1911 Encyclopædia Britannica/Number/Composition

37. Composition.&mdash;Considering $$\mathrm{X}, \mathrm{Y}$$ as given lineo-linear functions of $$(x, y), (x', y')$$ defined by the equations
 * $$\mathrm{X} = p_0xx'+p_1xy'+p_2x'y+p_3yy'$$
 * $$\mathrm{Y} = q_0xx'+q_1xy'+q_2x'y+q_3yy'$$

we may have identically, in $$x, y, x', y'$$,
 * $$(\mathrm{A}, \mathrm{B}, \mathrm{C})(\mathrm{X}, \mathrm{Y})^2 = (a, b, c)(x, y)^2\times(a', b', c')(x', y')^2$$

and, this being so, the form $$(\mathrm{A}, \mathrm{B}, \mathrm{C})$$ is said to be compounded of the two forms $$(a, b, c), (a', b', c')$$, the order of composition being indifferent. In order that two forms may admit of composition into a third, it is necessary and sufficient that their determinants be in the ratio of two squares. The most important case is that of two primitive forms $$\phi, \chi$$ of the same determinant; these can be compounded into a form denoted by $$\phi\chi$$ or $$\chi\phi$$ which is also primitive and of the same determinant as $$\phi$$ or $$\chi$$. If $$\mathrm{A}, \mathrm{B}, \mathrm{C}$$ are the classes to which $$\phi, \chi, \phi\chi$$ respectively belong, then any form of $$\mathrm{A}$$ compounded with any form of $$\mathrm{B}$$ gives rise to a form belonging to $$\mathrm{C}$$. For this reason we write $$\mathrm{C} = \mathrm{A}\mathrm{B} = \mathrm{B}\mathrm{A}$$, and speak of the multiplication or composition of classes. The principal class is usually denoted by $$1$$, because when compounded with any other class $$\mathrm{A}$$ it gives this same class $$\mathrm{A}$$.

The total number of primitive classes being finite, $$h$$, say, the series $$A, A^2, A^3, \And\!\!\!\!\mathrm{c}.$$, must be recurring, and there will be a least exponent $$e$$ such that $$A^e = 1$$. This exponent is a factor of $$h$$, so that every class satisfies $$A^h = 1$$. Composition is associative as well as commutative, that is to say, $$(\mathrm{A}\mathrm{B})\mathrm{C} = \mathrm{A}(\mathrm{B}\mathrm{C})$$; hence the symbols $$A_1, A_2, \dots A_h$$ for the $$h$$ different classes define an Abelian group (see ) of order $$h$$, which is representable by one or more base-classes $$\mathrm{B}_1, \mathrm{B}_2, \dots \mathrm{B}_i$$ in such a way that each class $$\mathrm{A}$$ is enumerated once and only once by putting
 * $$\mathrm{A}={\mathrm{B}_1}^x{\mathrm{B}_2}^y \dots {\mathrm{B}_i}^z \qquad (x \le m, y \le n, \dots z \le p)$$

with $$mn \dots p = h$$, and $${\mathrm{B}_1}^m = {\mathrm{B}_2}^n = \dots ={\mathrm{B}_i}^p = 1$$. Moreover, the bases may be so chosen that $$m$$ is $$n$$, $$n$$ of the next corresponding index, and so on. The same thing may be said with regard to the symbols for the classes contained in the principal genus, because two forms of that genus compound into one of the same kind. If this latter group is cyclical, that is, if all the classes of the principal genus can be represented in the form $$1, A, A^2, \dots A^{\nu-1}$$, the determinant $$\mathrm{D}$$ is said to be regular; if not, the determinant is irregular. It has been proved that certain specified classes of determinants are always irregular; but no complete criterion has been found, other than working out the whole set of primitive classes, and determining the group of the principal genus, for deciding whether a given determinant is irregular or not.

If $$\mathrm{A}, \mathrm{B}$$ are any two classes, the total character of $$\mathrm{A}\mathrm{B}$$ is found by compounding the characters of $$\mathrm{A}$$ and $$\mathrm{B}$$. In particular, the class $$\mathrm{A}^2$$, which is called the duplicate of $$\mathrm{A}$$, always belongs to the principal genus. Gauss proved, conversely, that every class in the principal genus may be expressed as the duplicate of a class. An ambiguous class satisfies $$\mathrm{A}^2 = 1$$, that is, its duplicate is the principal class; and the converse of this is true. Hence if $$\mathrm{B}_1, \mathrm{B}_2, /dots \mathrm{B}_i$$ are the base-classes for the whole composition-group, and $$\mathrm{A}={\mathrm{B}_1}^x{\mathrm{B}_2}^y \dots {\mathrm{B}_i}^z$$ (as above) $$\mathrm{A}^2 = 1$$, if $$2x = 0 \mbox{ or }m$$, $$2y = 0 \mbox{ or }n$$, &c.; hence the number of ambiguous classes is $$2^i$$. As an example, when $$\mathrm{D} = -1460$$, there are four ambiguous classes, represented by


 * $$(1, 0, 365), (2, 2, 183), (5, 0, 73), (10, 10, 39)$$;

hence the composition-group must be dibasic, and in fact, if we put $$\mathrm{B}_1, \mathrm{B}_2$$ for the classes represented by $$(11, 6, 34)$$ and $$(2, 2, 183)$$, we have $${\mathrm{B}_1}^{10} = {\mathrm{B}_2}^2 = 1$$ and the 20 primitive classes are given by $${\mathrm{B}_1}^x{\mathrm{B}_2}^y\ (x \le 10, y \le 2)$$. In this case the determinant is regular and the classes in the principal genus are $$1, {\mathrm{B}_1}^2, {\mathrm{B}_1}^4, {\mathrm{B}_1}^6, {\mathrm{B}_1}^8$$.

38. On account of its historical interest, we may briefly consider the form $$x^2+y^2$$, for which $$\mathrm{D} =-4$$. If $$p$$ is an odd prime of the form $$4n+1$$, the congruence $$m^2 \equiv -4 \pmod{4p}$$ is soluble (&sect; 31); let one of its roots be $$m$$, and $$m^2+4 = 4lp$$. Then $$(p, m, l)$$ is of determinant $$-4$$, and, since there is only one primitive class for this determinant, we must have $$(p, m, l)\sim(1, 0, 1)$$. By known rules we can actually find a substitution $$\begin{pmatrix} \alpha, & \beta \\ \gamma, & \delta \end{pmatrix}$$ which converts the first form into the second; this being so, $$\begin{pmatrix} \delta, & -\beta \\ -\gamma, & \alpha \end{pmatrix}$$ will transform the second into the first, and we shall have $$p = \gamma^2 + \delta^2$$, a representation of $$p$$ as the sum of two squares. This is unique, except that we may put $$p = (\pm \gamma)^2+(\pm \delta)^2$$. We also have $$2 = 1^2 +1^2$$ while no prime $$4n+3$$ admits of such a representation.

The theory of composition for this determinant is expressed by the identity $$(x^2+y^2)(x'^2+y'^2) = (xx'\pm yy')^2 + (xy'\mp yx')^2$$; and by repeated application of this, and the previous theorem, we can show that if $$\mathrm{N}=2^ap^bq^c \dots$$, where $$p, q, \dots$$ are odd primes of the form $$4n+1$$, we can find solutions of $$\mathrm{N}=(x^2+y^2)$$, and indeed distinct Solutions. For instance $$65 = 1^2+8^2=4^2+7^2$$, and conversely two distinct representations $$\mathrm{N} = (x^2+y^2) = (u^2+v^2)$$ lead to the conclusion that $$\mathrm{N}$$ is composite. This is a simple example of the application of the theory of forms to the difficult problem of deciding whether a given large number is prime or composite; an application first indicated by Gauss, though, in the present simple case, probably known to Fermat.