Every element of $\mathit H$ is representable as a product of $\mathit H$-irreducible elements I have the following exercise to do.
Let $\mathit H := \{4n+1 : n \in \mathbb N_0 \} \subseteq \mathbb N$. An element $p \in H \setminus \{1 \}$ is called $H$-irreducible if the following is true:

*

*if $p = a \cdot b$  where $a,b \in H$, then either $a = 1$ or $b = 1$.

Proove that every element of $\mathit H$ is representable as a product of $\mathit H$-irreducible elements.
As I understand it, I have to prove that every $q \in H$ can be written as $q = p\cdot r$ where $p,r$ are $H$-irreducible.
Before I start my attempt, a remark. I managed to prove that for $a,b \in H$ that $a \cdot b \in H$.
Let $p = a \cdot b$ and $r = c \cdot d$ with $a,b,c,d \in H$. Then $q = p \cdot r = a \cdot b \cdot c \cdot d$.

*

*case: $a = 1, c = 1$. Then $q = b \cdot d$ and $q \in H$.

*case: $a = 1, d = 1$. Then $q = b \cdot c$ and $q \in H$.

*case: $b = 1, c = 1$. Then $q = a \cdot d$ and $q \in H$.

*case: $b = 1, d = 1$. Then $q = a \cdot c$ and $q \in H$.

Is this a valid proof?
Also, if I calculate the first few terms of $H$ then I get:
$H = \{1,5,9,13,17,21,...\}$. How exactly can I get for example $13$ as a product of two elements of $H$ ?
 A: 
Proove that every element of $\mathit H$ is representable as a product of $\mathit H$-irreducible elements.As I understand it, I have to prove that every $q \in H$ can be written as $q = p\cdot r$ where $p,r$ are $H$-irreducible.

Your understanding as stated in the second sentence is not quite correct (as also indicated in Davide's question comment). The question is asking to prove that each element of $\mathit H$ can be represented as a product of any number (i.e., $0$ or more, so not necessarily exactly $2$) of $\mathit H$-irreducible elements. This is similar to as the statement that all positive integers can be represented as a product of primes means $0$ or more primes (e.g., $1$ is the only positive integer which is a product of $0$ primes).
Consider any element $h \in \mathit H$. It can be represented using some integers $m, n \ge 0$ by
$$h = \left(\prod_{i=1}^{m}p_i\right)\left(\prod_{j=1}^{2n}q_j\right) \tag{1}\label{eq1A}$$
where $p_i$ are primes congruent to $1$ modulo $4$ and $q_j$ are primes congruent to $3$ modulo $4$. The number of primes of the form $q_j$ must be even because an odd number of them would result in $h \equiv 3 \pmod{4}$.
Since the only factors of $p_i$ (for $1 \le i \le m$) in $\mathit H$ are $1$ and $p_i$, then $p_i = a \times b$ where $a, b \in \mathit H$ means $a$ or $b$ is $1$, i.e., $p_i$ is irreducible.
With $r = q_j q_k$ for any $1 \le j, k \le 2n$, then $r = a \times b$ (with $a$ and $b$ being positive integers) means there are $2$ possibilities. First, $a$ and $b$ are $q_j$ and $q_k$, in some order. However, neither $a$ nor $b$ would then be elements of $\mathit H$. The second possibility is $a$ and $b$ are $1$ and $r$ in some order, so since $1, r \in \mathit{H}$, this shows $r$ is irreducible.
Thus, rewriting \eqref{eq1A} as
$$h = \left(\prod_{i=1}^{m}p_i\right)\left(\prod_{j=1}^{n}(q_{2j-1}q_{2j})\right) \tag{2}\label{eq2A}$$
indicates how $h$ can be written as a product of the $\mathit H$-irreducible elements $p_i$ and $q_{2j-1}q_{2j}$.
