A standard argument of knot theory reveals that the Alexander polynomial of the sum of two knots $K= K_1 \# K_2$ is equal to the product of the Alexander polynomial of the two summands $K_1$ and $K_2$.
Suppose that the Alexander polynomial $\Delta_K(t)$ of a knot $K \subset S^3$ can be factorized as $\Delta_K(t)= \Delta_{K_1}(t) \cdot \Delta_{K_2}(t) $, where $K_1$ and $K_2$ are two knots in $S^3$ with non trivial Alexander polynomial (not a unit). Is it true that $K$ is a non prime knot?
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Another approach is to note that the Alexander polynomial is unchanged by "doubled delta moves." One way to visualize a doubled delta move is to tie a Borromean rings into three pairs of antiparallel strands. Given any knot, it is reasonable to expect that you can do doubled delta moves to make it prime. So any realizable Alexander Polynomial can be realized by a prime knot. In fact, it seems to be proven in the literature that every Alexander polynomial can be realized by a hyperbolic knot, which is even stronger than being prime.
No: let $K$ be $8_{15}$, let $K_1$ be the trefoil, and let $K_2$ be $7_2$. The Alexander polynomial for $K$ is $(t-1+t^{-1})(3t-5+3t^{-1})$, yet $K$ is prime.