Let $\lambda$ be an integer partition: $\lambda=(\lambda\_1\geq\lambda\_2\geq\dots\geq0)$. Denote its [conjugate partition](http://mathworld.wolfram.com/ConjugatePartition.html) by $\lambda'$. For example, if $\lambda=(4,3,1)$ then $\lambda'=(3,2,2,1)$.


Recall also the notation for the content of a cell $u=(i,j)$ in a partition is $c\_u=j−i$. 


I have made the following observation which seems interesting enough to ask here.



> 
> **QUESTION.** Is this true? It might even be known. Is it? Any reference?
>  $$\sum\_{i\geq1}\lambda\_i^2=\sum\_{u\in\lambda}(h\_u+c\_u).$$
> 
> 
> 


**REMARK 1.** The above identity implies 
$$\sum\_{i\geq1}(\lambda\_i^2+(\lambda\_i')^2)=2\sum\_{u\in\lambda}h\_u.$$


**REMARK 2.** It also implies that 
$$\sum\_{\lambda\vdash n}\sum\_{i\geq1}\lambda\_i^2=\sum\_{\lambda\vdash n}\sum\_{u\in\lambda}h\_u.$$



A link between hooks, contents and parts of a partition
Another approach is to notice that $$\sum\_{u\in\lambda}h\_u=\sum\_{u}d\_u$$
where $d\_u=i+j-1$ for $u=(i,j)$. 


**Proof:** The easiest way to see this is that both sides count the number of pairs $\{(i\_1,j\_1),(i\_2,j\_2)\}\in\lambda$ such that either $i\_1=i\_2$ and $j\_1\le j\_2$ or $j\_1=j\_2$ and $i\_1\le i\_2$. 




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Using this we have
$$\sum\_{u\in \lambda}(h\_u+c\_u)=\sum\_{u\in \lambda} (c\_u+d\_u)=\sum\_{(i,j)\in \lambda} (2j-1)=\sum\_{i\geq 1}\sum\_{j=1}^{\lambda\_i}(2j-1)=\sum\_{i\geq1} \lambda\_i^2$$
which gives us the equality we wanted.


