Knots with Signature 4 and Determinant 4k+1: An Answer to a Question of Shinohara

By a congruence of Murasugi, the determinant and the signature of a knot K satisfy det K ≡ (−1)^(σ(K)/2) (mod 4); moreover, a knot of determinant 1 has signature divisible by 8, because its symmetrized Seifert form is even and unimodular. In 1971 Shinohara realized all pairs (det K, |σ(K)|) allowed by these conditions, except those with det K ≡ 1 (mod 8) and |σ(K)| ≡ 4 (mod 8); he stated that the case det K ≢ 1 (mod 24) can be treated as well and left the case det K ≡ 1 (mod 24) open, expecting an affirmative answer. The question reappeared as Problem 12.21 of Ohtsuki's list of problems on invariants of knots and 3-manifolds: is there, for every n = 4k + 1 with k > 0, a knot with determinant n and signature 4? Stoimenow showed that every prime factor of a counterexample is ≡ 1 (mod 24) and at least 33049, and asked whether {det K : σ(K) = 4} = 5 + 4ℕ. We answer these questions affirmatively. For every n ≡ 1 (mod 4) with n ≥ 5 there is a knot of determinant n and signature 4; it can be taken of genus 2 unless n is a square all of whose prime factors are ≡ 1 (mod 4), and of genus at most 3 in that case. Consequently, an odd positive integer d and an even integer s are the determinant and the signature of some knot if and only if d ≡ (−1)^(s/2) (mod 4), where d = 1 requires s ≡ 0 (mod 8). The proof combines Seifert's realization theorem with an explicit family of genus-two Seifert matrices and a lemma based on Dirichlet's theorem on primes in arithmetic progressions and quadratic reciprocity. We also give explicit knots, among them a two-bridge knot of determinant 33049, and report exact computer checks. This is an unrefereed note.