56 ON THE ALGEBRAICAL SOLUTION OF 



and these appear to be along with (7) the three smallest 

 solutions which exist. 



QUESTION 4. Find n rational-sided right-angled triangles, 

 such that the algebraical sum of given rational multiples of 

 their area is zero. 



If the sides of the rth triangle are x r z +a r 2 , x r z a r 2 , 2x/i r , and 

 X r is the given multiple of it's area, we have evidently to 

 solve 



But this is evidently an indeterminate cubic in x t , x z , . . . 

 X T , . . . x n , having the particular solution o; 1 =a 1 , x z =a 2 , . . ., 

 x r =a r , . . ., x n =a n . Hence an algebraical solution can be 

 found. 



QUESTION 5. Solve the equation 



Xz 3 +[^ 3 =l, (1) 



knowing a particular solution say x=a, y=b. 



Putting x=XjT+a, y=yjT+b (2), equation (1) becomes on 

 expansion and rearrangement 



(*&!+ ( ^ 1 3 )r 3 +3(Xaz 1 2 + ( % 1 2 )r 2 +3(Xa 2 :r 1 + 1^ 2 </>=0 (3) 

 Hence, making the coefficient of r vanish by taking 



x^-d^/Xa 2 )^ (4) 



equation (3) is satisfied by taking 



r=-3(Xa 1 a + ! % 1 2 )/(Xz 1 3 + ivfl--9HtolQ*f-vVfa r (5) 

 Now since x- L = (\ib z l'ka z )y l , and equation (5) gives merely a 

 particular value of ry lt it is clear that the values of x and y 

 obtained from (2) will also be particular, viz. : 



z=a(l+n& 3 )/(xa 3 -n& 3 ), 2 /=-6(l+Xa 3 )/(Xa 3 -( A 6 3 ). (6) 

 In fact it is clear from the above and the solution of the 

 first part of Question 2, that no greater generality would have 

 been obtained by starting with the equation 



xZ 3 + ! x7 3 = I /Z 3 (7) 



knowing the particular solution X=a, Y=b, Z=c. 



