Determinants of Prime Alternating Achiral Knots: A Partial Answer to a Problem of Stoimenow

Stoimenow asked (Problem 12.25 of the problem list edited by Ohtsuki, 2002; Conjecture 4.2 of his paper in Comm. Anal. Geom. 13 (2005)) whether every odd sum of two squares \(n\notin\{1,9,49\}\) is the determinant of a prime alternating achiral knot. He proved this for squares and, with rational knots, for all \(n>1\) without prime factors \(q\equiv3\pmod 4\), and reported a check up to \(2000\). We give a partial answer; the question remains open. Gluing a plane network in a disk to its dual network, turned over by a half-turn of the sphere, is a form of a classical construction of self-dual maps; when the resulting graph is nonseparable and has an odd number of spanning trees, it is the checkerboard graph of a prime alternating achiral knot. For two and three boundary points this gives Stoimenow's determinant formulas back; for four it gives a new one. With it we show: (i) if \(n\) is the determinant of such a knot, so are \(s^4n\) for odd \(s\ge3\) and \(3^6n\); hence so is every \(n=j^4m\) with \(j\) odd and \(m>1\) composed of primes \(\equiv1\pmod4\), and the question reduces to \(n=d^2m\) with \(d>1\) a squarefree product of primes \(\equiv3\pmod4\); (ii) \(5q^2\) is the determinant of such a knot for every prime \(q\equiv3\pmod4\), which extends Stoimenow's remark on \(q\le11\); (iii) so is every number in an explicit range of the form \(d^2(A^2+B^2)\), by three-terminal graphs; (iv) so is every \(n\le10^6\) not covered by Stoimenow's results (\(16{,}703\) numbers), by explicit certificates checked by computer. For larger \(n\) the general case is open, and we explain why these families do not settle it. This is an unrefereed note.