The <conjugate partition> $\lambda'$ has the same hook lengths as $\lambda$ and the opposite contents. Applying part d(i) to both diagrams and adding yields
$$
\sum_i\bigl(\lambda_i^2+(\lambda'_i)^2\bigr)
=2\sum_{(i,j)\in\lambda}h_{i,j}.
$$

Now sum the first identity of part d(i) over all partitions $\mu\vdash n$. Conjugation is a bijection on those partitions, so the total of $\sum_i(\mu'_i)^2$ equals the total of $\sum_i\mu_i^2$. Dividing the summed displayed identity by two gives
$$
\sum_{\mu\vdash n}\sum_i\mu_i^2
=\sum_{\mu\vdash n}\sum_{(i,j)\in\mu}h_{i,j}(\mu).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 160 / 1 / d / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-160/1/d/ii/solution