Eisenstein norm ($N(a+b\omega )=a^2-ab+b^2$)

For an Eisenstein integer $z=a+b\omega$, with $\omega =(-1+i\sqrt{3})/2$, the norm is $N(z)=z\overline{z}=a^2-ab+b^2$. It is a positive integer for $z\ne 0$, since $N(z)=(a-b/2{)}^2+3b^2/4$, and is multiplicative by the multiplicativity of the complex modulus. An element is a unit exactly when its norm is one: one direction follows from multiplicativity and integrality, and the other from the inverse $z^{-1}=\overline{z}$.

Let $p\ge 3$ be a prime number dividing $n^2+3$, and set $\omega =(-1+i\sqrt{3})/2$. The case $p=3$ is $N(1-\omega )=3$. For $p>3$, useThe Eisenstein integer $p$ divides this product, but divides neither factor: divisibility by the rational integer $p$ in the basis $1,\omega$ requires both integer coefficients to be divisible by $p$, whereas the second coefficients are $2$ and $-2$. Thus $p$ is not a prime element of $\mathbb{Z}[\omega ]$.

The Euclidean norm on the Eisenstein integers makes this ring a principal ideal domain. Since irreducible elements of a principal ideal domain are prime, $p$ admits a factorization $p=uv$ into two nonunits. The positive integral Eisenstein norm is multiplicative, and its value is one exactly on units: when $N(u)=1$, its complex conjugate is an inverse. Thus $N(u),N(v)>1$ and $N(u)N(v)=p^2$, forcing both norms to be $p$. Writing $u=a+b\omega$ proves the displayed representation.