Irreducible in $2{\bf N}$ divides product of terms doesn't imply it divides a term Lemma $3$. if $p$ is prime and $p$ divides $bc$, then $p$ divides $b$ or $p$ divides $c$.
proof: we use corollary $8$ an application of Bezout's identity, namely: if $a$ divides $bc$ and $(a,b)=1$, then a divides $c$.
Let $2 \mathbb{N}$ denote the even integers > 0. Say that a number $a$ in  $2\mathbb{N}$ is irreducible if there are no numbers $b, c$ in $2 \mathbb{N}$ so that $a = bc$
question show that the anologue of lemma 3 fails in $2 \mathbb{N}$
this was 4 parts question. other 3 parts i understood and was able to finish them. but last part seem bit hard for me.

First make sure that you understand what’s being asked. The analogue of Lemma $3$ is:

if $a\mid bc$, and $a$ is irreducible, then $a\mid b$ or $a\mid b$.

You’re asked to show that this is false in $2\Bbb N$. To show this, you must find $a,b,c\in 2\Bbb N$ such that 


*

*$a$ is irreducible;  

*$a\mid bc$;  

*$a\nmid b$; and  

*$a\nmid c$.


So what elements of $2\Bbb N$ are irreducible? It might be easier to see what elements of $2\Bbb N$ are not irreducible. Suppose that $a\in 2\Bbb N$ is not irreducible; then $a=bc$ for some $b,c\in 2\Bbb N$. Since $b$ and $c$ are in $2\Bbb N$, there are $m,n\in\Bbb N$ such that $b=2m$ and $c=2n$, so $a=(2m)(2n)=4mn$. What members of $2\Bbb N$ are not multiples of $4$?
Now, under what circumstances can we have $a\nmid b$? If $a=2k$, and $b=2m$, then $a\mid b$ if and only if $k\mid m$, so we need to arrange matters so that $k\nmid m$. Similarly, if $c=2n$, we need to arrange matters so that $k\nmid n$. But on the other hand we do want to have $a\mid ab$, i.e., $2k\mid 4mn$; that’ll certainly be the case if $k\mid mn$. Can you find a way to choose $k,m$, and $n$ so that $k\mid mn$, $k\nmid m$, and $k\nmid n$?
