Math Help - Further Factorable Primes

1. Further Factorable Primes

I found something interesting, but it is possible it was discovered by somebody else and I am not try to plagerize (however the word is spelled). Given a prime number it is not possible to factor non-trivially, for this is the definition of a prime number. Let us make things more challenging:
A "pure factorization" is one in which a positive integer was factored into the form (a+bi)(c+di) where abcd=/0. Thus non of the integers a,b,c and d can be zero. Now this is the interesting result that those and only those prime numbers of the form 4k+1 can be "purely" factored. I am too tired to post my proof it happens to be an elementary proof, based on the fact that if p=a^2+b^2 then it can be expressed as p=(a+bi)(a-bi) and those prime numbers of the form 4k+1 can be expressed as a sum of two squares uniquely by one of Fermat's theorems.

2. Originally Posted by ThePerfectHacker
I found something interesting, but it is possible it was discovered by somebody else and I am not try to plagerize (however the word is spelled). Given a prime number it is not possible to factor non-trivially, for this is the definition of a prime number. Let us make things more challenging:
A "pure factorization" is one in which a positive integer was factored into the form (a+bi)(c+di) where abcd=/0. Thus non of the integers a,b,c and d can be zero. Now this is the interesting result that those and only those prime numbers of the form 4k+1 can be "purely" factored. I am too tired to post my proof it happens to be an elementary proof, based on the fact that if p=a^2+b^2 then it can be expressed as p=(a+bi)(a-bi) and those prime numbers of the form 4k+1 can be expressed as a sum of two squares uniquely by one of Fermat's theorems.
This says that an ordinary prime p is a Gaussian Prime iff p = 3 mod 4, a well-known fact. Good for you that you discovered this independently.