**Notation**

$P_k[n]=\{$multilinear polynomials in $\Bbb R[x_1,x_2,\dots,x_{n-1},x_n]$ of total degree exactly $k\}$.

$k=1$ is just linear polynomials.

**QUESTION**

Is there a triplet $(p,f,g)\in (P_{k}[4],P_1[4],P_1[4])$, with $k\in\{2,3,4\}$ such that $\forall s\in\{0,1\}^4,\mbox{ }$ following four conditions are satisfied? $$(I):p(s)\in\{0,1\}$$ $$(II):f(s),g(s)\in\Bbb R$$ $$(III):p(s)=0\iff{f(s)=0}$$ $$(IV):\mbox{ }p(s)=1\iff{g(s)=0}$$

Note above conditions imply $$\forall s\in\{0,1\}^4:p(s)=\frac{f(s)}{f(s)+g(s)}$$

Hence it seems above question at $k=2$ is same as following question:

Is there a $4$ variable total degree $2$ multilinear polynomial that agrees with ratios of a pair of total degree $1$ linear polynomials over $\{0,1\}^4$ with each evaluation on $\{0,1\}^4$ evaluating to $\{0,1\}$?

$\underline{\text{Conjecture: Answer to above question is negative with cases }k\in \{2,3,4\}}$.

I am unable to find an example in $(P_{2}[4],P_1[4],P_1[4])$, $(P_{3}[4],P_1[4],P_1[4])$, $(P_{4}[4],P_1[4],P_1[4])$. Examples in these three cases (Of these easiest seems $k=2$) will be interesting.

Is it possible to extend following attempt to prove no triplet exists in $(P_{2}[4],P_0[4],P_1[4]),(P_{2}[4],P_1[4],P_0[4])$ to above?

Note that $\forall s\in\{0,1\}^4$, $p(s)=\frac{f(s)}{f(s)+g(s)}$.

Case $(P_{2}[4],P_1[4],P_0[4])$: $g$ is constant function. So $f$ will be $0$.

Case $(P_{2}[4],P_0[4],P_1[4])$: $f$ is constant function. So $g$ will be $0$.

$p$ will be degree $0$.

This will prove no triplet exists in $(P_{2}[4],P_0[4],P_1[4]),(P_{2}[4],P_1[4],P_0[4])$.

Is there a similar approach to $(P_{2}[4],P_1[4],P_1[4])$, $(P_{3}[4],P_1[4],P_1[4])$, $(P_{4}[4],P_1[4],P_1[4])$?

Through tedious calculations if I possibly could show triplet in $(P_{2}[4],P_1[4],P_1[4])$ cannot exist provided if $2$ coordinates will be $0$, degree $2$ polynonmial still remains degree $2$, then possibly we will be done.

Example: If $p\in P_{2}[4]$ will be of form $$\sum_{i,j=1,i\neq j}^4a_{i,j}x_ix_j+\sum_{i=1}^4b_{i}x_i+c$$ then at $x_3,x_4=0$, $p$ reduces to $$a_{1,2}x_1x_2+\sum_{i=1}^2b_{i}x_i+c$$ then there could be a tedious path to show non-existence of triplets of said property.

It seems every $p\in P_{2}[4]$ will remain degree $2$ with some projection $(x_i,x_j)=(0,0)$.

I seem to have tedious path which is very inelegant. Is calculations only approach possible to question which is short?