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254 Mr Mordell, On the representation of algebraic numbers 



If c is divisible by 2 but not by 4 put 



c = 2C, a = 2^, 6 = 2B, v = ^P, q = 2Q 



then p/i =-2Q- P\ a/i = iC {A + P) ~ (Q - Bf. 



If now ^ is even, take Pand Q — B both odd. Then since C is odd, 



a/4 = 3 (mod 8). 



Should, however, pj^ be of the form M, in this case 



p/4 = 7 (mod 8) 



for these values of P and Q, then since the only restriction on Q is 

 that Q — B should be odd, the change of Q into Q + 2 changes 

 /)/4 into a number pj/4, such that 



Pi/4 = 3 (mod 8) 

 while u is changed into a number a-^ still satisfying the congruence 



(Ti/4 = 3 (mod 8). 

 If however A is odd, take P even and Q — B odd. Then 



tT/4 = 3 (mod 8). 



The only restriction on P is that it should be even and from 

 § (2) we can take it so that p is not of the form M. Hence H holds 

 if c is divisible by 2 but not by 4. 



H holds also if c is divisible by 4 but not by 8. For putting 



X = c/2^ 

 we find 



y^ - Ibf + lacij - lc2 = 0, 



where all the coefficients are integers and c^/8 is divisible by 2 but 

 not by 4. Hence H holds for y and also for x. 



Suppose now that c is divisible by 8 or say 2'^ but not by 2«+i. 

 Then put again 



X = c/2y 

 so that as before 



y^ - |6?/2 + lacy - ^c^ = 0. 



Hence if 6/2 is odd, H holds for y and hence for x. If however 6/2 is 

 even, put 



a = 2A, b = iB, c = 8C 



so that the equation (1) becomes 



x^ - 2Ax^ + ABx -8G = 0, 

 or, putting x = 2z 



z^- Az^ + Bz-C = 0. 



But C is divisible now by 2"-^ and not by 2"-^. Hence if this 

 process be continued, we shall arrive at an equation of the form 



x^ — ax^ + 6a; — c = (1) 



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