H. F. TALBOT ON FERMAT S THEOREM. 405 



the same n factors. And it cannot possibly have any other. Therefore, since it 

 has the factor b + c, this factor must itself be either = a, or divisible by a. 



But we shall now proceed to show that b + c cannot possibly be divisible by a, 

 and therefore the original hypothesis, viz., that a n =b n + c\ must be impossible. 



In order to show this, we will first observe that (& + cf is greater than &" + c", 



since it exceeds it by the quantity nb n ~ l c + — ^ — b n ~ 2 c 2 + &c. 

 But &" + c"=a K by hypothesis. 



.*. & + c|"^>«", and b + cya. 

 On the other hand, since evidently 



b S a and c<^a, Hcwa. 



But since b + c is greater than a and less than 2 a, it cannot possibly be divisible by 

 a. Which was to be shown . And it therefore follows that the equation a" = b n + c" is 

 impossible, if n is an odd number ^> 1, always supposing, however, that a is a 

 prime number. 



Demonstration of Theorem II. 



Let us suppose, if possible, that a n ~b n — c. Then since V-c is always divi- 

 sible by b — c, let the quotient be Q. Therefore a"— b—c . Q. Now since a is a 

 prime number, the first side of this equation is the product of the n factors 

 ay. ay. ax. &C. 



Consequently the second side of the equation is the product of the same n fac- 

 tors, and it cannot possibly have any other. Therefore, since it has the factor 

 &— c, this factor must itself be either = a or divisible by a. But, on the other 

 hand, it can be shown, as follows, that it is not divisible by a. 



Since a n =b"—c n , therefore & n =a n + c n , and b is the greatest of the three numbers. 

 Now since a+ ~c[ />a n + c", and a" + c n = b n , .■-. a + if ^> b n , and a + c ^> b, and a ^> b - c. 



Since, therefore a is greater than b—c, it cannot possibly divide it. And there- 

 fore the original hypothesis that a n =b '— c n is impossible, if n is any number 



^> 1, always, however, on the supposition that a is prime . 



But in reviewing this demonstration, we find a case of exception ; for it will 

 be seen that we assert that b — c can have no factor different from a. This is cor- 

 rect in one sense, but not so in another, since it may have the factor unity, which 

 is usually disregarded, though it is here a consideration of the greatest import- 

 ance. And since we have shown that b—c cannot have either the factor a or any 

 other factor, it follows that it can have no factor except unity, that is to say, it 

 must be itself equal to unity, and can have no other value. 



The above two theorems form together a conclusive demonstration of Fek- 

 mat's theorem, in the case of a prime number. 



