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 #=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 y. lt y. z , . . ., \t. n are given coefficients and V v F 2 , . . ., 

 F n variables, since Fj= F 2 = . . . = F n =0, x=a, y=b, zc is, 

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

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

 from (4) except that the term SnA 3 would have appeared in 

 the coefficient of r 3 . Hence, subtracting 27x 2 c 6 (2[A/ s 3 ) from 

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

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

 V s =-9ab(a-b)(a?-ab+b 2 )X 2 t s , 



>x 



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

 algebraical. 



Thus if we put n1, ^=5, the equation 



which has the obvious particular solution F = 0, x = 3, y = 2, 

 2=1, has the algebraical solution 



F= -31 



z= -20X 3 -4860f. 

 Thus X = 4, t = 1 gives the solution 



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

 QUESTION 3. Solve the equation 



xy^-^^tf + W, (1) 



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 i - b 2 ) = u z + 2tf (2) 



