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Full text of "History Of The Theory Of Numbers - I"

CHAP. IV]                 RESIDUE OF (U*~l-I)/p MODULO p.                     107
where ^i+...+v = p, v>0. The term given by r = p is ap. For p a prime, the left member is =a (mod p) and we have Fermat's theorem. By induction on r,
Taking r=p, we have
D. Mirimanoff12 wrote a0 for the least positive integer making aQp+l divisible by the prime r<p, and denoted the quotient by rebi, where 61 is prune to r. Similarly, let a be the least positive integer such that atp+6t- = r*bi+i. We ultimately find an n for which 6n = l. Then &+; = &;. By (2),
Let r belong to the exponent co modulo p and set e<o = p  1. Then Set-=&>, while l,&i,..., 6n_i are the distinct residues of the eth powers of the integers <r and prune to r. Thus
n-lfl>
#r=e2) ~ (mod p).
t0 0{
The formula obtained by taking r a primitive root of p is included in the following, which holds also for any prime r:
p-ia. ^^^(modp),
ai being the least positive integer for which aip-f-ft=0 (mod r). Set A = p-5, p'p=l (mod r), 0<;p'<r. Then a^p'6-1 (mod r),
jfcj- being the least positive residue modulo r of k.   Whence Sylvester V
statement.
J. S. Aladow13 proved that (1) has at most (p=pl)/4 roots if p = 4m=1=L A. Cunningham13" listed 27 cases in which rp~~l=^l or ^==1 (mod p'),
r<p*~1, where Z is a divisor of p  1.    For the 11 cases of the first kind,
p = 5, 7, 17, 19, 29, 37, 43, 71, 487.
W. Fr. Meyer14 proved by induction that, if p is a prime, a;15""1  1 is
divisible by pk (l^k<ri), but not by pfc+1, for exactly pn-l~k (p-1)2 posi-
tive integers x<pn and prime to p, and is divisible by pn for the remaining
p  1 such integers.   Set
"Jour, fttr Math., 115, 1895, 295-300.
"St. Petersburg Math. Soc. (Russian), 1899, 40-44.
lsaMessenger Math., 29, 1899-1900, 158.    See Cunningham128, Ch. VI.
"Archiv Math. Phys., (3), 2, 1901, 141-6.