216 HlSTOBY OF THE THEORY OF NUMBERS. [CHAP. VII
when p is an odd prime, and a quadratic non-residue g of p is known. Set p = 2sa-hl, where s^l and a is odd. Then 7 = apx~1 is odd, and 4>(px) = 2s7. Tonelli's earlier work for modulus p now holds for modulus px and we get x= ± geyc(y+1)/2. If s = 1, then e = 0 and the root is that given by Lagrange if X=l. If s=2, whence p=4a-f l=8Z+5, the expression for x is given a form free of e = e0:
A. Tonelli196 expressed the root a; in a form free of e for every s:
r±l ^±^-V28-3...^3^2C2,
where the v's are given by the recursion formula
t^-crSTa**1'. . .it£i+* (* = 2, 3, . . .)
Here k is an existing integer such that k+1 is a quadratic residue of p, and A; 1 a non-residue. Thus, if s = 3,
where we may take &= 2 if a is not divisible by 3, but k = 4 if a is divisible by 3, while neither a nor 4a+l are divisible by 5.
N. Amici86 proved that x2k=b (mod 2"), b odd, k^v2, is solvable only when b is of the form 2k+2h-\-l and then has 2k+l roots, as shown by use of indices. For (2m)2*= 5, the same condition on b is necessary; thus it remains to solve xm=f3 (mod 2") when m is odd. If /3 = 8/c+l or 8/c+3, it has an index to the base 8^+3 and we get an unique root. If ft = 8k 3 or Sk 1, then xm= ft has a root a by the preceding case, and a is a root of the proposed congruence .
Jos. Mayer197 found the number of roots of x3==a (mod pn), for the primes 2, 3, £ = 6^=*=!. If ab a2, . . . are residues of nth powers modulo p} and if q is the g. c. d. of n and p 1, then axa2. . . = +1 or 1 (modp), according as p' = (p l)/g is odd or even. If p' is even, we can pair the numbers belonging to the exponent p' so that the sum of a pair is 0 or p ; hence there exists a residue of an nth power s 1 (mod p) ; but none if p' is odd.
K. Zsigmondy87 obtained by the use of abelian groups known theorems on the number, product and sum of the roots of xs=l (mod m).
G. Speckmann198 considered x2=a (mod p), where p is an odd prime. Set P = (p-l)/2. When they exist, the roots may be designated P-k, P+l+k, whose sum is p. The successive differences of P2, (P+l)2, (P-f2)2,. . . arep, p+2, p+4,. . .. The sum of z = s+l terms of 2, 4, 6,. . . is $2+3s+ 2 = 22+2. Adding to the latter the remainder r obtained by dividing P2 by p, we must get pn+a. Hence in pn-J-a r we give to n the values
198Atti R. Accad. Lincei, Rendiconti, (5), 2, 1893, 259-265.
197Ueber nte Potenzreste und binomische Congruenzen dritteu Grades, Progr., Freising, 1895.
198Archiv Math. Phys., (2), 14, 1896, 445-8; 15, 1897, 335-6.