= Solution
For
$$
R=\{(a,b)\in\mathbb Z^2:a\equiv b\pmod2\},
$$
the parity condition is preserved by componentwise addition, negation, and multiplication, and $(1,1)$ is the identity. Thus $R$ is a <ring>.
It is not an <integral> domain, since $(2,0)$ and $(0,2)$ are nonzero elements of $R$ whose product is zero. It is also not a product of two nontrivial <rings>. Indeed, an <idempotent> in $\mathbb Z^2$ has each coordinate in $\{0,1\}$, and the parity condition leaves only $(0,0)$ and $(1,1)$. The proved equivalence then excludes a nontrivial product decomposition.
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