722 



MR EDWARD SANG ON THE THEORY OF COMMENSURABLES. 



deduce others, as 6, 8, 10 ; 30, 40, 50, by multiplication. Hence it is almost 

 enough to consider only those cases in which the three numbers are prime to 

 each other. For convenience we shall indicate that the solution is in the lowest 

 terms, by the use of the small letters ; solutions in general being indicated by 

 capitals. 



2. Any odd prime number may represent the side of a rational right-angled 

 trigon, but of only one. 



For the equation A^ + B^ = C^ may be put under the form A^ = C^— B^ = (C + B) 



(C— B), while A^ may be regarded as the product of unit by A^ ; wherefore we 



may put 



A2 = C + B; 1 = C-B, whence 



B =i(A2-l); C=i(A2 + l) 



so that if Abe an odd number, A"— 1 and A^ + 1 are both even, and therefore 

 B and C both integers. 



Representing prime numbers by the Greek letters, if a = a, we have J = | 



(a-^ — 1) ; c = ^ («^ + 1), whence the solutions 



3, 4, 5 ; 5, 12, 13 ; 7, 24, 25 ; 11, 60, 61, &c. 



Since a is a prime number, the only decompositions of a' are «' x 1 and 

 ax a. The former of these has just been considered ; the latter would give 

 a = C— B, a = C-l-B, whence B = o, C = a, showing that the trigon has collapsed 

 into a straight line. 



3. Any odd number which is the product of two unequal prime factors may 

 represent the side oifour distinct rational right-angled trigons : of these two are 

 reducible and two are in their lowest terms. 



If A be the product of two primes, a, ^ (neither of which is 2), we may put 

 A'^ as the product of two factors in five ways, viz., a^jS^ x 1 ; a'^^S x /3 ; a^'^ x a ; 

 a: X |8^ ; and a^xa(3; the last of these can give no trigon, so that there remain the 

 four solutions 



&=i(a2/32-l) 





B=A (al3^-a) 

 C=i (a/52 + a) 



5=1 (a'-^-'') 



In the second of these, each side is divisible by (3, in the third by a, so that 

 these two solutions are included among those of the preceding article : but in the 

 first and last, the sides are all prime to each other. 



Hence the solutions 15, 112, 113 ; 15, 8, 17 ; 33, 544, 545 ; 33, 56, 65, &c. 



4. Any odd composite number of the form a^ . /3^ . 7'' = A may be the side of 

 half as many distinct trigons as there are units in tlie product (l + 2p) {l + 2q) 

 (l-l-2r) . . . less one. 



For the continued product 8^ = a^^ . ^'"'^ . y'^''. . . &c., has, inclusive of unit and 

 A"^ itself, divisors to the number (1 -1- 2p) (1 -1- 2q) (] + 2r) ... of which A itself is 



