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Consider the set $\mathcal H$ of Hilbert numbers (numbers of the form $4n + 1$, for $n \ge 0$). Define a Hilbert prime as any number $h$ in the Hilbert set satisfying $h \neq 1$ and if $h \mid ab$ where $a \in \mathcal H$, $b\in \mathcal H$ then $h \mid a$ or $h \mid b$. Define a member of the Hilbert set $q$ as Hilbert irreducible as if and only if $q \neq 1$ and $q$ cannot be expressed as the product of two smaller Hilbert numbers.

I am trying to determine if Hilbert prime implies Hilbert irreducible, however I am not particularly strong at number theory. I have been unsuccessful in finding a counterexample, and I am starting to believe the implication holds. The textbook I pulled this example from for self study (*Rings, Fields, and Groups: An Introduction to Abstract Algebra* by Reg Allenby) has answers in the back of the text. However, the solution only says that Hilbert primes are also primes in $\mathbb{Z}$. Any help would be greatly appreciated, I can’t see to find any proof regarding this concept anywhere!

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Suppose $h$ is not Hilbert irreducible. Then there are $m,n\in\mathbb{N}$ such that $m,n\neq 0$ and $h=(4m+1)(4n+1)$. Then $h|(4m+1)(4n+1)$, but certainly $h\nmid (4m+1)$ and $h\nmid (4n+1)$, so $h$ is not Hilbert prime. Therefore, by the contrapositive, any Hilbert prime is Hilbert irreducible.

Since Hilbert primes are also primes in $\mathbf{Z}$, and since primes in $\mathbf{Z}$ are irreducible in $\mathbf{Z}$, then Hilbert primes must be Hilbert irreducible.

More explicitly, suppose $h$ is a Hilbert prime which is not Hilbert irreducible. Then we can write $h=pq$ where $p,q$ are smaller Hilbert numbers. But Hilbert numbers are integers, so we’ve just found a non-trivial factorization of the $\mathbf{Z}$-prime $h$, which is a contradiction.

It’s easier to prove that a Hilbert irreducible is not necessarily prime. Choose positive primes $p \equiv q \equiv 3 \pmod 4$; these are clearly in $\mathbb Z$ but not in $\mathcal H$. Then $pq$, $p^2$, $q^2$ and $p^2 q^2$ are all numbers in $\mathcal H$, and all four but the last are also irreducible.

From this we readily find that $pq \mid p^2 q^2$ but $pq \nmid p^2$, $pq \nmid q^2$ either. Example: choose 3 and 7, and thus we find 21, 9, 49, 441; the first three are irreducible but not prime.

But what you really want to prove is that all primes in $\mathcal H$ are irreducible. If $p$ is prime, then, as you well know, $p \mid ab$ means $p \mid a$ and/or $p \mid b$.

If $p$ is prime but reducible, we should be able to find numbers $m$ and $n$ in $\mathcal H$ such that $mn = p$. And $m \neq 1$, $n \neq 1$, to be perfectly clear. And it goes without saying that $-1 \not \in \mathcal H$.

So, if $mn = p$, then $p \mid m^2 n^2$ but $p \nmid m^2$ and $p \nmid n^2$ either, contradicting the assertion that $p$ is prime. As $p$ being prime but reducible leads to a contradiction, this must mean that if $p$ is prime, it is also irreducible.

For example, let’s say that 45 is prime but not irreducible (in fact it’s neither, but please play along). We have $5 \times 9 = 45$, and we readily find that $45 \mid 2025$ but $45 \nmid 25$ and $45 \nmid 81$ either, contradicting the assertion that 45 is prime.

Maybe the foregoing feels like sophistry. Remember that in $\mathcal H$, the irreducibles are congruent to 1 modulo 4. If an irreducible is not divisible by any prime congruent to 3 modulo 4 in $\mathbb Z$, it must also be prime, in both $\mathcal H$ and $\mathbb Z$.

To prove that a Hilbert prime is a prime number, show that a composite member of $H$ is not a Hilbert prime, as follows:

If $h\in H$ is composite, there exist odd $c,d,$ each greater than $1,$ with $h=cd$ and $\gcd(c,d)=1.$ Let $a=c^2$ and $b=d^2.$

Then $a,b\in H$ (because $c,d$ are odd and greater than $1$),and $h|ab=h^2$. But $h$ does not divide $a$ or $b.$ (As $a/h=c/d$ and $b/h=d/c$ are not integers, because $c, d$ are greater than $1$ and are co-prime.)

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