Problem. Prove that in the field $F, \text{char }F\neq2$ every element $a$ can be decomposed in the following way: $a = a_1\cdots a_k$ and $a_1 + \dots +a_k = 0$.
Attempt 1.
For the fields in which $x^2 = \pm a$ has at least one solution I can find this decomposition directly (suppose $x^2 = -a$ has solution). Let's take $a_1 = \sqrt{-a}, a_2=-\sqrt{-a}$.
But there are some problems in the fields in which this property does not hold. For example $\mathbb{F}(x)-$ ratioanl functions. I don't know decomposition for instance for $x$ over any field $\mathbb{F}$.
Attempt 2. Let's look at the field like the vector space over $\mathbb{Q}$ if char $F =0$ and over $\mathbb{Z}_p$ if char $F \neq0$ (denote it in each case as $\mathbb{V}$). The idea is to find an element $b$ such that $-b^2a \in \mathbb{V}$. Then we can decompose it directly. But it is also not seems to be true.
If $a=-1$ take $a_1=1$, $a_2=-1$. Otherwise $a+1\neq 0$, so you can take
$a_1=\frac{a}{a+1}$, $a_2=\frac{1}{a+1}$, $a_3=-1$, $a_4=a+1$, $a_5=-(a+1)$.
We have $\sum a_i=0$ while $\prod a_i=a$.