Question:
It's an open problem whether or not the Jones polynomial distinguishes the unknot from all other knots. That is, the following problem is unsolved.
Does there exist a knot $K$ which is not the unknot, such that $V(K) = 1$?
I was wondering if a solution to the following related problem is known:
Does there exist a knot $K$ which is not the unknot, such that $V(K)$ is a unit in $\mathbb{Z}[t,t^{-1}]$? (Explicitly, the units are of the form $\pm t^k$ for $k \in \mathbb{Z}$.)
My attempts:
Running my eyes through some knot tables, I couldn't find any examples for knots with at most 10 crossings.
I then attempted to prove that this question is equivalent to the open problem concerning "when does $V(K) = 1$?" using the Skein relations. However, the Skein relations may convert knots to links, which is a problem because the open problem above strictly applies to knots. I also thought that I could work backwards from non-trivial links $V$-indistinguishable from the unlink (by again using the Skein relations) to exhibit a knot with unit Jones polynomial, but I couldn't figure out how to do this either.
Unnecessary background:
The reason I'm interested in this question is because I was wondering if it was possible to prove that knots decompose into connected sums of primes by using the Jones polynomial instead of the genus (but following a similar proof outline). I realised that the analogous proof fails in the crucial step where we require "$K$ is non-trivial" to imply that "$V(K)$ is non-trivial". In this context, we want to use that integral Laurent polynomials form a UFD together with $V(K+P) = V(K)V(P)$. Therefore "non-triviality" for a polynomial corresponds to whether or not it's a unit. (Of course this proof is doomed to fail because of the openness of "when is $V(K) = 1$?", but nonetheless it made me think about the general question with units.