578 
PROBLEMS AND SOLUTIONS. 
[485 
I have somewhere made the remark that, on the left-hand side, the terms which 
belong to the odd and the even values of a + b+...(= p) are equal, and that we have 
therefore 
which is a theorem having a curious analogy with that demonstrated above. 
[Yol. vii. pp. 99—102.] 
2286. (Proposed by W. H. Laverty.)—If we have (n - 2) sets of n quantities 
each, (cq, a, ... a n ), (A> /3 2 ... /3 n ), ... (A a , A, ... A w ), connected with the n quantities 
(r 1} r. 2 ...r n ) by \n(n — 1) equations of which the type form is 
(a* - az) 2 + (A - A) 2 + ... (A* - A*) 2 = r£ + rf ; 
then show that 
where P is any one of the quantities a, ¡3, 7 ... A. 
Solution by Professor Cayley. 
Consider the case n = 4; we have between (a,, a 2 , a 3> a 4 ), (A, /3,, /3 3 , AX... (r 1} r 2 , r 3 , r 4 ) 
six equations, such as the equation 
(a x - a 2 ) 2 + (A - A) 2 = r i + ; 
(12) 
and it is in effect required to show that these equations give 
~ s : 4 : A : X = (234) : “< 341 > : < 412 > : -< 123 >< 
M '2 '8 ' 4 
(123)= a lf A, 1 , &c., 
«2, A, 1 
«3, A, 1 
viz. considering (a x , AX ( a 2> AX (®3> AX ( a 4> A) as the rectangular coordinates of four 
points in a plane, then (123) is the area (taken with a proper sign) of the triangle 
formed by the points 1, 2, 3; and the like for (234) &c. 
Combining the equations as follows, 
(12)+ (34)-(13)-(24), 
the r’s disappear, and we have an equation 
(®i ^4) (®a 83) + (A ~ A) (A — A) — 9,