Mr. J. Cockle on the Method of Symmetric Products, 173 

 Then 



and 



(1, 3, 3) = (1, 3, 4) = (1, 3, 5) = (1, 4, 5) = 3, 

 and 



(l,3,5) = (l,3,4)=:~3. 

 33. Let 



and, for a reason derived from (26), let 



^ + ff=f[m,n)=f{r-l,s-\). 

 Then, placing the suffix of the square to the left, 



(r,l,.)=S. ^1.72(^3 + ^4)=^. ^+/K'*)=-l+/(^-l.^-l). 



33. Let ^(^) = a^ + a-^, 

 then 



/(r-1, 5-l)=(^(5-r)</)(3r--3) +^(35~3?-)(/)(r-l), 

 whence we find 



/(I, 3)= -3, /(],3) = 3=:/(l,4). 



34. The nature of a is such that 



/(I, 2)=/(2, 4)=/(3, l)=/(4, 3) = 3-5, 

 /(I, 3) =/(2, 1) =/(3, 4) =/(4, 2) =3, 

 /(I, 4) =/(2, 3) =/(3, 2) =/(4, 1) = 3. 



35. By the cycle of (30) we find 



(^,r,5) = (^--l,r-l,5-l), 



where, when zero occurs among the symbols, 5 is to be substi- 

 tuted for it. And we now see that 



(3, 3, 4) = (1,3, 3), (3, 3, 5) = (1,3, 4), (3, 4, 5) = (1, 3, 4), 



(3, 3, 4) = (3, 1,3), (3, 3, 5) = (3, 1,4), (4, 3, 3) = (3, 1, 3), 



(4, 3, 5) = (3, 1,4), (5, 3, 3) = (4, 1,3), (5, 3, 4) = (4, 1,3). 



36. By putting ^^^ Y^ in place of Y^ in the original system, 

 we arrive at 



(^,r,5) = (^-2,r-2,5--2); 

 and, so, at 

 (3, 4, 5) = (1,3, 3), (4, 3, 5)= (3, 1,3), (5, 3, 4) = (3, 1, 3), 



results also obtainable from (35) by a double reduction. 



37. We are thus conducted to 



^ • 2/1^2/2^3=22: . 2/1^^22/3-52:'. 2/1^(2/22/5 + 2/32/4). 

 S' being such that each value of y'^ and oiy^y^ occurs in it once 

 and once only. It has what may be termed a partial symmetry. 



38. Combining the last equation with {g), {h), {i) and {j), 



