INDETERMINATE QUARTIC EQUATIONS 63 



As a particular case put x 1 x 2 =ac=l, 6=2, so that 



ar 3 =-5/7, r= -21/13. 

 Hence substituting in (2) the foregoing values we obtain 



8 4 +28 4 +47 4 =34 4 +34 4 +41 4 . 



In the same way we may derive other algebraical solutions 

 by putting d=b, e=c,f=a, etc. 

 Second method. In the identity 



replace a by x^ and 6 by 



Then 



t.e. 



4\4 /T_ 4 _2?/. 4 \ 4 fx 2 \ 4 

 'l 2 



4 \ 4 



4 \ 4 



4 _2 4\4 f~ 2\4 



(1) 



Taking the first only of these equations and integralising 

 we have 



(2) 



- 

 which is an identity of the kind required. Thus we have 



of which the former is the second smallest solution which exists, 

 the smallest being 2 4 +4 4 +7 4 =3 4 +6 4 +6 4 . 



Cor. In equation (2) replace x lf y lt x z , y z by their recipro- 

 cals and multiply each root by x-^-y^-x^-y^. We thereby 

 obtain 



