576 
PROBLEMS AND SOLUTIONS. 
[485 
Introducing an arbitrary constant 0 o , and putting the equation under the form 
(a? + y-f + Oi + u 0 - 0„) (x 2 + if) + 6 0 (x 2 + y 2 ) + v, + v 1 +v 0 =0, 
this may be identified with 
O 2 + if + }h +Po) (^ 2 + f + <h + Ço) + n + n = 0 ; 
viz. the conditions in order to this identity are 
(Pi ~ tfi) 2 = *fi 2 - 4v 3 - 40 o (ix 2 + y 2 ), 
where the right-hand side is a quadric function {x, y) 2 , which, when the discriminant 
thereof is put = 0, (that is, when 0 O is determined as the root of a quadric equation,) 
is a perfect square, p x — q x is then a known linear function, and p x + q x being equal to 
the linear function u x , we have p x and q x as linear functions of (x, y). We may take 
for the constants p 0 and q 0 any values satisfying the equation p 0 + q 0 = u 0 — 0 O ; and we 
then have 
r 1 =v 1 -p 1 q 0 -p 0 q 1 , r 0 = v 0 -p 0 q 0 , 
which completes the determination ; the form 
(x 2 + y 2 + pi + po) {x 2 + y 2 + #i + 2o) + n + r 0 = 0 
is of course the same as the proposed form SS' — Jc 3 L = 0. 
Cor. A somewhat more convenient form is UU'—k 2 V= 0, where 77=0, U' = 0 
are the equations of two evanescent circles (pairs of imaginary lines), V= 0 the equation 
of a circle ; in fact the original form SS' — k 3 L = 0 may be written (S — a) (S' — a') 
+ (aS'+ a'S— aa' — k 3 L) = 0, which, when a, a' are so determined that S— a = 0, S'—a'= 0 
may be evanescent circles, is of the required form UU' — k 2 V = 0. The equation JJU'= 0 
is that of the two pairs of tangents to the curve at the circular points at infinity 
respectively; in fact, writing U=pq, U' =p'q', each of the lines p = 0, q = 0, p' = 0, q =0 
meets the circle V=() in one or other of the circular points at infinity, and therefore 
only in a single point not at infinity; hence each of these lines meets the curve 
TJU' — k 2 V — 0 three times in one of the circular points at infinity, that is, the line 
in question is a tangent to one of the two branches through the circular point at 
infinity. 
[Yol. vu. pp. 87, 88.] 
2309. (Proposed by Professor Cayley.)—Show that for n things 
1 - (no. of partitions into 2 parts) + 1.2 (no. of partitions into 3 parts).... 
+ 1.2.3.. (n — 1) (no. of partitions into n parts) = 0.