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CHAP. VIII]                              CUBIC CONGKTJBNCES.                                        255
D. Mirimanoff139 noted that the results by Arnoux112* 137 may be combined by use of the discriminant D= — 4b3— 27a2= — 3*62.K in place of R, since —3 is a quadratic residue of a prime p = 3&4-1, non-residue of p = 3k — 1, and we obtain the result as stated by Voronoi.132
To find which of the values 1 or 3 is taken by v when D is a quadratic residue, apply the theorem that if /(x)=0 (mod p) is an irreducible congruence of degree n and if XQ is one of its imaginary roots (say one of the roots of the equation f(x) = 0), the roots are
XQl           Xl = Z0P, . . . ,
Hence a function unaltered by the cyclic substitution (x^. . .xn_i) has an integral value modulo p. Take n=3, jD=d2, a a root F^l of 2?=! (mod p), and let
If p==l (mod 3), a is an integer, and M is an integer if » = 1, while M is the cube of an integer if i> = 3. Thus we have Arnoux' s criterion:112 j>=3 if M or -I (— 9a+ V— 3d) is a cubic residue modulo p. If p== —1 (mod 3) ?=3 if and only if Mk=^l (mod p), where fc = (p2— 1)/3.
For quartic congruences, we can use (x0— Xi+x2— z3)2.
R. D. von Sterneck140 noted that if p is a prime >3 not dividing A, and if k = 3AC— B2^Q (mod p), then the number of incongruent values taken by Ax3+Bx2+Cx+D is i{2p+(-3/p)}; but, if fcsO, the number is p if p = 3n— 1, (p+2)/3 if p = 3n+L Generalization by Kantor.181
C. Cailler141 treated x3+px+qz==Q (mod 1), where Z is a prune >3. By the algebraic method leading to Cardan's formula, we write the congruence in the form
(1)                            x3-3dbx+db(a+b)=Q (mod I),
where a, 6 are the roots of 22+3g2/p—p/3s=0 (mod 0, whence
z== (z0+azi-fa2a;2)3/(9;p)>          a2+a+l==0 (mod T).
Let A = 4p3+27#2.   If 3A is a quadratic residue of Z, a and b are distinct and real.    If 3A is a non-residue, a and b are Galois imaginaries where N is any non-residue.   For a root x of (1),
Use is made of a recurring series S with the scale of relation [a+b, — ab] to get 7/0, 2/i,. - .-   Write Q = (3A/Z).   If Z = 3m-l, Q = l, then
If Z = 3m+l, Q==l, the congruence is possible only when the real number a/6 is a cubic residue, i. e., if z/m=0 in S; let a/b belong to the exponent 3ju=pl modulo Z-, whence
139L'enseignement math., 9, 1907, 381-4.
""Sitzungsber. Ak. Wiss. Wien (Math.), 116, 1907, JIa, 895-904.
ulL'enseignement math., 10, 1908, 474-487.