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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.

Solved by gpt-5.6-sol high.