258 
Proceedings of Royal Society of Edinburgh. [sess. 
we see that g 2 - x 2 must be a factor of the left-hand side. Now 
1%! . 02 > 0 3 ) = (01 - 0s) + (03 - 030l)02 + (03 - 1 )02. 
= A +B g 3 + Cgl , say : 
and D(</j , g 3 , g 3 ) = (l ~9i9 3 ) + ( 2 0i03“0i “ 0s)02> 
= A' +BV 2 , say. 
The above equation thus becomes 
A + B<7 2 + Cg% 
A + Bx 2 + Qx\ 
A' + B 'g 2 
A' + B'x 2 
which on the removal of g 2 - x 2 is easily reducible to 
B + C(£ 2 + z 2 ) B' 
A - C g 2 x 2 A' 
or 
B B # 
A A' 
+ 
C 
02 + ^2 
" 02*2 
B' 
A' 
= 0 . 
From this the further factor C may be removed because 
B B' 
B + A B' + A' 
A A' 
A A' 
= C(9l9s + 9i9l~9i-9l)i 
consequently our equation takes the final form 
(20103 - 01-03)02*2 + (l-0103)(02 + *2) + (0?03 + 010f ~ 01 “ 0s) = °> 
and is thus seen to be (1) symmetrical with respect to g 2 and x 2 , 
(2) symmetrical with respect to g 1 and y B , (3) linear in each of the 
two former, (4) a quadratic in the two latter. Solving for g 2 and 
x 2 wo obtain, as we ought, 
02 = $(*2>03>0l)> 
*2 “ $(02 ’ 03 > 0l) * 
so that we have the very interesting proposition : — 
V <£(gp g 2 > g 3 ) = <£(gi> x 2 gs) and g 2 and x 2 he unequal, 
then g 2 = <j>(x 21 g 3 , gj) and x 2 = <j>(g 2 , g s , g 3 ) . 
Arranging the equation as a quadratic in we may write it 
(l-0s)-0f + D '(0«. 02. *2)-01 + (0302*2 - 02- *2 + 0a) = °> 
and if the result of solution he 
01 ~ */h(03 ’02’ *2) 01 ^2(03’ 02’ ^2) 
