92 
ON THE BINOMIAL EQUATION X p — 1 = 0. 
[731 
In this equation the only terms of the second order are — h(h + k), which contain the 
factor h; the terms of the third order contain this same factor h, and throwing it out, 
and reducing, the equation is found to be 
(9 ~ k) 2 + (h -/)- = h + k, 
or, as it may also be written, 
g 2 + k 2 — 2 hf — h + (h 2 +f 2 — 2gk — k) = 0 ; 
and the foregoing values of l, to are 
7 _(g 2 +k 2 - 2hf- h) - (gh - kf) 
g+h-k-f 
k 2 — lr + gk — hf 
m = g+h-k-f 5 
and by means of these three equations all the foregoing equations are satisfied. 
We have 
FiFi 3 = (X - Z) 2 + (Y - W) 2 
= X 2 + Y 2 + Z 2 + W 2 — 2 (XZ+ YW) 
= — (a + b + c + d) + 2 (l + m); 
or, substituting for ci, b, c, d, this is 
= 1 + 2 (/+ g + h + k) + 4<(l + m), 
viz. it is 
= ^(p + l) + 4i(l + m); 
or, substituting for l + m its before-mentioned value, then, according as jp = l or 5 (mod. 8), 
the value is = p or — p; that is, we have 
FiFi 3 = (—) 4 p. 
Again, we have 
(Fi) 2 = (X + iY — Z — iW) 2 
= X 2 - Y 2 + Z 2 - W 2 - 2XZ + 2YW + 2i(XY- YZ + ZW- WX) 
= {a — b + c — d + 2 (to -0 + 2 (/- g + h — k) i}(X -Y+ Z - W) 
■=(A+Bi)F(- 1), 
where 
A = a — b + c — d + 2 (to — l), = — 1 + 4 (m — l), 
B = 2(f-g + h-k)- 
or, since X — Y + Z— W = F (— 1), this equation is 
(Fi) 2 = (A + Bi) F (- 1): 
(Fi 3 ) 2 = (A - Bi) F(- 1). 
and similarly