prime and irreducible elements $\equiv 1$ modulo $4$ Consider the set $L= n \in \mathbb{N}$, such that $n \equiv 1$ modulo $4$.
1) What is the set of prime or irreducible elements in $L$?
2) Is prime and irreducible the same (in $L$)?
3) Is there a unique factorization of irreducible elements $\forall n \in L$?
I know that a prime number in $\mathbb{N}$ has to be prime in $L$, too. But are there any more?
I think you can find irreducible elements which are not prime, but didn't manage to do so, yet.
Could you please help me with this problem? I'm stuck..
Are Hilbert primes also Hilbert irreducible ? Furthermore, are Hilbert primes also primes in $\mathbb{ Z}$?
 A: This is one of those questions where the OEIS can really help you out; it certainly helped me write this answer. The OEIS is like Google, but for sequences of integers.
To make sure that we're on the same page: a number is irreducible if it's divisible only by units (just $1$ in this case) and associates (trivially itself in this case). But a number $p$ is prime if whenever $p \mid ab$, either $p \mid a$ or $p \mid b$. If $p \mid ab$ but $p \nmid a$ and $p \nmid b$, then $p$ is not prime.
Then the irreducible numbers in your $L$ (I prefer $\mathcal S$) are $$5, 9, 13, 17, 21, 29, 33, 37, 41, 49, 53, 57, 61, 69, 73, 77, 89, 93, 97, \ldots$$ (look this up in the OEIS for a longer listing). As you can see, this list contains some numbers that are composite in $\mathbb Z$, like $9$ and $21$.
Now notice that $9 \times 49 = 21^2 = 441$. Obviously $9 \mid 441$. However, $9 \nmid 21$, nor $21$ either (I'm writing that twice on purpose). Therefore $9$ is irreducible but not prime. This shows that factorization in $\mathcal S$ is not always unique. We can also conclude that all primes are irreducible, but not all irreducibles are prime.
P.S. One more example that factorization in $\mathcal S$ is not always unique: $693 = 9 \times 77 = 21 \times 33$. Together with $441$, that should be enough to find more numbers like that in the OEIS.
