What are the uses of Gaussian integers?

Answer 1

Gaussian integers are complex numbers of the form a+bi where a and b are integers. The collection of Gaussian integers has many properties similar to those of the ordinary integers such as factorization into primes, a Euclidean algorithm, etc. They are interesting in their own right, but can also be used to prove things about ordinary integers. For example, the fact that every prime of the form 4k+1 is the sum of two squares is often proven by use of the Gaussian integers.

Answer 2

Strange Integers
Gaussian Integers
Z[3] is not the only algebraic construct for which Euclid's Algorithm and the Fundamental Theorem of Arithmetic (uniqueness of the prime factorization) make sense. The very first result in this spirit was obtained by Gauss who considered the ring Z[i] = {a+bi: a,bZ, i = (-1)}. This is the set of complex numbers with integer coefficients. In his honor the numbers are called the Gaussian Integers.

Extension of Euclid's Algorithm to Z[3] was made possible with the introduction of the function N and the establishment of an inequality that ensured convergence of Euclid's Algorithm.

For A = a+ bi from Z[i], define A' = (a+bi)' = a-bi and N(A) = AA' = a2 + b2. For Z[i], N(A) is never negative and is only 0 for A = 0, i.e. when a = b = 0. For thus defined N, the following holds

(1) For any given pair A and B from Z[i], B0, there exist in Z[i] numbers Q and R such that.

A = BQ + R, and N(R) < N(B)

The proof follows that for Z[3] but is a little simpler. The inequality (4) is replaced with a more straightforward

(2) |(x-r)2 + (y-s)2| (½)2 + (½)2 = ½ < 1

Thus it follows that Euclid's Algorithm is applicable to the Gaussian integers. And, as a consequence, the prime factorization is unique in Z[i]. Let's briefly consider a few examples.

Z[i] has just four unities: ±1 and ±i.

2 is composite (not prime) in Z[i].
Indeed, 2 = (1 + i)(1 - i), and N(1 + i) = N(1 - i) = 2, so none of the factors is a unity.

3 is prime in Z[i].
Assuming 3 = AB we first get N(3) = 9 = N(AB) = N(A)N(B). If neither A nor B is unity, N(A) = N(B) = 3. Let's show that the equation N(A) = 3 has no integer solutions. Unless a = b = 0 (mod 9), (a2 + b2) (mod 9) is one of the sequence 1, 4, 7, (1+1)=2, (1+4)=5, (1+7)=8, (4+4)=1, (4+7)=2, (7+7)=5 (mod 9). But 3|a and 3|b implies 9|(a2 + b2) and ,hence, 9|3. A contradiction.

5 is composite (not prime) in Z[i].
Indeed, 5 = (1 + 2i)(1 - 2i), and N(1 + 2i) = N(1 - 2i) = 5, so none of the factors is a unity.

Solve x2 + y2 = z2 in integers.
To solve the Pythagorean equation, first assume that Pythagorean triples x,y,z exist. We may then restrict our search to triples with gcd(x, y, z) = 1. This is the same as gcd(x, y) = 1. Factor the left hand side into (x + yi)(x - yi). I leave this without a proof that gcd(x,y) = 1 implies gcd(x + yi, x - yi) = 1. It would be nice to conclude from (x + yi)(x - yi) = z2 that for some Gaussian integers A and B, x + yi = A2 and x - yi = B2. But this may not be the case. In addition, we must consider, say, x + yi = iA2 and x - yi = -iB2 as valid possibilities.
In the first case, let A = u + vi, then x + yi = (u2 - v2) + 2uvi, and x = u2 - v2, y = 2uv. Substitution yields z = u2 + v2. In the second case, the roles of x and y are reversed.


Remarks and Examples

Divisor of zero
U is a divisor of zero iff there exists V0 such that UV = 0. We found that neither Z nor Z[m], where m is not a complete square of an integer, have no divisors of zero. But some algebraic structures do. Zm is a usual notation for the set of residues modulo m. Then, for example, in Z6, 2*3 = 0. So, in Z6, both 2 and 3 are divisors of zero.

Function N
In the theory of numbers function N is called a norm. It is hard to overestimate the importance of this function. Function N has been used in virtually every single proof so far. Why was it needed? For m>0, Z[m] is dense on the real line. The proof uses the Pigeonhole principle in a manner similar to that of a lemma we proved when shredding the torus. This is the reason that we had to use the norm in Euclid's algorithm. Had we used the magnitude of a number, the algorithm would have never stopped.

For m = -1, N(AB) = N(A)N(B), when written explicitly, takes the form

(ac - bd)2 + (ad + bc)2 = (a2 + b2)(c2 + d2),
A = a+ib, and B = c+id. This is the two-squares identity. It asserts an interesting fact: a product of the sum of two squares is itself the sum of two squares. If A = B, then N(A2) = N(A)2. On the left we have the sum of two squares. On the right - a square of an integer. This is the way to generate Pythagorean triples.

For m positive, N(AB) = N(A)N(B) also has a nice form:

(ac + bd)2 - m(ad + bc)2 = (a2 - mb2)(c2 - md2)
This is similar to the two-squares identity. The product of two numbers in the form "a square minus a square time an integer" is an integer in the same form.

Unity
A unity (or just a unit) is a number that divides all other numbers. This is equivalent to saying that it divides 1. If U is a unity, there exists V such that UV = 1. So that U has an inverse. The opposite is also true. Every invertible element is a unity. In a field, all elements, besides 0, are invertible. In a field division becomes ubiquitous and, therefore, not interesting.

In Z6, 2 and 3 are divisors of zero and, therefore, are not invertible. (For if UV = 1 but UW = 0 with V0, then UVW = 0*V = , or 1*W = W = 0. Contradiction.) On the other hand, 5*5 = 1 (mod 6). Therefore, in Z6, 5 is a unity. The remaining non-zero element 4 is a divisor of zero. For 4*3 = 12 = 0 (mod 6). Also, 4*4 = 4 (mod 6). Such elements are called idempotents. By induction, in Z6, 4n = 4. This is true for any positive n.

We may try to consider extensions of finite rings, like Zm. Z7[2] = Z7 because, e.g., 32 = 2 (mod 7). On the other hand, if, as usual, i is taken to be a square root of -1, then Z5[i] = Z5 because 22 = 4 = -1 (mod 5).

Answer 3

1

Answer 4

The set of all Gaussian integers can be made into an Euclidean Domain by specifying the appropriate function d(z). Then by studying the algebraic nature of this domain we can solve number theory problems. you can see one such application in the text book "Topics in Algebra" by I.N. Herstein in which Fermat's theorem has been proved via so called algebraic component Gaussian integer.

Answer 5

http://en.wikipedia.org/wiki/Gaussian_integers

Answer 6

http://mathworld.wolfram.com/GaussianInteger.html

http://web.usna.navy.mil/~wdj/book/node56.html