406 
PROFESSOR CAYLEY ON THE PROBLEM OF 
<pa=X 4 ( + 1) 
+X 3 ( 2a 3 — 18a 2 + 52a- 46) 
+X 2 ( — 18a 3 +162a 2 — 420a+221) 
+X( 52a 3 — 420a 2 + 7 04a + / ) 
+ x 4 — 46a 3 +221a 2 + lx 
4-1 f X 2 ( - 9) 'l 
j 4-X( - 12a+135) > 
^ — 9a 2 4- 135a + ^ 
where the constants l, X have to be determined ; I have in the first instance written 
/(X+a) + Af, instead of LX+/a+X$, thus introducing two constants only, since it is 
clear from the symmetry in regard to a, X that we must have 1= L. We must have 
<pa=0, when the curve is a conic or cubic. Writing a=2, we have 
<pa=X 4 +2X 3 -115X 2 +144X+532+5(-9X 2 +lllX+234)+/(2+X)+^, 
and then for the conic, X = 2, |=6. 
Writing a = 3, we have 
<pa=X 4 +2X 3 -67X 2 -264X+828+|(-9X 2 +99X+324)+/(3+X)+M, 
and then for the three cases of the cubic X=6, £=18 ; X=4, f = 12 ; and X = 3, | = 10. 
We have thus the four equations 
2912 + 4/4- 6X=0, 
9252 + 9/+18*=0, 
579G + 7/+ 12l=0, 
4968+6/+10?i=0, 
all satisfied by /= + 172, a= — GOO. Whence, finally, 
<px= X 4 ( + 1) 
+X 3 ( 2a 3 — 18a 2 + 52a- 46) 
+ X 2 ( — 18a 3 + 162a 2 — 420a+ 221) 
+X( 52a 3 — 420a 2 + 704a+ 172) 
+ a 4 -46a 3 +221a 2 + 172a 
+sr x 2 ( - 9)i 
• +X( - 12a + ] 35)r 
— 9a 2 +135a— GOO ^ 
but on account of the symmetry the number of triangles is =one sixth of this expression. 
