742 MR EDWARD SANG ON THE THEORY OF COMMENSURABLES. 



divisible by 3, and therefore the subtense AB and the side AF being both divi- 

 sible by 3, the other side BF must also be so, and consequently the trigon could 

 not have been in its lowest terms. 



53. The subtense of 120°, in its lowest terms, cannot be even. 



For if both a and h were odd, we might put a = 2a + 1, & = 2/5+ 1 which give 



which is inconsistent with c being even. 



54. If one of the sides of 120° be even, it must be divisible by 4. 



For if we put a — la, b=2f3+l, we have 4 (a" + a/3-|-/3-) + 2a + 4/3+l = c^ ; 

 now, in all cases of c being odd, c'— 1 is divisible by 4, wherefore 2a, that is a, 

 must be divisible by 4. 



55. The subtense of 120°, in its lowest terms, cannot be a multiple of Jive. 

 For if c were divisible by 5, the side a must belong to some of the forms 



5a ± I, 5a ± 2, while b must also be of some one of these forms. Now if we con- 

 join the supposition that a = 5a + l, with each of the four b = 5^+l, b = 5(3 + 2, 

 Z) = 5/3 -I- 3, ft = 5/3 + 4, we find that the sum a^ + ab + b'^ is of the form 5w + 3 in the 

 first and third case, and 5n + 2 in tlie second case, neither of which is a possible 

 form for a square number; but in the fourth case we find c^ ^i?=i 57z+ 1, which is 

 possible; but then a being 5a + 1, and b, 5/3 + 4, their sum is divisible by 5, so 

 that (No. 49) the trigon cannot have been in its lowest terms. Next, combining 

 the form a = 5a-|-2, with b = o[3+2, 5jS + 3, 5/3+4, we find that the first and last 

 give the forms 5^ + 2 and 5w + 3, which are impossible, while the intermediate 

 case gives the possible form 5/^—1, but then as a = 5a + 2, J = 5/3 + 3, a + b would 

 be a multiple of 5. Combining now the form 'a = 5a + S, with J = 5/3 + 3, and 

 Z* = 5/3 + 4, we find the forms 5n + 2, while lastly, combining 5a + 4, with 5/3 + 4, 

 we obtain the form 5w + 3, which is impossible for a square number. 



Hence in no combination of numbers for a and b not divisible by 5, can we 

 obtain a form possibly a square, for a^ + ab + b'^. 



56. If a and b be two numbers prime to each other, a^ + ab + b^ is either 

 divisible by 3 or of the form 6n+l. 



In reference to the divisor 6, all numbers leave the remainders 0, 1, 2, 3, 4, 5, 

 and the only possible combinations of two of these, not giving numbers having 

 2 or 3 for a common division are,— 



a=6a + 0, 6=6/3 + 1 P 5^, 6?i + l 



6=6/3 + 5 P ^^ 6n+l 



a=6a + 2, 6=6jS + l P ^. 6» + l 



b=z6(3 + S P v« 6n + l 



6=6/3 + 5 P ^ 6n + 3 



a=6a + 4, 6=6^+1 P ^ 6?i + 3 



6=6^ + 3 P ^ Sn+l 



b=6(3 + 5 P 5fej 6n+l 



