= Solution
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> satisfies
$$
(H_1(Y),\lambda_Y)\cong(H_1(Y),-\lambda_Y).
$$
Fix 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 $P$ satisfying
$$
P^TAP=-A.
$$
Taking determinants gives
$$
(\det P)^2=(-1)^d.
$$
If $p\equiv3\pmod4$, then $-1$ is not a square in $\mathbb F_p$, so every such $d$ is even. The sum of these graded dimensions is the exponent
$$
\nu_p|H_1(Y)|=\nu_p|\Delta_K(-1)|.
$$
It is therefore even, as required.
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