108 History of the Theory of Numbers. [Chap, iv 
If k is the least index for which fXk^Xk, M/»=^/. (mod p) for h<k, then 
A'~^ — l is divisible by p*", but not by p^"*"^. 
A. Palmstrom and A. Pollak^^ proved that, if p is a prime and n, m are. 
the exponents to which a belongs modulo p, p~, respectively, then a"" — ! 
is divisible by p^, so that m is a multiple of n and a divisor of np, whence 
m = n OT pn. Thus according as a^~^ is or is not =1 (mod p^), m = n or 
m = np. 
Worms de Romilly^^" noted that, if co is a primitive root of p^, the incon- 
gnient roots of x^^=l (mod p^) are co^^{j = l,. . ., p — 1). 
J. W. L. Glaisher^^ proved that if r is a positive integer <p, p sl prime, 
r''-^ = l+^iP+K^i'-^2)p'+|(^i'-3^i^2+2^3)p'+. . ., 
where Qn is the sum of the nth powers of 
1 _2_ r-1 . 1 2 r-1 . 1 
a [2(t]'" '' [{r-l)a]' r+cr' r+[2(T]'" '' r+[(r-lV]' 2r+(r'* ' '' 
a being the least positive residue modulo r of — p. If /i^ is the least positive 
solution of ani=i (mod r), viz., p/i<H-i=0, then 
Ml , M2 , I Mr-l , Ml , M2 I , Mr-1 , Ml , 
12 r-1 r + 1 r+2 ' ' ' ' ' 2r-l ^ 2r+l ' " 
Set /Ur = Oj Mi+jr =Mi- Then 
g^=i\^)\ s'^'^O(modp). 
Sylvester's corrected results are proved. From (1 + 1)^, 
op_2 1 / 1 \ 
=l-^+i- • . • 7^2(l+K . . . + -) (mod p). 
p p — 1 \ p — 2/ 
For r' = r+A:p, let m/ be the positive root of p/x/+^=0 (mod r'). Then 
It is shown that, for some integer t, 
k k k^ 2k k^ 
hi-gi+- = tp, h2-g2= -2---2+^t-y9i-:^ (mod p), 
Glaisher,^^ using the same notations, gave 
^'-'-l+p(^+f +...+^l) (modp=). 
"L'intermddiaire des math., 8, 1901, 122, 205-6 (7, 1900, 357). 
^Ibid., 214-5. 
'•Quar. Jour. Math., 32, 1901, 1-27, 240-251. 
"Messenger Math., 30, 1900-1, 78. 
