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 ∉ {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 ≡ 3 (mod 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 s⁴n for odd s ≥ 3 and 3⁶n; hence so is every n = j⁴m with j odd and m > 1 composed of primes ≡ 1 (mod 4), and the question reduces to n = d²m with d > 1 a squarefree product of primes ≡ 3 (mod 4); (ii) 5q² is the determinant of such a knot for every prime q ≡ 3 (mod 4), which extends Stoimenow's remark on q ≤ 11; (iii) so is every number in an explicit range of the form d²(A² + B²), by three-terminal graphs; (iv) so is every n ≤ 10⁶ 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. Unrefereed preprint released for independent mathematical scrutiny. Publication on Zenodo does not constitute peer review. AI-assisted tools supported research, computation, proof development, and manuscript preparation. The author remains responsible for all claims and the final text. Corpus identifier: AMR-103-0230 (Ohtsuki (ed.), Problems on invariants of knots and 3-manifolds, Geom. Topol. Monogr. 4 (2002), Problem 12.25, A. Stoimenow).
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