= Solution
An element $a$ is a <prime element> when it is nonzero and not a <unit>, and $a\mid bc$ implies $a\mid b$ or $a\mid c$. If $a=bc$ and $a$ is a <prime element>, assume for example $b=ad$. Cancellation in $a=adc$ gives $dc=1$, so $c$ is a <unit>. This proves every <prime element> is an <irreducible element>.
In a <Euclidean domain>, each nonzero <ideal> has a generator of least Euclidean value: division of another member by that generator leaves a remainder in the <ideal> of smaller value, so the remainder must vanish. Let $a$ be an <irreducible element> and suppose $a\mid bc$ but $a\nmid b$. Write $(a,b)=(d)$. Since $d\mid a$, irreducibility says $d$ is a <unit> or is associate to $a$. The second case would give $a\mid b$, so $d$ is a <unit>. Hence $1=ra+sb$ for some $r,s$. Multiplying by $c$ shows $a\mid c$. Thus \b[irreducibles in a <Euclidean domain> are prime].
For $R=\mathbb Z[i\sqrt7]$, use the multiplicative integer <field norm> $N(m+ni\sqrt7)=m^2+7n^2$. <Units> have <field norm> one and are exactly $\pm1$. If $2=uv$ were a product of nonunits, their norms would be integers greater than one multiplying to four, hence both would have <field norm> two. But $m^2+7n^2=2$ has no integer solution: $n\ne0$ makes the left side at least seven, and $n=0$ would give $m^2=2$. Therefore \b[2 is an <irreducible element>].
On the other hand,
$$
(1+i\sqrt7)(1-i\sqrt7)=8
$$
is divisible by 2, while neither factor is, because <divisibility> by 2 requires both integer <coefficients> to be even. Thus \b[2 is not a <prime element>] in this <ring>. It is also \b[not a <prime element> in $\mathbb Z[i]$]: $2\mid(1+i)^2=2i$, but $2\nmid1+i$. In fact $2=(1+i)(1-i)$ is already a nontrivial <factorization> there.
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