Assume that you are familiar with the imaginary number *i*, which
is sqrt(-1).

Gaussian integers are numbers of the form *x*+*yi*, where
*x* and *y* are integers. Formally, the set

*R* = { *x*+*yi* : *x* and *y* integers }

defines Gaussian integers. (Nothing is special about choosing the letter
*R*.)

Examples are -3, 4, 0, *i*, -*i*, 3+4*i*, -5+2*i*,
2-2*i*.

The units of *R* are 1, -1, *i*, and -*i*. It is easy
to verify that they are indeed units. To see that they are the only ones,
suppose *u*+*vi* is a unit, so (*u*+*vi*)(*s*+*ti*)
= 1 for some integers *s* and *t*, which means *us* - *vt*
= 1 and *ut* + *vs* = 0, and also *u*(*s*^2 + *t*^2)
= *s*. If *s*=0, then *t*!=0, *u*=0, and *v*=1
or -1. If *s*!=0, then |*s*|=1 and *t*=0, otherwise |*u*(*s*^2
+ *t*^2)| = |*s*| would not hold; so *v*=0 and *u*=1
or -1.

In *R*, 2 = (1+*i*)(1-*i*), so 2 is no longer a prime
in *R*. This is not the only case: 5 = (1+2*i*)(1-2*i*)
too. Fortunately, the new rules for primes and composites are still simple:

- Composties in the integers remain as composites in
*R*. - A prime
*p*in the integers with*p*mod 4 = 3 remains as a prime in*R*. - A prime
*p*in the integers with*p*=2 or*p*mod 4 = 1 becomes a composite in*R*. However, you can write*p*= (*a*+*bi*)(*a*-*bi*), and*a*+*bi*and*a*-*bi*are primes in*R*. - All primes in
*R*can be obtained from the above rules.

So for example, 3, 7, 11 are primes in *R*, while 2 and 5 are not,
as discussed; but then their factors, 1+*i*, 1-*i*, 1+2*i*,
and 1-2*i* are primes in *R*, so the factorizations of 2 and
5 shown is finished.

The Gaussian integers *R* obeys a unique-factorization theorem
analogous to that of the integers.

**Theorem**: In the Gaussian integers *R*, each non-zero, non-unit
member can be factorized into a product of primes; furthermore, the factorization
is unique up to ordering and associates.

For example, 3 is itself a prime, so 3 = 3 is a prime factorization.
As another example, (1+2*i*)(1-2*i*) is a prime factorization
of 5. Now, we seems to have another factorization: 5 = (2+*i*)(2-*i*),
but notice that 1+2*i* and 2-*i* are associates, and so are 1-2*i*
and 2+*i*: (1+2*i*)*(-*i*) = 2-*i* for example. Thus
we are having the same factorization essentially. This is in the same principle
as recognizing that 3*3 and (-3)*(-3) are essentially the same factorization
of 9.

As a final example, 10 = 2*5 = (1+*i*)(1-*i*)(1+2*i*)(1-2*i*).