CONTINUANTS RESOLVABLE INTO LINEAR FACTORS. 351 
(12) There is a special case of the theorem (VII) in § 9 which deserves particular 
attention, viz. the case where o vanishes, 7.e. where 5d=2b+c. In this case the s™ 
factor 
= {a+2(n—-1)b} - 2(s—1){2b-3c} + (n-s+2)(s -— 1){2b - 3c} 
= {a+2(n—1)h} + (n—s)(s - 1)(26 -3¢) ; 
and the (n—s+1)™ factor 
= {a+2(n—1)b} — 2(m—-s){2b-3c} + (s+1)(n—s){2b - 3c} 
= {a+2(n—1)b} + (n-s)(s—1)(2b- 3c). 
This means that when 5d=2b+e w the continuant of § 9 the s” factor from the 
beginning 1s the same as the s factor from the end, and consequently that an even- 
ordered continuant of this kind 1s a square. . ; A DS) 
(13) The question of the generalisation of the theorem of § 3 may be investigated 
in a manner perfectly similar to that followed in the preceding paragraphs with regard 
to the theorem of § 2. The essential point of difference is to be found in the new 
set of column-multipliers, which are now all of the form C,,,, ,_; instead of 20,1, 04 9-1. 
It will suffice merely to enunciate the results. The first is— 
If the contunuant 
| a (n—1)B, c ee wae 
| (m+2)y, atp (n—2)6, doen ar 
(n+ 3)y5 a+¢g (i= 3) By 2,2 28 3 
(n+4)y, GEEHNID A we nes fs Si 
be resolvable into linear factors by means of the set of column-multipliers 
(eR ISE TK. he ate Ct; 
eae OOO Ate Cu 
ipa SRO sR eee Cx 
iN ase ee Cae 
jot ean (Caen 
then (1) every four consecutive 6's are connected by a linear relation, viz. 
3(B, — 28,+Bs)= 9B, — 283+ ,) , 
ne a ere oe ae 26,+ ay ) 
thus making all the B's Preble m terms of any three ; (2) all the y's ure expressible 
m terms of the same three 3's because of the fact that for all values ofm By+%m= 
— 9B, + 258,—148,; and (3) p,q, 7, . . . are also so expressible because of the mode 
of removing the first factor from the ee ak 0.6) 
It should be noted that the linear relation connecting the fing: four consecutive @’s 
is that which in § 8 connects the second four,—that, in fact, the relation here is 
(2r+1){B,- 2B,41+ Bpzo} =(27+7){ Bra — 2Bry2 + Bros} 
whereas in § 8 it is 
(2r - 1){Br - 26,41 + B42} = (27+ 5){Bri1— 2B, 49+ Bes} - 
TRANS. ROY. SOC. EDIN., VOL. XLI. PART II. (NO. 13). 52 
