Full text: The collected mathematical papers of Arthur Cayley, Sc.D., F.R.S., late sadlerian professor of pure mathematics in the University of Cambridge (Vol. 8)

490] ON A PROBLEM OF ELIMINATION. 
Suppose this equation has equal roots, then we have 
23 
Disct. Resit. (AP + /xQ, U, V) = 0, 
the discriminant being taken in regard to A, ¡x. This is of the form 
(A, ...)2 (in'll— 1 ) _ 0 ; 
It is moreover clear that the nilfactum is a combinant of the functions P, Q; 
and the form of the equation is therefore 
mn (mtt-i) 
Now the equation in question will be satisfied, 1°. if the curves U = 0, V= 0 touch 
each other; let the condition for this be V = 0. 2°. If there exists a curve 
AP + /xQ = 0 passing through two of the intersections of the curves U= 0, V = 0; let 
the condition be il = 0. There is reason to think that the equation contains the 
factor il 2 , and that the form thereof is il 2 V = 0. 
Assuming that this is so, and observing that V, the osculant or discriminant of 
the functions U, V, is of the form 
V = (a,, ...)»(«+2771—3) (fa yn (m+2n—3) 
we have 
Ikn (n—i) (2m—l) + (fc—l) n (n+im—S) ^ 
|km (m—i) (271—1) + (k—i) m (m+m—3) 
and consequently 
( a yk n (71—1) k (2m—1) +J (*—1) n (71+2771—3) x 
(Jj (771—1) k (27i—i) + J (k—i) m (m+2?i—3) 
which is the solution of the proposed question. Suppose for instance n— 1, then 
;—i) (m—l) (fo ' ^m (m—i)k+i (k—i) (m—1)_ 
If moreover k= 1, then
	        
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