= Solution
Since $\omega^2=-1-\omega$, products of integer linear combinations of $1,\omega$ remain of the same form. Thus the <Eisenstein integers> $E=\mathbb Z[\omega]$ form a subring of $\mathbb C$, and hence an <integral domain>. Their <Eisenstein norm> is
$$
N(a+b\omega)=(a+b\omega)(a+b\overline\omega)=a^2-ab+b^2
=\left(a-\frac b2\right)^2+\frac34b^2.
$$
It is a positive <integer> on nonzero elements and is multiplicative, because the <complex modulus> is multiplicative.
For $\alpha,\beta\in E$ with $\beta\ne0$, write $\alpha/\beta=u+v\omega$ with real $u,v$. Choose integers $m,n$ with $|u-m|,|v-n|\leq1/2$, and let $q=m+n\omega$. With $s=u-m$ and $t=v-n$,
$$
\left|\frac\alpha\beta-q\right|^2=s^2-st+t^2
\leq s^2+|st|+t^2\leq\frac34<1.
$$
Consequently $r=\alpha-q\beta$ belongs to $E$ and satisfies $N(r)\leq\tfrac34N(\beta)<N(\beta)$ when $r\ne0$. This proves \b[$E$ is a <Euclidean domain> with Euclidean function $N$], the <Euclidean norm on the Eisenstein integers>.
\Image[/past-exam-of-the-rings-and-modules-course-of-the-university-of-oxford/2017-eisenstein-division.png]
{title=Rounding the coordinates in the <Eisenstein integer> lattice gives a Euclidean remainder}
{description=After subtracting the chosen lattice point, the error lies in the shaded parallelogram. Its squared modulus is at most $3/4$, strictly smaller than one. The parallelogram is a rounding cell, not the hexagonal nearest-point cell.}
For the representation of the prime $p$, first note that \b[$3=N(1-\omega)$]. Now suppose $p>3$ and $p\mid n^2+3$. The factorization
$$
n^2+3=(n+1+2\omega)(n-1-2\omega)
$$
shows that the <Eisenstein integer> $p$ divides a product. It divides neither factor: an element $a+b\omega$ lies in $pE$ exactly when both integer coefficients are divisible by $p$, and the second coefficients here are $2$ and $-2$. Hence $p$ is not a <prime element> of $E$.
To justify the necessary factorization step, in any <principal ideal domain> an <irreducible element> $\pi$ is prime. If $\pi\nmid a$, write $(\pi,a)=(d)$. Since $d\mid\pi$, irreducibility makes $d$ a unit or an associate of $\pi$. The latter would give $\pi\mid a$, so $d$ is a unit and $u\pi+va=1$ for some $u,v$. If $\pi\mid ab$, multiplication of this identity by $b$ gives $\pi\mid b$. This proves <irreducible elements of a principal ideal domain are prime> without assuming unique factorization.
By part (a), $E$ is a <principal ideal domain>. The integer $p$ has norm $p^2>1$, so it is a nonzero nonunit. Since it is not prime, the result just proved implies that it is reducible: $p=uv$ with $u,v$ both nonunits. A nonzero element of $E$ is a <unit> exactly when its norm is one, since $N(z)=1$ gives inverse $\overline z$ and multiplicativity proves the converse. Thus
$$
1<N(u),N(v),\qquad N(u)N(v)=N(p)=p^2.
$$
The only possible positive integer factors exceeding one are $N(u)=N(v)=p$. Writing $u=a+b\omega$ gives
$$
\boxed{a^2-ab+b^2=p.}
$$
This is the <prime representation by the Eisenstein norm> argument.
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