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An odd 3-strand pretzel knot $K=P(p,q,r)$ has $\Delta_K(t)=1$ if $pq+pr+qr=1$. This fact, along with a theorem of Fintushel and Stern (every odd 3-pretzel knot with trivial Alexander polynomial is not smoothly slice) generates an infinite amount of topologically slice knots that are not smoothly slice. I am trying to look at the same thing for an arbitrary $n$-stranded odd Pretzel knot. Does there exist a general formula, such that an odd Pretzel knot $K=P(p_1,p_2,…p_n)$ has $\Delta_K(t)=1$ if some relation on $p_1,…,p_n$ is satisfied?

Edit: Is there any information on what that relation is? I am curious to see if it remains homogeneous as $n$ increases.

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  • $\begingroup$ Isn't the answer tautological? Are you asking if the solution set is an affine algebraic variety? $\endgroup$
    – Ryan Budney
    Aug 8, 2022 at 18:46
  • $\begingroup$ Yes, that is a part of what I am asking! It tautologically is an affine algebraic variety, but I’m interested in what that precise variety is. To better rephrase above, I am asking if there is a sufficient condition on $p_1,…,p_n$ that ensures the knot will have a trivial Alexander polynomial. For instance, in 3 dimensions we have $p_1p_2+p_1p_3+p_2p_3=1$. I am interested to see if there is a pattern to the $n$ dimensional case. $\endgroup$
    – KnotEnthusiast
    Aug 8, 2022 at 18:54

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