CONTINUANTS RESOLVABLE INTO LINEAR FACTORS. 345 
and the resulting factors greater by 
respectively. 
(4) Changing a into a+ 2c, n into n—1, b into b+c in § 3 we have 
a+ 2c (n — 2)(b +c) : 
(n+ 1)(6 — 2c) a+8e (n — 3)(b + 2c) : tye ae 
(n+ 2)(b — 3c) a+18c (n—4)(6+3c).... 
: (n + 3)(b — 4c) a+ 32¢ 
= {a+2(n —2)b+2(n—1)c} - {a+ 2(n —4)b + 6(n — 2)e} 
-{a+2(n-6)b+10(n—3)e}...... 
But the continuant here is the complementary minor of the element in the place (1, 1) 
of the continuant in § 2. Consequently by division we obtain 
a = = ERD (a 2)ln-+ 1Y(b— 20)(-+0) 
ae a+ 2.27-¢ ae 
“, 1-(2n-2)(b—ne + c)(b +n - 2c) 
TP) G22 122 
_ {a+ 2(n —1)d} {a+ 2(n—3)b + 2(2n — 3)c} {a+ 2(n—5)b+4(2n—5)c} . . . - (IL) 
{a+ 2(n—2)b4+ 2(n-l)e}{a+2(n—4)b+6(n—2)c} .... 
(5) If wm the results of §§ 2, 3 we annex f as a factor to every term on both sides 
that 1s independent of a, the identity is not interfered with. — . o TCLS) 
For, taking (in the fourth order, for shortness’ sake) the continuant dealt with in 
§ 2, and putting a/f for a we have 
ae 3b 
6(b — 3c) = + 6¢ 2(b +e) 
7(b — 4c) a l6c b+2c 
8(b-5e) “§+30¢e 
+ (4 + 6b (4 pe Me)(4 ae 200)(f aioe 18e) 
whence on multiplying both sides by f* there results 
a 3bf 0 
6(6-3c)f atb6cef W(b+c)f 
T(b—4ce)f a+l6cf (b+2c)7 
: 8(b-5e)f at30ef | 
= (a+ 6bf)(a + 2bf + L4cf)(a — 2bf + Wef)(a — 6hf + 18c/), 
as asserted. 
