= Solution
Because $\lambda$ is self-conjugate, every off-diagonal hook of even length is paired with its transpose, while every diagonal hook has odd length. Suppose even hooks exist and let $L$ be their maximum length. Apply the <Murnaghan–Nakayama rule> to cycle types beginning with $L$ and complete the remaining cycle type with the principal hooks of the residual diagram. The <principal-hook character value of a symmetric group> makes each surviving residual character equal to $1$ or $-1$.
The assumed vanishing forces cancellation among the removable $L$-hooks. The standard maximal-hook comparison shows that the only possible cancellation is one transposed pair: the hooks must be $H_{1,j}$ and $H_{j,1}$ for a single $j>1$. Any further hook of length $L$, or a maximal hook with both indices greater than one, can be isolated by the residual principal-hook cycle type and would give a nonzero value. Thus either there are no even hooks or the maximum even length occurs exactly at that pair.
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