526 
on richelot’s integral of the 
[956 
we can get rid of these terms, and so bring each side to contain only terms in 
6 2 , 6, 1; viz. writing 
□ = — 2e6 2 — dd — c + G, 
where G is a new arbitrary constant, the equation becomes 
] 
+ 6 [—2 e xy (x + y) — d xy 
+ [ e x 2 y 2 
-(G-c)(x + y) + b] 
+ (G-c)xy + a], 
which still contains the two arbitrary constants 6, C. 
But this gives the three equations 
(x - yf 
= e (x + y) 2 + d (x + y) + C, 
= - 2exy(x + y)-dxy-(G -c) (x +y) + b, 
= e x 2 y 2 + (G — c)xy + a. 
The first of these is Lagrange’s integral containing the arbitrary constant G; and 
it is necessary that the three equations shall be one and the same equation; viz. 
the second and third equations must be each of them a mere transformation of the 
first, equation. 
It is easy to verify that this is so. Starting from the first equation, we require, 
first the value of 
WX-^Y)(yYX-xYY) n 
{x-yy 
-2 
for a moment. 
We form a rational combination, or combination without any term in this is 
(a , (VX-yry _ 2 WX-^Y)(yYX-x^Y) 
K y O - yf O - yf 
= e(x + yy + d (x + y y + C(x + y) + O, 
where the left-hand side is 
(x-y)(X-Y) 
(x - yf 
X-Y 
x~y 
which is 
= e(x 3 + x 2 y + xy 2 + y 3 ) + d (;v 2 + xy + y 2 ) + c(x + y) + b, 
and we thence have for 
2 WX-,/Y)(y YX-xYY) 
the value given by the second equation.