54 ON THE ALGEBRAICAL SOLUTION OF 



gives a second solution when one is known. Thus if X=6, then 

 x=3, y=2, 2=1 is a known solution, and (B) gives another 

 solution x= 128, y=5, 2=20, from which we may derive a third, 

 and so on ad infinitum. 



If, on the other hand, the original equation had been 



in which pj, |x 2 , . . ., p w are given coefficients and F 1? F 2 , . . ., 

 F n variables, since V 1 = F 2 = . . . = F w =0, #=, 2/=6, 2=c is, 

 by hypothesis, a particular solution, the substitutions (3) and 

 V,=t s r would have led to an equation in no wise differing 

 from (4) except that the term 2 ^ A 3 would have appeared in 

 the coefficient of r 3 . Hence, subtracting 27x 2 c 6 (^M s 3 ) from 

 the denominator of r in (6), it is clear that the final solution of 

 (I'), omitting the common denominator, would be 



where X is written for bx 1 ax z , and this solution is obviously 

 algebraical. 



Thus if we put n=l, |^ 1 =5, the equation 



5 F 3 + xy (x y ) = 6z 3 



which has the obvious particular solution F = 0, # = 3, 2/ = 2, 

 2=1, has the algebraical solution 



F= - 



2= -20Z 3 -4860f. 

 Thus X = 4, t = I gives the solution 



F = 1512, x = 3597, y = 382, 2 = 1535. 

 QUESTION 3. Solve the equation 



xy(x z -y z ) = u 2 + 2tf, (I) 



knowing a particular solution say x = a, y = b, u = c, v = d. 



This equation is of the fourth degree in x, y, u, v, but by 

 putting y = b it becomes bx(x z 6 2 ) = u 2 + 2v* (2) 



