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CHAP, vii]                      BINOMIAL CONGRUENCES.                               217
0, 1, 2, ... until we reach a number of the form #+z (found by extracting the square root). Then k~z, so that the roots P k, P+l+k are found. N. Amici199 proved that if neither m nor 6 is divisible by the prime p, and if a is a given root of xm=b (mod p), and if ft, q are (existing) integers such that
then apX~V is a root of xm=b (mod px). Hence we limit attention to the case \ = 1. Consider henceforth x2k=-b (mod p), where p=2ah+l is an odd prime, h being odd, and b not divisible by p. First, let k^s. Then 6^=1 (mod p) is a necessary and sufficient condition for solvability and #===*= ft5 are roots, where q is such that 2kq 1 is divisible by h. If g is a quadratic non-residue of p, all 2* roots are given by b9ghe, where e = et-h2e2 + . . . +2S~2S_!, the i taking the values 0 and 1 independently. Finally, let k<s. Then two roots =*= are determined by the method of Tonelli, while all the roots are given by
i,          .=0 or 1.
R. Alagna200 considered a prime p=4fc+l for which & is a prime. Since 2 is known to be a primitive root of p, it is easy to write down those powers of 2 which give all the roots of ofel (mod p), where d is one of the six divisors 2i or A of p  1, likewise of xd=N, since N must be congruent to an even power of 2. For the modulus px, we may apply the first theorem of Amici or proceed directly. The same questions are treated for a prime 4fc+3 for which 2k +1 is a prime.
A. Cunningham201 treated at length the solution of xl=l (mod N1), where AT is a prime, and gave tables showing all incongruent roots when J = 1, 2, JV^ 101, Z any admissible divisor of N 1; also for a few additional ^s when N is small.
Cunningham2010 treated ap==l (mod g2) and 3.2*==*= 1 (mod p). He2016 treated the problem to find b"s=+l or ==a, given a*=l, ax==b (mod p), where  is odd and , a;, ?? are the least values of their kind; also given a*=l, ax=bj az=^=Cj to find the least ft and 7 such that &^=c, cT=6 (mod p).
W. H. Besant202 would solve y2 = ax-\-b by finding the roots s of s2s=6 (mod a). Then t/ = ar+s, x = ar2+2rs-f-(s2  6)/a.
G. Speckmann203 replaced ofefc (mod p) by the pair of congruences a^-irrr^ xr=k (mod p). In np+k give to n the values 0, 1, 2, ... until we find one for which np+k = rx such that, by trial, xn~l= r. The method is, of course, impractical.
"'Rendiconti Circolo Mat. di Palermo, 11, 1897, 43-57.
""Rendiconti Circolo Mat. di Palermo, 13, 1899, 99-129.
201Messenger of Math., 29, 1899-1900, 145-179.   Errata, Cunningham228, p. 155.   See 13a of
Ch. IV.
*laMath. Quest. Educ. Times, 71, 1899, 43-4; 75, 1901, 52-4. Ml6/6td., (2), 1, 1902, 70-2. MMath. Gazette, 1, 1900, 130. 2MArchiv Math. Phys., (2), 17, 1900, 110-2, 120-1.