multiple of p. Then each term 

 of the series P, ;• P, /•, P, &c. will have one and only one equiva- 

 lent in a series of equal extent M p + 1, M ;> + r, M.p +r,, &c. 

 so that the two series are identical. Putting S = 1 x r x r, x r,, 

 . . «. and taking // to indicate the number of terms in each of 

 the two series just described, we shall have for the product of 

 all the terras P" S = M p + S. Wherefore P" S - S = M j?; 

 that is (P" — 1 ) S = M p. But S is entirely composed of factors 

 prime top; and is .*. prime to p (1). Consequently it must be 

 the other factor P" — 1 that is divisible by p. — Q. E.D. 



Cor. — 1. Let m be the number of terms incommensurate with 

 p,\n the natural series 1, 2, 3 .... p. Then (P" — 1) x P'" = 

 pp _ p" being divisible by p, it follows that P*" and P"', when 

 divided hy p, leave the same remainder. 



2. If p is a prime number, n is = p — 1, and we have 



pp-' _ 1 



an integer ; which is the theorem of Fermat. 



