MR EDWARD SANG ON THE THEORY OF COMMENSURABLES. 741 



the corresponding function of (p—6hjSL change of sign, as we can in orthogonal 

 trigonometry. 



49. If a number be of the form a'^ + ab + b'\ its square is of the same form. 



In order to decompose P"^ into A^ + AB + B'^, we have only to suppose that 

 a, 8 of the preceding article are equal to a and b ; whence, 



P2 =(a2-&2)2 + (^2_52) (2aZ> + a2) + (2a& + «2)2 

 = (62_a2) +(62_^2-) (2a6 + 62) + (2a& + 62-)2 



only one of which belongs to the prescribed form, the other belonging to 

 A2-AB4-B^ 



50. The number 3 is of the form 1^ + 1.1 + ^1, it is the only number of that 

 form whose square cannot be put in the same form. 



If we make, in the preceding formulae, a = 1, 5= 1, we obtain 



9=02+0-3 + 32 

 9=02-0-3 + 32 



Our previous demonstration proceeded on the assumption that a and b are prime 

 to and different from each other. 



In the present case the trigon is collapsed into a line, while the conjugate 

 trigons take the forms 0, 3, 3, and 3, 3, 3, the one being a line and the other an 

 equilateral trigon. 



51. When the number P = a^ + «J + 5^ is multiplied by 3, we have, on substi- 

 tuting 1 and 1 for a, /5 of No. 48, 



3P=(2a + &)2+(2a + 5) (6-a) + (6-a)2 

 of which the two conjugate forms are 



3P=(a + 2&)2-(a + 2&) (&-a) + (6-a)2 

 3P=:(a + 2&)2-(a + 26) (2a+ 6) + (2a+ 6)2 



52. When the sides of a trigon having one angle 120° are in their lowest 

 terms, the subtense cannot be a multiple of 3. 



If the subtense were divisible by 3, the sides must necessarily be of the forms 

 3w =t 1 ; these admit of three distinct classes of combinations. First, a and b may 

 be both of the form 3?^^- 1 ; secondly, they may be one of the form 3/i-f- 1, the 

 other of the form 3w— 1 ; and thirdly, they may be both of the form 3w— 1. 



In the first case, if we put « = 3a + 1, J = 3/5 + 1, we have, 



9(a2 + a/5 + |S2) + 9(a + /3) + 3=c2 



now, no square can be divisible by 3 without being also divisible by 9, therefore 

 this combination is impossible. 



The same argument applies to the third case, for if a = 3a— 1, 6 = 3j5— 1, we 



have, 



9(a2+a/3 + /52)_9(a+/3) + 3=c2. 



In the second case, if we put a = 3a + l, J = 3/3—1, the sum of a and b is 



VOL. XXIII. PART III. 9 N 



