584 
Proceedings of the Royal Society 
Again, we may write 
Hence ^3 = 2, and 3(g^,...,^,_3) + (g5,...,^,.3) = (^4,...,^,_2) + 
(g5,.,.,g'^_2) is a partial solution. The second part of this, how- 
ever, is advantageously carried farther. Thus we have from it 
3 {g' 4 v 5 2 z- 3 ) + ( 9'55 •••,^ 2 - 3 ) = gz- 2 (^ 4 » ••• + ( 2 ' 4 »-*-» 9 'e -) 
+^z-^[q5, g 2 - 3 )+(g 5 J ••• > 
ana .-. q,_ _ 3(g4-- g^-3) + (g5. •••> g^-3) - (?4 ••• - (gs 
^ (g'4 ... 2'z-3) + fe-‘- 2'3-3) 
How this fraction cannot = 1, for then we should have 
2(^4) • • • ) Qz -s) ~ (2'4’ • • • • j S'z-i) + (S'S’ • • • ) 
which is impossible, for 2(^^, . . . . , g^_3) is greater than either of the 
continuants on the right. Trying if it may equal 2, we put the 
result in the form 
^ _ 0 , fe • • • » 9z-^) - fe • • • R-s) - (^4 • • • R-i) -(%••• R-i) 
[/z-2 f - - I - - , 
04--'2.-3) + (?5- ••.S.-s) 
from which it is at once manifest that the only solution is 
2,.2=2 and O4 . . . 5.-3) = O4. • . ?.-4) + (?6- • • + -l.-t)- 
Lastly, let us try if can be equal to 2 + a. We should then 
have 
(3 + «) (^4 ... gz-3) + (2s ••• $z-3) = (l + a)(g4 ••• 9'^-2) + (9'5 ••• ^z-l) , 
= (1 4-a)5'^-2(g'4 ... g'2_3) + (l + oc) {q^...qz-^ 
+ qz- 2 {q 5 qz- 2 ) + {q 5 -• qz -^) ; 
.-. {3 + a-(l + «)^3_2}(g4 ••• gz-3) = (!+«) (g4 ••• ^2-4) + (^z-2-l)(g'g ... qz-z) 
This necessitates 
( 3 + ) - (1 + a)2'2_2 >0 ; 
a -f- 3 
< 
< 1 H- 
a -f 1 a + 1 
and g ^_2 = l 
Putting qz-^ = 1, we see that we must further have 
2(24. • • • , 2. a) = (1 + “)(24. • • • . 2 .- 4 ) + ( 25 . • • • . 2 .- 4 ) 
^.e., 
= (1 -f-a, ^4, . . ., 
as was to be shown. 
