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

In a Young-diagram hook, the cell $(i,j)$ has ${\lambda }_i-j$ cells to its right and ${\lambda }_j^{\prime }-i$ below it. Summing the arm lengths row by row and the leg lengths column by column givesAdding the content $c_{i,j}=j-i$ turns the summand on the right into $2j-1$. Since ${\sum }_{j=1}^m(2j-1)=m^2$, summing each row proves

The conjugate partition ${\lambda }^{\prime }$ has the same hook lengths as $\lambda$ and the opposite contents. Applying part d(i) to both diagrams and adding yields

Now sum the first identity of part d(i) over all partitions $\mu ⊢n$. Conjugation is a bijection on those partitions, so the total of ${\sum }_i({\mu }_i^{\prime }{)}^2$ equals the total of ${\sum }_i{\mu }_i^2$. Dividing the summed displayed identity by two gives

We use mathematical induction on $n$. The identity is immediate for the empty partition. Add a removable corner $(r,s)$ to a partition of $n-1$, and put $c=s-r$. Only the new hook and the hooks to its left in row $r$ or above it in column $s$ change. Each old affected hook length increases by one. If $S$ is the sum of those old hook lengths, direct substitution in the hook formula givesThere are $r+s-2$ affected old hooks, so the increase in the sum of squared hook lengths isThe increase in $n^2+\sum c_{i,j}^2$ is likewise $n^2-(n-1{)}^2+c^2=2n-1+c^2$. The induction closes and proves