= Solution
Let $\mathcal F$ be the proper nonprincipal ideals. A chain in $\mathcal F$ has a nonprincipal union: if its union were $(a)$, then $a$ would belong to one member of the chain, forcing that member to equal the union and be principal. The union is also proper. Thus <Zorn lemma> gives a maximal member whenever $\mathcal F$ is nonempty.
The family satisfies the condition from part (a). Indeed, suppose
$$
I+(a)=(b),\qquad (I:a)=(c).
$$
Write $a=bd$ and $b=i+ra$, where $i\in I$. Every $x=bs\in I$ has $as=dx\in I$, so $s\in(c)$ and $I\subseteq(bc)$. Conversely $ac\in I$ because $c\in(I:a)$, while $ic\in I$ because $i\in I$. Thus $bc=ic+rac\in I$, proving $(bc)\subseteq I$. Hence $I=(bc)$ is principal.
If a nonprincipal ideal existed, part (a) would therefore produce a nonprincipal prime ideal, contrary to the hypothesis. Every ideal is principal, so the <integral domain> is a <principal ideal domain>.
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