= Solution
<ideals> are coprime when $A+B=\mathcal O_K$, equivalently $1=a+b$ for some $a\in A$, $b\in B$. Always $AB\subseteq A\cap B$. If $x\in A\cap B$, then $x=xa+xb\in AB$, giving $\boxed{AB=A\cap B}$ in the coprime case. For nonzero integral <ideals>, the <Chinese remainder theorem> gives $\mathcal O_K/(AB)\cong\mathcal O_K/A\times\mathcal O_K/B$, and taking finite cardinalities yields $\boxed{N(AB)=N(A)N(B)}$.
For a nonzero <prime ideal>, the chain of finite-index inclusions gives
$$
N(P^e)=[\mathcal O_K:P]\prod_{i=1}^{e-1}[P^i:P^{i+1}]=N(P)^e
$$
by the given hypothesis. Distinct nonzero <prime ideals> are maximal and coprime, as are their positive powers. <Unique factorization of ideals in a number field> therefore gives $N(A)=\prod_PN(P)^{v_P(A)}$. Exponents add under <ideal> multiplication, proving multiplicativity for arbitrary nonzero integral <ideals>. This is the standard domain of the finite <ideal norm> used here.
For the <Dedekind factorization theorem>, choose an integral primitive element $\theta$ of $K$ with monic <minimal polynomial of an algebraic element> $f\in\mathbb Z[X]$, and assume the rational prime $p$ does not divide the index $[\mathcal O_K:\mathbb Z[\theta]]$. Factor
$$
\bar f=\prod_{i=1}^g\bar f_i^{e_i}
$$
over $\mathbb F_p$, with distinct monic irreducible factors. For <integer> <polynomial> lifts $f_i$, the distinct <prime ideals> above $p$ are $P_i=(p,f_i(\theta))$, and
$$
\boxed{(p)=\prod_iP_i^{e_i},\quad\mathcal O_K/P_i\cong\mathbb F_p[X]/(\bar f_i),\quad N(P_i)=p^{\deg\bar f_i}.}
$$
Thus the <residue degree> is $\deg\bar f_i$, the <ramification index> is $e_i$, and $\sum_i e_i\deg\bar f_i=[K:\mathbb Q]$. The index restriction is essential to this direct <factorization> formula.
For $K=\mathbb Q(\sqrt{-2})$, the <ring of integers> is $\mathbb Z[\sqrt{-2}]$, so the index is one and the theorem applies to every prime. Modulo two, $X^2+2=X^2$, giving one <prime ideal> of <ideal norm> two. For an odd rational prime, a <prime ideal> has <ideal norm> $p$ if the <polynomial> splits, and <ideal norm> $p^2$ if it is irreducible. In the former case the supplied condition forces $p\equiv1$ or $3\pmod8$; in the latter case every odd square is one modulo eight. Consequently
$$
\boxed{N(P)=2\quad\text{or}\quad N(P)\equiv1,3\pmod8.}
$$
The odd part of any <ideal norm> is thus one or three modulo eight, because $3^2\equiv1$. Multiplication by a power of two leaves possible residues $1,3$ for exponent zero, $2,6$ for exponent one, $4$ for exponent two, and $0$ for exponent at least three. Hence $\boxed{N(I)\not\equiv5,7\pmod8}$ for every nonzero <integral ideal>.
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