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MATHEMATICS: D. N. LEHMER 
Apk-i — ( l) p 1 A pk _ l , 
A pk =(-l) p W pk + MA pk _ x ) 
for p = 1, 2, 3, ... . 
Theorem II. With AjB m defined as in Theorem I, and A"jB" m de- 
fined as the (w + l)st convergent of 
(K, a k - { a k _ 2 , .... a 2 , a { , L, a k _ u a k _ 2 , • • • • a 2 , a u — a k m — aM), 
m = 2, 3, 4, 5, . . . . , 
where Z = (B k _ { - A k _ 2 ) / A k _ x 
and L = -a k -2 {B 2 k _ x + B k _ 2 A k _,) / A k _, B k _ t ; 
a u a 2 . .. . . . a*, and M as before, and Z^.! not zero, then: 
Bp*.! - (- 
5p, =(-D p {A^ + MA^.d 
for p = 1, 2, 3, ... . 
These two theorems are proved by complete induction. The proof of 
each proceeds in an unusual and interesting way by assuming both equa- 
tions of the theorem to hold for p^n, then proving that the first of the two 
equations must be valid for p = n -f- 1, and then using this result to show 
that the second holds also for p = n + 1. 
The three fractions considered in these two theorems are then discussed 
with respect to a modulus n prime to 2a k , A k _i and Bk-i. It is possible to 
find an even value of m, say m = 2X, which shall be less than 2n, and which 
shall satisfy the congruence 
a k m + 2M = 0 (mod n) 
This value of m will provide a partial quotient in the second fraction of 
Theorem I, which shall be congruent to zero modulo n. The two continued 
fractions of this theorem are then seen to be inverse modulo n as far as this 
partial quotient. From the properties of inverse continued fractions, we 
have then ^2*x-i = A^kX-i- But by Theorem I itself A 2kX -i = — A^kX-i 
so that we derive A 2 k\-i — 0 (mod n). 
It is possible, using only Theorem I, to show also that A 2k \ = B 2 k\-i = 
(- J) (X( * +1) . To prove, however, that B 2k \ = M(-l) Hk+l) it is necessary 
to use Theorem II. 
Certain interesting palindromic relations develop in the course of the 
proof, such as 
