138 RESOLUTION OF ALGEBRAICAL EQUATIONS. 
A A 5 
tiplied bya power of the other; as, Y by U et ¥ by un 3 Y by: 
6 : : 
U *, and so on; the pairs of equations, 
AXN=khs+B, 
AN=qs+ fi, tesseseseeesesseseces (3) 
HvA=Ks+6, 
j - @ 
Hi =Qs+h, § 7 
and so on, subsisting; where A, H, 4, K, g, Q, &c., are whole nume 
bers, each less than s 
To prevent misunderstanding, we may instance the function, 
fe) =pr > pyr + {6+p" tanll p janlnedt (2+ p)'}? v 
as one in which the two surds pt and (2 + ar are similarly involved. 
For, calling the former Y and the latter U, we have s=o=7, A=5, 
uM = 1, B=3, fi=2, 5=1, and 3,3. Consequently equations 
(3) and (4) become, 
5h=7k +3, 
D7 i: 2: 
5H=7K +1, 
isloa 7 (Qa) s 
where integral values of #, H, &c., less than 7, can be found: 
V2, k= 1,'¢= 0, 
ee Ts ean eer 
We proceed with the proof of the Proposition Let z, be ao™ 
root of unity, distirict from unity; and when U is changed into 2,U , 
let the terms m @) heen es 
Fe(2)s 4.(2), of UC) Sale ae MG ACS Buf eS (5) 
Since U disappears from F,,, (x), the continued product of the terms 
in (2) is not affected when we replace U by 2, U. Therefore 
2 s 2 s 
F,(a)X Fe(@) x... X Fe(@)=fe(@) X fe(®) x ... X fe @). 
Hence, either F. (Z) is equal to one of the terms in (5), or it has with 
a . 
one of them, as fe (Z), a common measure, of less dimensions, as 
respects Z, than Fe(%). Suppose, if possible, that F,(@) is not equal 
to any term in (5); and that L is its H. C. M. with “fe (z). The ex- 
