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In fact the conclusion holds for every amphichiral knot; the hypothesis on the Arf invariant of a knot is unnecessary. Let $Y={\Sigma }_2(K)$ be the two-fold branched cover of a knot. Amphichirality gives an orientation-reversing self-homeomorphism of $Y$, so its linking form of a branched cover satisfiesFix an odd prime $p$ and pass to the $p$-primary subgroup. The standard filtration by powers of $p$ decomposes its linking form into nonsingular symmetric forms over ${\mathbb{F}}_p$. On a graded piece of dimension $d$, an anti-isometry has a matrix $P$ satisfyingTaking determinants givesIf $p\equiv 3(\mathrm{mod}4)$, then $-1$ is not a square in ${\mathbb{F}}_p$, so every such $d$ is even. The sum of these graded dimensions is the exponentIt is therefore even, as required.

A genus-one Seifert matrix for the Stevedore knot isIts Alexander polynomial iswhose roots are $2$ and $1/2$. There are no unit roots, so Question 1(a) proves that every Levine-Tristram signature of the Stevedore knot vanishes.

On the other hand,has Smith normal form $\mathrm{diag}(1,9)$. ThereforeIf a knot is doubly slice, the linking form on the first homology of its two-fold branched cover of a knot is hyperbolic: it has two complementary metabolizers, arising from the two sides of the unknotted sphere. A cyclic group of order nine has a unique subgroup of order three, so its linking form of a branched cover cannot have two complementary metabolizers. The Stevedore knot is consequently not doubly slice, despite its identically vanishing signature function.