Show that primes of the form are of the form .
(Note : since primes of the form are of the form , the difficulty of Lagrange's four square theorem is thus reduced to primes of the form .)
Alright, so nobody has answered this problem so I'll give my answer.
For , both are quadratic residues; therefore is a quadratic residue.
For , neither are quadratic residues; therefore is a quadratic residue.
Therefore we can find such that , i.e. . Since is a unique factorization domain, and since neither factor on the left is divisible by , we deduce that is not prime in , and admits a factorization . Taking norms we get , so .
There is an inequality about congruences that you need to know in order to proof this result. Let be an integer so that where is an odd prime. Then you can solve the congruence with the inequality such that .
Let us prove this congruence inequality. Define and consider for . There are such choices for and , therefore, for by pigeonholing modulo . Thus, . Set . Thus, we get that . It must be the case that because otherwise this would force which would be a contradiction. Thus, but they also satisfy by construction. Thus, we have found a solution for which satisfies this inequality.
If then and if then . Thus, in either case there exists such that . By above there exists a solution to with . Then, . Thus, . This means, by definition, there exists such that . However, . This forces, . If then and the proof is complete. Otherwise, . But then this forces to be even, so . This implies . In either case it is always possible to express like so.
It follows from the fact that at most two prime ideals can lie over a prime in a quadratic extension. Since are two ideals lying over , there can be no other such ideals. This implies that there can be no other element of norm : any other such element would generate another principal ideal of norm , which is impossible.
There is just a little bit more work here to be done. First we need to know that but this is easy because otherwise and so which will force because . Also, if then is a prime ideal and so . Thus, . So it is still possible to generate the same ideals with different elemetns, it is just that these elements differ by only a multiple of unit. Thus, .
I asked the question about uniqueness of expression because I seen lots of people simply say, "uniquness in factorization implies uniquness of expression" but a lot of times this kind of argument is not so easy. I have seen expression problems which do not even have a uniquness up to , they were more complicated. Usually, if there are are many units then the express is "less unique".