Prove that exists a function $f$, such that $\sum_{i=0}^{n} i (c_i - 1) = 0 \Leftrightarrow \sum_{i=0}^{n} f(i) (c_i - 1) \neq 0$

Question

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 $i^2$?

Answer 1

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.