1911 Encyclopædia Britannica/Number/Complex Numbers (Gaussian)

42. Complex Numbers.&mdash;One of Gauss's most important and far-reaching contributions to arithmetic was his introduction of complex integers $$a+bi$$, where $$a, b$$ are ordinary integers, and, as usual, $$i^2 = -1$$. In this theory there are four units $$\pm 1, \pm i$$; the numbers $$i^h(a+bi)$$ are said to be associated; $$a-bi$$ is the conjugate of $$a+bi$$ and we write $$\mathrm{N}(a \pm bi) = a^2+b^2$$, the norm of $$a+bi$$, its conjugate, and associates. The most fundamental proposition in the theory is that the process of residuation (&sect; 24) is applicable; namely, if $$m, n$$ are any two complex integers and $$\mathrm{N}(m) > \mathrm{N}(n)$$, we can always find integers $$q, r$$ euch that $$m = qn+r$$ with $$N(r)\le\tfrac{1}{2}\mathrm{N}(n)$$. This may be proved analytically, but is obvious if we mark complex integers by points in a plane. Hence immediately follow propositions about resolutions into prime factors, greatest common measure, &c., analogous to those in the ordinary theory; it will only be necessary to indicate special points of difference.

We have $$2 =-i(1+i)^2$$, so that $$2$$ is associated with a square; a real prime of the form $$4n+3$$ is still a piime, but one of the form $$4n+1$$ breaks up into two conjugate prime factors, for example, $$5 = (1-2i)(1+2i)$$. An integer is even, semi-even, or odd according as it is divisible by $$(1+i)^2, (1+i)$$ or is prime to $$(1+i)$$. Among four associated odd integers there is one and only one which $$\equiv 1 \pmod{2+2i}$$; this is said to be primary; the conjugate of a primary number is primary, and the product of any number of primaries is primary. The conditions that $$a+bi$$ may be primary are $$b \equiv 0 \pmod{2},\quad a+b-1 \equiv 0 \pmod{4}$$. Every complex integer can be uniquely expressed in the form $$i^m(1+i)^na^\alpha b^\beta c^\gamma \dots$$, where $$0 \le m<4$$, and $$a, b, c, \dots$$ are primary primes.

With respect to a complex modulus $$m$$, all complex integers may be distributed into $$\mathrm{N}(m)$$ incongruous classes. If $$m = h(a+bi)$$ where $$a, b$$ are co-primes, we may take as representatives of these classes the residues $$x+yi$$ where $$x = 0, 1, 2,\dots \{(a^2+b^2)h-1\}; y=0, 1, 2, \dots (h-1)$$. Thus when $$b = 0$$ we may take $$x = 0, 1, 2, \dots (h-1); y = 0, 1, 2,\dots (h-1)$$, giving the $$h^2$$ residues of the real number $$h$$; while if $$a+bi$$ is prime, $$1, 2, 3,\dots (a^2+b^2-1)$$ form a complete set of residues.

The number of residues of $$m$$ that are prime to $$m$$ is given by
 * $$\phi(m) = \mathrm{N}(m)\Pi\left(1-\frac{1}{\mathrm{N}(p)}\right)$$

where the product extends to all prime factors of $$m$$. As an analogue to Fermat’s theorem we have, for any integer prime to the modulus,
 * $$x^{\phi(m)}\equiv 1 \pmod{m}, x^{\mathrm{N}(p)-1} \equiv 1 \pmod{p}$$

according as $$m$$ is composite or prime. There are $$\phi\{\mathrm{N}(p)-1\}$$ primitive roots of the prime $$p$$; a primitive root in the real theory for a real prime $$4n+1$$ is also a primitive root in the new theory for each prime factor of $$(4n+1)$$, but if $$p=4n+3$$ be a prime its primitive roots are necessarily complex.

43. If $$p, q$$ are any two odd primes, we shall define the symbols $$\left(\frac{p}{q}\right)_2$$ and $$\left(\frac{p}{q}\right)_4$$ by the congruences
 * $$p^{\tfrac{1}{2}\{\mathrm{N}(q)-1\}}\equiv\left(\frac{p}{q}\right)_2, p^{\tfrac{1}{4}\{\mathrm{N}(q)-1\}}\equiv\left(\frac{p}{q}\right)_4,$$

it being understood that the symbols stand for absolutely least residues. It follows that $$\left(\frac{p}{q}\right)_2=1\mbox{ or }-1$$ according as $$p$$ is a quadratic residue of $$q$$ or not; and that $$\left(\frac{p}{q}\right)_4=1$$ only if $$p$$ is a biquadratic residue of $$q$$. If p, q are primary primes, we have two laws of reciprocity, expressed by the equations
 * $$\left(\frac{p}{q}\right)_2=\left(\frac{q}{p}\right)_2, \left(\frac{p}{q}\right)_4\left(\frac{q}{p}\right)_4=(-1)^{\tfrac{1}{4}\{\mathrm{N}(p)-1\}\cdot\tfrac{1}{4}\{\mathrm{N}(q)-1\}}$$

To these must be added the supplementary formulae
 * $$\left(\frac{i}{p}\right)_2=(-1)^{\tfrac{1}{4}\{\mathrm{N}(p)-1\}},\qquad \left(\frac{1+i}{a+bi}\right)_2=(-1)^{\tfrac{1}{8}\{(a+b)^2-1\}}$$,


 * $$\left(\frac{i}{a+bi}\right)_4=i^{\tfrac{1}{2}(a-1)},\qquad \left(\frac{1+i}{a+bi}\right)_4=i^{\tfrac{1}{4}\{a+b-(1+b)^2\}}$$,

$$a+bi$$ being a primary odd prime. In words, the law of biquadratic reciprocity for two primary odd primes may be expressed by saying that the biquadratic characters of each prime with respect to the other are identical, unless $$p \equiv q \equiv 3+2i \pmod{4}$$, in which case they are opposite. The law of biquadratic reciprocity was discovered by Gauss, who does not seem, however, to have obtained a complete proof of it. The first published proof is that of Eisenstein, which is very beautiful and simple, but involves the theory of lemniscate functions. A proof on the lines indicated in Gauss’s posthumous papers has been developed by Busche; this probably admits of simplification. Other demonstrations, for instance Jacobi’s, depend on cyclotomy (see below).