Consider all the solutions to $p_1, p_2\in P[x_1,x_2]$ for the identity: $(x_1+1)(x_2+1)=p_1(x_1+1)+p_2(x_2+1),$
where $P[x_1,x_2]$ is the the space of all polynomials over $x_1,x_2$ with natural coefficients.
It is not hard to see that there are only two solutions; one with $p_1\equiv 0$, and the other with $p_2\equiv 0$.
With a little work I was able to show that in all the solutions for $p_1, p_2, p_3\in P[x_1,x_2, x_3]$ to the identity:
$(x_1+1)(x_2+1)(x_3+1)=p_1(x_1+1)+p_2(x_2+1)+p_3(x_3+1),$
at least one of $p_1, p_2, p_3$ is zero.
My conjecture is that for all $n$, in any solution $(p_1,\ldots,p_n)\in P[x_1,\ldots,x_n]^n$ for the identity:
$\prod_{i=1}^n(x_i+1)=\sum_{i=1}^n p_i(x_i+1),$
at least one of the $p_i$'s is zero.
The arguments I used for $n=3$ get too involved even for $n=4$.
Any idea on what might work to prove this conjecture will be highly appreciated.
UPDATE: THE CONJECTURE IS FALSE
Here is a counter example for $n=4$:
$\prod_{i=1}^4(x_i+1)=(x_4+1)(x_1+1)+x_1x_3(x_4+1)(x_2+1)+x_2(x_4+1)(x_3+1)+(x_3+x_1x_2)(x_4+1).$
What about $$p_{k}(x_1,\dots,x_n)=\frac{1}{n(x_k+1)}\prod_{i=1}^n(x_i+1)?$$ This satisfies $\prod_{i=1}^n(x_i+1)=\sum_{i=1}^n p_i(x_i+1)$.