Which primes in Z are also prime in Z[i]?

Question

Which integer primes are also primes in the Gaussian Integers?

I'll be quite impressed if somebody who hasn't seen this question before works out not just the correct answer, but a solid proof for it as well.

I'm still holding out on Best Answer until somebody exhibits a proof for the 1 mod (4) part.

But yes -- the answers are correct. Primes congruent to 1 or 2 (mod 4) are composite in Z[i], while those congruent to 3 mod(4) are prime.

For that matter, I don't know we've seen a clean proof yet that primes congruent to 3 (mod 4) are prime in Z[i]. The proof I favor goes like this:

1. If p has a divisor d, then N(d)|N(p).
2. N(p) = p^2, so if p has non-trivial divisors they have norm p.
3. But no Gaussian integer can have a norm congruent to 3 mod(4), since a norm is just a sum of integer squares, and all integers squares are congruent either to 0 or 1 (mod 4).

Answer 1

Any prime p = 1 (mod 4) can be written as a sum of squared integers (since -1 is a quadratical residue modulo p iff p = 1 (mod 4)) ==> we can write p = a^2 + b^2, with a,b integers, and thus p = (a + bi)(a - bi) ==> p is NOT a prime in Z[i] since it is not irreducible (Z[i] is an euclidean domain, of course).
On the other side, primes = 3 (mod 4) can NOT be written as a sum of two integer squares and are thus prime in Z[i], since:
p = (a+bi)(c+di) ==> N(p) = N(a+bi)N(c+di) ==> p^2 = (a^2+b^2)(c^2+d^2) <==> either p = a^2 + b^2 or p = c^2 + d^2 (or both)...contradiction.

Also, 2 = (1+ i)(1 - i) ==> not a prime.

Regards
Tonio

Answer 2

The integer primes congruent to 3 mod 4 are also primes in Z[i], while those congruent to 1 mod 4 are not--thanks to Fermat.

Edit: You already know the answer--why are you playing games with our hearts?

Answer 3

Hmmm... (a+bi)(c-di) = (ac+bd) + i(ad-bc)

So all primes that we can express as (ac + bd), where (ad - bc) = 0, are composite of Gaussian integers.

Certainly if a=c and b=d, we get all primes of the form a^2 + b^2, so all primes that are the sum of two squares are composite in the Gaussian integers.

I will have to think a bit more it see if these are the only ones...