LINEAR GROUP WITH QUADRATIC INVARIANT. 



177 



is a square in the field or else zero. 1 ) Eliminating /3 between the 

 two equations , we find 



or 

 or 



(1 + a + a 2 + 2s) 2 = 4s(l + 

 Hence will s be a square. 2 ) Solving, we find 



The linear substitution (J 2 i 4 i 5 ) expressed in the indices Y {j takes 

 the following form, say V: 



2 2 



J_ JL^ 



T Y 



2 r 2 



5 



21 6> 



s 



The product Fi7 will be simpler than F, if we take as E: 



which is recognized as the second compound of 



E'= 



1 -a /3 ( 

 0100 

 0010 

 -3 -a 1 



1) The case s = requires 1 -}- a -f- a*= 0, and may thus be avoided. 



2) For # n = 3, 7 or 11, there exist solutions of a*-|- (3 2 = 1 for which 

 is an arbitrary square in the field. Is this always true? 



DlCKSON, Linear Groups. 12 



