MR EDWARD SANG ON THE THEORY OF COMMENSURABLES. 743 



a=6a + l,b = 6/3 + l P -^^ 6» + 3 



6=6/5 + 3 P.d?H6w+l 



6=6/3 + 5 V-^Qn + l 



a=6a + 3, 6=6^5 + 5 F -isP- 6n + l 



57. The subtense of 120°, when the three sides are in the lowest terms, is of 

 the form 6n+l. 



For, according to the preceding theorem, every number P is of one of the two 

 forms 6n+l, 6n + 3 ; now it cannot be of the form 6n + S when the three sides 

 are prime to each other ; wherefore, the subtense must always be of the form 

 6n+l. 



58. If a prime number a divide any number P of the form a^ + ab + b^, a and 

 b, being prime to each other, another number of the same form, but less than a% 

 may be found also divisible by a. 



If a and b be greater than a, we may put them under the forms a = na± c, 

 b — taz^d, in which c and d are each less than the half of a. Inserting these 



values in the equation, 



P = a2+a6 + &2 



we find, — 



P = w2a2 ±29^ca + c2+?^m2 :±itcaz^ndar±zcd-\-t'^a'^ T±z2tda + d'^ 



wherefore, if P be divisible by a, c^ z^cd+d"^ must also be divisible by it; now 

 c^ ±cd+d'' ^ |a^ 



59. If a and b be prime to each other, no prime number of the form Qn—\ 

 can be a divisor of a^ -\-ab-\-h — P. 



For if some prime number a divide P, we may find some other c^ ■\-cd + d'^, 

 less than fa^, which is divisible by a. Now, we have shown that c^ ^-cd^d"^ is 

 either divisible by 3 or of the form 6w + 1 ; if it be divisible by 3, its third part is 

 also of the same form, so that eventually we arrive at a number of the form 

 Qn+1. But no number of this form can be divided by a 'isp^Qn—Y unless the 

 quotient be of the form Qn—l ; hence if c^ + cd+d^ ts?>. «, the quotient /3 less than 

 fa must be of the same form. By proceeding in the same way, we can show 

 that some other number 7, less than |/3, must be a divisor of some number of 

 the form e^ + ef-vf- ; and so we can continue. By this process we must at last 

 arrive at the number 5, the least of this class : now, we have shown that 5 

 cannot be a divisor; wherefore no number of the form 6w— 1 can divide the 

 subtense of 120°. 



60. The subtense of 120°, when the numbers are in their lowest terms, can 

 only be a prime number of the form 6w + 1, or the product of two or more of 



' such primes. 



This follows immediately from the preceding theorem. Hence, in order to 

 tabulate trigons of this kind, we must decompose the primes 7, 13, 19, &c., 

 into three parts, a^ ab^ b'^. Having found, for example, that 7=1 + 2 + 4, we 

 obtain 7^ = 3^ + 3.5 + 5^ from the formula given in article 49. 



