Combinatorial proof for formulas involving partitioning a multiset into 2 equal sums

Question

$A$ is a non-empty multiset that contains $m$ integers $a_{1},a_{2},...,a_{m}$. $P(X,n)$ is the number of ways to partition $X\cup [n]$ into 2 equal sums (where $X$ and $[n]$ are multisets and $n$ is an integer), when there is no $n$, we denote $P(X)$. Here is an example: $A=[1,2,3]$, $P(A)=1$ and
for $n=1$, $A\cup[n]=[1,1,2,3]$ then $P(A,n)=0$;
for $n=2$, $A\cup[n]=[1,2,2,3]$ then $P(A,n)=1$
These are the formulas I found: $$ 2^{m-1}-P(A)=\sum_{n=1}^\infty P(A,n) $$
$$ 2^{m-1}\sum_{i=1}^m a_{i}^{2}=\sum_{n=1}^\infty n^{2}P(A,n) $$
$$ 3(2^{m-1})\left(\sum_{i=1}^m a_{i}^{2}\right)^{2}\ -2^{m}\sum_{i=1}^m a_{i}^4=\sum_{n=1}^\infty n^{4}P(A,n) $$ A lot more formulas can be derived using the same technique. However, they were derived via calculus, not combinatoric. I tried to sketch out, for example: $\sum_{n=1}^\infty n^{2}P(A,n)=\frac{1}{2}\sum(\pm a_{1}\pm a_{2}\pm a_{3}\pm...\pm a_{m})^{2}$ (due to the fact $P(A,n)=0$ if $n\ne \pm a_{1}\pm a_{2}\pm...\pm a_{m}$) then I am not sure what to do next.
Could you help proving these formulas using combinatoric? Thanks in advance.

Answer 1

1. For the first one first notice that $P(A,0)=P(A)$ and notice that there are $2^{m-1}-1+1=2^{m-1}$ ways to partitiona set of $m$ elements into one or two blocks. Let $\pi _1,\pi _2$ be the blocks(we can have $\pi _2=\emptyset$) and assume $\sum _{x\in \pi _1}x\geq \sum _{x\in \pi _2}x,$ then let $n=\sum _{x\in \pi _1}x- \sum _{x\in \pi _2}x,$ and so this will be counted in $P(A,n).$ That is a bijection (why? to return just check where is the $n$.)
2. For the second one, you are in the right track, just notice that $$(b_1+\cdots +b_m)^2=\sum _{i = 1}^mb_i^2+2\sum _{i<j}b_ib_j,$$ so consider $v\in \{+,-\}^m$ then your expression in the RHS is $$\frac{1}{2}\sum _{v\in \{+,-\}^m}\left (\sum _{i = 1}^ma_i^2+2\sum _{i<j}v_iv_j\cdot a_ia_j\right ),$$ but there are $2^m$ such $v,$ so you get $\frac{2^m}{2}\sum _{i=1}^ma_i^2$ and notice that for a fix $i,j$ you have $v_iv_j=\pm$ and notice that out of the $4$ options, you have $2$ options giving you $+$ and $2$ options giving you $-$ and so they will cancel out, so $\sum _{v\in \{+,-\}^m}\sum _{i<j}v_iv_j\cdot a_ia_j=0.$
3. Seems to me that the last one can be done using the same as you propose and I have pointed out for the second one.