Consider the symmetric <Laurent polynomial>
$$
f(t)=t^2-1+t^{-2}.
$$
It satisfies $f(1)=1$, so the <Alexander polynomial realization theorem> gives a <knot> $K_3$ with $\Delta_{K_3}(t)\doteq f(t)$. The polynomial is not a <unit in a ring>[unit] of $\mathbb Z[t^{\pm1}]$, whereas
$$
\det K_3=|\Delta_{K_3}(-1)|=|1-1+1|=1.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 112 / 1 / iii / Solution

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Every Laurent polynomial $f(t)\in\mathbb Z[t^{\pm1}]$ satisfying $f(t^{-1})=f(t)$ and $f(1)=1$ is the <Alexander polynomial of a knot> up to multiplication by a unit $\pm t^k$.

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 Incoming links: Alexander polynomial realization theorem