INDETERMINATE CUBIC EQUATIONS 49 



assumed solution, in this way an infinite number of fractional 

 solutions is found. The reason why the method fails to give 

 an algebraical solution in this latter case is because equation (6) 

 gives a linear and homogeneous relation between x l and a; 2 , 

 and when it has been satisfied equation (5) is no longer an 

 indeterminate one between r and x l but gives a unique 

 determinate value for the single variable formed by their 

 product XjT. Thus both values of the variables X lt X 2 in (4) 

 are particular. (See Question 5 below.) 



Again, if an indeterminate equation of any degree can, by 

 regarding certain of the variables as constant, be considered 

 as an indeterminate cubic in two or more variables of which 

 a particular solution is known, then another solution can be 

 found, but since it will present the roots as functions of the 

 variables which were for the time regarded as mere coefficients, 

 it is clear that the solution will always be algebraical. 1 



Finally, if <(.X\, X 2 , . . ., X n )=0 be an indeterminate 

 cubic of which a particular solution is known, namely, 



and if ^(Y lt Y 2 , . . ., Y r ) be a function of the third degree, 

 not necessarily homogeneous but containing no constant 

 term, of r variables Y lf Y 2 , . . ., Y r , none of which are identi- 

 cal with any of X 1} X 2 , . . ., X n except those whose value 

 for the particular solution of </>=0 is zero, then the equation 



<Kr lf Y 2 , . . ., 7 r )+4>(X lf x 2 , . . ., x n )=o 



can be solved. For it is evidently an indeterminate cubic in 

 n+r variables having the particular solution 



YI= Y 2 = . . . = Y r =0, X l =a 1 , X 2 =a 2 , . . ., X n =a n . 

 An example of this is given in Question 2 below. 



5. It is to be further remarked that if <=0 is a quadratic 

 indeterminate, then equation (5) does not contain A 3 r 3 , and 

 solutions are obtained by simply taking r=A 1 /A 2 , and 



1 See Part II, Section I : On the Algebraical Solution of the Equation 

 Question 1. 



