Prove that exists a function $f$, such that $\sum_{i=0}^{n} i (c_i - 1) = 0 \Leftrightarrow \sum_{i=0}^{n} f(i) (c_i - 1) \neq 0$ I would like to prove that exists a function $f$ such that $\sum_{i=0}^{n}  i (c_i - 1) = 0 \Leftrightarrow \sum_{i=0}^{n} f(i) (c_i - 1) \neq 0$ $\forall \: n >  0$.

*

*This must hold $\forall \: c_i \geq 0$, with $c_i$ being an integer, and with $\sum_{i=0}^{n} c_i = n$

*We have at least one $c_i \neq 1$

*$f(i)$ cannot be a function of $c_i$
I thought this should work if $f(i) = i^2$, but I don't know how to prove it. Also, is there any other function $f$ for which that holds and that grows slower than $i^2$?
 A: Such a function $f$ cannot exist.
Answer for original version of the question: (new version below)
Let us consider the case that $n=1$.
For $c_1=2$ we have $\sum_{i=0}^n i(c_i-1)\neq0$ and therefore
$$
f(0) (c_0-1)+f(1)(c_1-1)=0.
$$
For $c_0=1$ the above implies $f(1)=0$,
and for $c_0=2$ we can conclude $f(0)=0$.
However, for $c_1=1$, $c_0=0$
we have $\sum_{i=0}^n i(c_i-1)=0=\sum_{i=0}^n f(i)(c_i-1)$,
which violates the required equivalence.
Answer for new version:
Let $L$ be the left-hand side and $R$ be the right-hand side
of the equivalence.
Let us consider the case that $n=2$.
If $c_0=0$, $c_1=c_2=1$ then we have $L=0$.
Then $R=f(0)(-1)$ and $f(0)\neq0$ follow.
On the other hand, we can also consider the cases
where $c_i=2$ for some $i\in\{0,1,2\}$ and
the other $c_i$ values are $0$.
In these cases we always have $R\neq0$.
Then $L=0$ follows.
In other words, we have
$$
0=f(0)-f(1)-f(2)
=-f(0)+f(1)-f(2)
=-f(0)-f(1)+f(2).
$$
Some simple linear algebra leads to $f(0)=0$,
which is a contradiction to the result above.
Thus, such a function $f$ cannot exist.
