406 H. F. TALBOT ON FERMAT S THEOREM. 



Extension of Fermat's Theorem. 



Always supposing that a is a prime number, and that b-c is greater than 

 unity, the theorem a n =b n — c n (impossible), maybe extended to a much more ge- 

 neral theorem, viz., that a m =b n — c n is impossible, provided that m is less than n. 



Demonstration. Let a m =k", where we no longer suppose k to be an integer. 

 Therefore since mSn, a must be greater than k. But k n —b n —c n by hypothesis; 

 therefore b n =c n + k n , which is less thanc + &|"; therefore b is less than c + k, and 

 b-c is less than k. A fortiori, b— c is less than a. 



But in the given equation a m = b n -c n , since a is prime, and b-c divides 6 n -c", 

 therefore b — c=a, or else is divisible by a, a number which we have shown to be 

 greater than itself, which is impossible. Therefore a m =b n —c n is impossible un- 

 less m ^> n. But if m ^> n, it is possible. 



Example. 3 3 =6 2 -3 2 , where a is prime and b—c greater than 1, but m\n. 



By an analogous method we obtain the extended theorem No. II. 



If n is an odd number, a m =b n + c" is impossible, provided that a is prime, and 

 m<(n. 



We have hitherto supposed a to be prime, whereas Fermat's theorem has no 

 such limitation ; it remains, therefore, to enquire how far the present extended 

 theorems are true when a is not a prime number. 



In conclusion, we may oberve that the ancients themselves had discovered the 

 possibility of the equation a 2 = b 2 + c 2 . 



But from what precedes, we may deduce the following theorems concerning it. 



1. If a 2 = b 2 + c 2 , and c is a prime number, then a-b is always =1. 



2. If a 2 = b 2 + c 2 , b and c cannot both be prime numbers. For because c is prime, 

 it follows that b=a—l. 



And because b is prime, therefore c=a-l, therefore 6=c, and a 2 =2 b 2 . 



But this is impossible, since one square cannot be double of another, in integer 

 numbers. 



Examples. 5 2 =4 2 + 3 2 , and 3 being prime, we have 5—4=1. 



Again, l3 2 =12 2 + 5 2 , and 5 being prime, we have 13-12=1. 



Again, 25 2 =24 2 + 7 2 , and 7 being prime, we have 25-24=1. 



The converse, however, is not true. For if a 2 = b 2 + c 2 , and a -6=1, it by no 

 means follows that either b or c is a prime. For example, 221 2 =220 2 + 21 2 , none 

 of which numbers are primes. 



