60 ON THE ALGEBRAICAL SOLUTION OF 

 SECTION I On the algebraical solution of the equation 



1. Before attempting to solve this equation for all values 

 of n and r, it will be convenient first to give solutions for a 

 few particular values of n and r. 



QUESTION 1. Solve in integers the equation 



P^+P^PY+PY. (1) 



Let P 1 =s 1 +2 3 , P 2 =z 2 24, P\=z l z Bi P' 2 =2 2 +2 4 and the equa- 

 tion becomes 



^1 2 3 "+ 2 r^3 == ^2 Zf^-ZqZ^ (2) 



This is an indeterminate cubic in z x and z 2 and if any par- 

 ticular solution is known, another can be found, but as it 

 presents the new roots as functions of the coefficients z 3 and 

 2 4 , it will be an algebraical solution. Thus, putting z 1 = 

 Z 2 =x 2 r+y 2 , (2) becomes 



or, on expansion and rearrangement, 



4 3 ) =0. (3) 



To make the term independent of r vanish, we must have 



= (y 2*24 



which is equivalent to knowing a particular solution of (1) 

 since it may be written 



and to satisfy this we may evidently take 



2/i-f z 3 =-2/ 2 -z 4 , 2/i-2 3 = 2/ 2 -^ -c. 2/i=-^ 2/2=-% (5). 

 In order to make the coefficient of r in (3) vanish, we must take 



t.e. ^2 



=[2 3 (3z 4 2 +2 3 2 )/z 4 (3z 3 2 + 24 2 )]^ (6) 



on substituting for y lt y 2 their values given by (5). 



