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Which primes in Z are also prime in Z[i]?
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).
3 Answers
- BertrandoLv 41 decade agoFavorite Answer
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
- Anonymous1 decade ago
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?
- Phineas BoggLv 61 decade ago
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...
