Proposition: A polynomial has a root if and only if divides .
Corollary: A polynomial of degree over a field has at most roots.
Proposition: A finite subgroup of the multiplicative group of a field is cyclic.
Proof: Let be such a subgroup. By the fundamental, theorem of abelian groups, is the product of its Sylow -subgroups. Let be such a subgroup. If were not cyclic, then and hence would have more than elements that are solutions to the equation . But has at most roots. Since is cyclic for each dividing the order of , itself is cyclic.
Corollary: The group of units is cyclic.
Generators of this group are called primitive roots mod .
Back to the Gaussian integers
The irreducibles in are:
with
where for and .
A positive integer is a sum of two squares if and only if it factors where the and the and all the are even.
The proof follows from the question of when is for some .
Algorithm for Fermat’s theorem
Suppose . To write , find a solution to the congruence . Then use the Gaussian Euclidean algorithm to find a generator for the ideal in . This generator divides so its norm is a divisor of . If its norm were, then would be an associate of and this would mean divides , which it visibly does not. If its norm were , then the ideal would be all of and so we would have in . But in that case, multiplying by would be and since divides the left side we’d have dividing , which is not true. So therefore and so if we have .