TITLE: A link between hooks and contents: Part II
QUESTION [8 upvotes]: This is a question in the spirit of an earlier problem. 
Let $\lambda$ be an integer partition: $\lambda=(\lambda_1\geq\lambda_2\geq\dots\geq0)$. 
Recall also the notation for the content of a cell $u=(i,j)$ in a partition: $c_u=j−i$. Further, let $h_u$ denote the hook-length of the cell $u$.
The below identity is rather cute for which I don't remember a reference or a proof, so here I ask.

Identity. Given a partition $\lambda\vdash n$, it holds that
  $$\sum_{u\in\lambda}h_u^2=n^2+\sum_{u\in\lambda}c_u^2.$$

For example, if $\lambda=(4,3,1)\vdash 8$ then the hooks are $\{6,4,3,1,4,2,1,1\}$ and the contents are $\{0,1,2,3,-1,0,1,-2\}$. Hence
\begin{align}
LHS=6^2+4^2+3^2+1^2+4^2+2^2+1^2+1^2&=84 \\
RHS=\mathbf{8^2}+0^2+1^2+2^2+3^2+(-1)^2+0^2+1^2+(-2)^2&=84.
\end{align}

REPLY [4 votes]: Similarly to the other question, one can give a direct proof, but it requires a few more identities. Given a cell $u\in \lambda$ with coordinates $(i,j)$ we can check that $h_u+c_u=\lambda_i+\lambda'_j-2i+1$ and $h_u-c_u=\lambda_i+\lambda'_j-2j+1$
so we can compute
$$\sum_{u\in \lambda} \left(h_u^2-c_u^2\right)=\sum_{(i,j)\in \lambda}\left(\lambda_i+\lambda'_j-2i+1\right)\left(\lambda_i+\lambda'_j-2j+1\right)$$
$$=\sum_{(i,j)\in \lambda}\left(\lambda_i^2+(2-2i-2j)\lambda_i+\lambda_j'^2+(2-2i-2j)\lambda_j'\right)+2\sum_{(i,j)\in \lambda}\lambda_i\lambda_j'+\sum_{(i,j)\in\lambda}(2i-1)(2j-1).$$
In order to simplify this we can compute the summands individually as follows:
$$\sum_{(i,j)\in \lambda}(\lambda_i^2-2i\lambda_i)=\sum_i (\lambda_i^2-2i\lambda_i)\sum_{j=1}^{\lambda_i}1=\sum_i (\lambda_i^3-2i\lambda_i^2)\tag{1}$$
$$\sum_{(i,j)\in \lambda}(\lambda_j'^2-2j\lambda_j')=\sum_{j}(\lambda_j'^3-2j\lambda_j'^2)\tag{1'}$$
$$\sum_{(i,j)\in \lambda}(2j-2)\lambda_i=\sum_i\lambda_i\sum_{j=1}^{\lambda_i}(2j-2)=\sum_{i}(\lambda_i^3-\lambda_i^2)\tag{2}$$
$$\sum_{(i,j)\in \lambda}(2i-2)\lambda_j'=\sum_j (\lambda_j'^3-\lambda_j'^2)\tag{2'}$$
$$\sum_{(i,j)}\lambda_i\lambda_j'=\sum_i \lambda_i\sum_{j=1}^{\lambda_i}\lambda_j'=\sum_i i\lambda_i^2+\sum_{i_1<i_2}\lambda_{i_2}\lambda_{i_2}$$
$$=\sum_{i}i\lambda_i^2+\frac{1}{2}(n^2-\sum_i \lambda_i^2) \tag{3}$$
$$\sum_{(i,j)}\lambda_i\lambda_j'=\sum_{j}j\lambda_j'^2+\frac{1}{2}(n^2-\sum_j \lambda_j'^2) \tag{3'}$$
$$\sum_{(i,j)\in \lambda}(2i-1)(2j-1)=\sum_i(2i-1)\sum_{j=1}^{\lambda_i}(2j-1)=\sum_i (2i-1)\lambda_i^2 \tag{4}$$
$$\sum_{(i,j)\in \lambda}(2i-1)(2j-1)=\sum_j (2j-1)\lambda_j'^2 \tag{4'}$$
Substituting these in our expression we get
$$\sum_{u\in \lambda} \left(h_u^2-c_u^2\right)=n^2-\sum_i(i-\frac{1}{2})\lambda_i^2-\sum_j(j-\frac{1}{2})\lambda_j'^2+\sum_{(i,j)\in \lambda}(2i-1)(2j-1)$$
$$=n^2-\frac{1}{2}\sum_{(i,j)\in \lambda}(2i-1)(2j-1)-\frac{1}{2}\sum_{(i,j)\in \lambda}(2i-1)(2j-1)+\sum_{(i,j)\in \lambda}(2i-1)(2j-1)$$
which is simply $n^2$, as desired.