appertains to (I); for, writing p and q instead of

α, β

23-0a, and 28-0y, we have |a, 8 |=

α, 1⁄2 (p+0a)

a,

Y, 1⁄2 (q + Oy), α, Y, P, q (of which α and γ are relatively prime) being four numbers which satisfy the system

(3); i. e. a, appertains to (I), an improper automorphic of ƒ.

б

It follows from (B) that, if we calculate the ambiguous forms and Ф appertaining to every improper automorphic off, we shall obtain all the ambiguous forms to which fis equivalent; it remains to see how many of these ambiguous forms are different from one another. If (I) =\";is any given improper automorphic off, all its similar automorphics are contained in the four formulæ

μ,

1, 0

× (I),

(T)2×(1), (T)2+1× (I), (T)2×│5,1 o_1 | ×(

where & is any positive or negative number, and (T)= -u1b, -u1c

= [t1, u,] representing a fundamental solution of the equation t2- Du2=1. =Similarly, if (J) represent the four transformations, appertaining to (I), by which passes into p or -, all the proper transformations off into or - are included in the formula (T) × (J). We shall now show that the four transformations included in the formula (T)*x (J) appertain to the improper automorphic (T)2 × (I). Writing

ax=(tx—bux)a—cuμ¿Y›

Px=(tx-bux)p-cuzl

Yk=auza+(tx+bux)y, qk=auxP+(tx+bux)q,

|, (T)2 × (I)= | μ2k, —λ2k

|

Also attending to the equations (2) and (3), and to the relations

we obtain, after substitution and reduction,

Pk az=λ2k Pk Y2x=2x-1,

i. e. (T)* × (J) appertains to (T)2 × (I), if (J) appertains to (I). It follows from this result that the ambiguous forms appertaining to (I)

and to (T) × (I) are the same; for fis transformed into the same forms by (J) and (T) ×(J); and conversely, if the ambiguous forms appertain

ing to two different automorphics (I) and (I') are identical, an equation of the form (I')=Tx (I) will subsist; for if (J) and (J') are the transformations appertaining to (I) and (I'), since by hypothesis (J) and (J) transform ƒ into the same form, we must have an equation of the form (J') = (T)′′ × (J); but (J') appertains to (I'), and (T)*× (J) to (T)TM × (I) ; therefore (I')=(T)2 × (I), by what has been shown above (A).

2k

2k

If then we calculate the eight ambiguous forms appertaining to the four improper automorphics

these eight forms will be the only ambiguous forms equivalent to f. Thus every uneven ambiguous class contains eight ambiguous forms.

Combining this result with the preceding we obtain the Theorem, "The number of uneven ambiguous classes is one half of the whole number of assignable generic characters."

The number of semieven and even ambiguous classes is determined by the two following Theorems :—

"When D= ±1, mod 4, there are as many even as semieven ambiguous classes."

"When D=1, mod 2, there are as many semieven as uneven ambiguous classes, or only half as many, according as there are altogether as many semieven as uneven classes, or only half as many."

2k

То prove the first of these theorems, let D=i", mod 4, and let

it is evident from the principles of the composition of forms that if (4) is a given semieven ambiguous class, the equation (2) × (q)=(1+i) × (ƒ) is satisfied by one and only one even ambiguous class (ƒ); in addition to this we shall now show that, if (ƒ) is a given even ambiguous class, the same equation is satisfied by one and only one semieven ambiguous class (); from which two things the truth of the theorem is manifest. First, let the whole number of even classes be equal to the whole number of semieven classes*; then the equation

(E) × (p)=(1+i) × (f)

* That if D=+1, mod 4, there are either as many semieven as even classes, or else three times as many, is a theorem of M. Lipschitz (Crelle, vol. liv. p. 196), of which it is worth while to give a proof here. The number of even classes is to the number of semieven classes, as unity to the number of semieven classes satisfying the equation

(E)×(p)=(1+i)×(ƒ),

f representing any given even form. To investigate the semieven classes satisfying this equation, apply to ƒ a complete system of transformations for the modulus 1+i, for example, the transformations

is satisfied by only one semieven class (p); and this class is ambiguous, for the equation is satisfied by the opposite of (p) as well as by (4) itself; therefore (4) and its opposite are the same class, or (4) is an ambiguous class. Secondly, let the number of semieven classes be three times the number of even classes; then the equation

(E) × ($)=(1+i) (ƒ)

is satisfied by three and only three different classes (4); but it is also satisfied by the opposites of these classes; therefore one of them is necessarily an ambiguous class. Let that class be (4.); the other two are defined by the equations

(1+i) (p,)=(0,) × (p), (1+i) (42)=(02) × (4%), and cannot be ambiguous classes; for by duplication we find (4) × ($,)=(1+i) (o2), (42)×($,)=(1+i) (o,); whereas every semieven ambiguous class produces (1+i)o, by its duplication *.

The second theorem may be proved as follows. Let

f=([1+i]p, q, [1+i]r)

be a semieven form of determinant D; and let

。。=((1+i), 1, − 1 = 1);

we suppose that p is uneven. The equation (o) × (p)=(ƒ) is satisfied by one uneven class (4.), or by two (4) and (4), according as the forms =(p, q, 2ir), and 4,=(2ip, q, r), if r is uneven, or the forms =(p, q, 2ir), and 4,=(2ip, [1+i]p+q, p+[1−i]q+r), if r is even, are or are not equivalent. If any one of the forms f, 4., 4, is ambiguous, the others are so too; the same thing is therefore true for the classes (ƒ), (4%), (4.). Thus the number of semieven ambiguous classes is equal to or

and divide the resulting forms by 1+i; of the quotients, one, or three, will be semieven, according as D=+1, or +5, mod (1+i). It will be found that each of these semieven forms satisfies the equation Exp=(1+i)׃; and, conversely, every semieven form satisfying that equation is equivalent to one of these forms; for, from any transformation of (1+i)ƒ into Σ×p, we may (by attributing to the indeterminates of Σ the values 1, 0) deduce a transformation of modulus 1+i by which ƒ passes into (1+i)ø ; i.e., is equivalent to one of the forms obtained by the preceding process. It only remains to show that when there are three of these forms, they constitute either one or three classes, but never two. For this purpose it is sufficient to consider the three semi

even forms = =(1+i, 1, −1), σ, and σ,, obtained by the preceding process from

the form Σ. These forms satisfy the equations ×σ=(1+i)σ„, σ1×01=(1+i)σ2, •2×0,=(1+i)01, σ1×σ2=(1+i)σ; from which it follows that any one of the suppositions σ12, 6=0, 0=0, involves the other two.

* For the definition of the classes (a), (σ1), (σ2) see the preceding note.

↑ The forms and 4, are obtained by applying to ƒ a complete set of transformations of modulus 1+i, dividing the resulting forms by 1+i, and retaining only those quotients which are uneven forms,

is one half of the number of uneven ambiguous classes, according as the classes (4) and (,) are identical or not; i. e., according as the whole number of semieven classes is equal to or is one-half of the whole number of even classes.

The demonstration in the Disquisitiones Arithmetice,' that the number of genera of uneven forms of any determinant cannot exceed the number of uneven ambiguous classes of the same determinant, may be transferred without change to the complex theory. We thus obtain a proof (independent of the law of quadratic reciprocity and of the theorems which determine the quadratic characters of i and 1+i) of the impossibility of one-half of the whole number of assignable generic characters; and from that impossibility, as we shall now show, the quadratic theorems are themselves deducible.

(1) If p is an uneven prime =1, mod 2, there are two genera of uneven forms of determinant p: of these one is the principal genus, and has the complete characters 1, y=1; the other, containing the form

(i, 0, +ip), has the particular character y=-1; whence it follows that every uneven form of determinant p, which has the character y=+1, is

p+1)

a form of the principal genus, and has the character []=+1. Again, fp=1,mod 4, the form (2i, i, ( is an uneven form of determinant p; this form has the particular character y=-1, because-?+1=i, mod 2;

is 2i

2i

it is therefore not a form of the principal genus; but it has the character =1, because 2i is a square; therefore, if p=1, mod 4, every uneven

form of determinant p has the character

=+1.

(2) There is but one genus of forms of determinant i, and its complete character is a=+1; there is also but one genus of forms of determinant 1+i, and its complete character is ẞ=+1.

(3) Let p and q be uneven primes of which the imaginary parts are even; to prove the law of reciprocity, it will suffice to show that if if? =1,

[2] then [2]=1. The equation [2]=1 implies the existence of a con

gruence of the type w2-p=0, mod q, and consequently of an uneven

form of determinant p, and of the type (q, w,pe (4%, w, w2 —p). This form has the

character y=+1, because q=1, mod 2; it therefore has the character |

=1; i. e.
· [p] =1.

(Np-1)

(4) To prove the equation [+]=(-1) (NP), in which we may sup

pose that the uneven prime p is primary, it will suffice to show (i) that

(Np-1)

=+1, then (-1)' =1; (ii) that if (—1)' =1. (i) Let

=1; then, if w2-i=0, mod p, (p, w,

(p, w, w2 —3)

is a form of determinant i; it therefore has the character a=1, i. e.

(−1)*(Np-1)=1. (ii) Let (-1) (Np-1=1; then p=1, mod 4, and the form (i, 0, ip) is an uneven form of determinant p; it therefore has

the character

(分)

= +1; whence

=+1.

(5) Similarly, if p=p,+ip, is an uneven and primary prime, to prove

(i) Let

the type (p, w,

Р

=1; then there is a form of determinant 1+i and of

fore (-1) 8

=

; this form has the character ẞ=+1; there

+1; then p is either 1-2i, or=1, mod (1+i); if p=(1+i)k+1—2i, ([1+1]3, i, 1-2ki) is an uneven form of determinant p; this form has the character y=+1, and consequently it also has the character

+1; if p=(1+i) k+1, one or other of the

forms ([1 +i]3, 1, −k), and ([1+i]3, 1+[1+i]3, 1−k) is an uneven form

of determinant p, having the character

IV. The representation of Binary Forms of the principal Genus by Ternary Forms of Determinant 1.

The solution of the general problem, "To find the representations (if any) of a given binary by a given ternary quadratic form," depends, in the case of complex as of real numbers, on the solution of the problem of equivalence for ternary forms. Extending the methods of Gauss to the complex theory, we find the necessary and sufficient condition for the primitive* * If a matrix of the type

transforms a ternary into a binary quadratic form, the representation of the binary by the ternary form is said to be primitive when the three determinants of the matrix are relatively prime.