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Let $\{e_k\}_{1\le k\le 16}$ denote the standard basis of $\mathbb{R}^{16}$, and $x:=(x_1,\dots,x_{16})$. Let's consider the difference operator in the $k$-th variable, $\delta_k:\mathbb{R}[x]\to\mathbb{R}[x ]$, that is $\delta_kp(x):=p(x+e_k)-p(x)$. So $$\delta_{13}\delta_9 \delta_5 \delta_1p(x)=\sum_{\epsilon } (-1)^{|\epsilon|_1}p(x+\epsilon_i ),$$ the sum being extended over all $\epsilon\in\{0,1\}^{16}$ with support in the set $S:=\{1,5,9,13\}$: it vanishes if and only if $p=\sum_{k\in S} p_k$ with $\deg_k p_k=0 $ for any $k\in S$. Now if $g\in \mathbb{R}[x]$ verifies the assumptions, $\delta_{13}\delta_9 \delta_5 \delta_1g(0)=g(e_1+e_5+e_9+e_{13})\neq0$, proving that $g$ contains a monomial of positive degree in all variables $x_1, x_5, x_9,$ and $x_{13}$ (and for the same reason, it must also contain any term of the expansion of $f$).

Let $\{e_k\}_{1\le k\le 16}$ denote the standard basis of $\mathbb{R}^{16}$, and $x:=(x_1,\dots,x_{16})$. Let's consider the difference operator in the $k$-th variable, $\delta_k:\mathbb{R}[x]\to\mathbb{R}[x ]$, that is $\delta_kp(x):=p(x+e_k)-p(x)$. So $$\delta_{13}\delta_9 \delta_5 \delta_1p(x)=\sum_{\epsilon } (-1)^{|\epsilon|_1}p(x+\epsilon_i ),$$ the sum being extended over all $\epsilon\in\{0,1\}^{16}$ with support in the set $S:=\{1,5,9,13\}$: it vanishes if and only if $p$ is of the form $p=\sum_{k\in S} p_k$ for some $p_k\in\mathbb{R}[x]$ with $\deg_k p_k=0 $, for any $k\in S$. Now if $g\in \mathbb{R}[x]$ has $g^{-1}(0)\cap\{0,1\}^{16}=\mathcal{Z},$ we have $\delta_{13}\delta_9 \delta_5 \delta_1g(0)=g(e_1+e_5+e_9+e_{13})\neq0$, proving that $g$ contains a monomial of positive degree in all variables $x_1, x_5, x_9,$ and $x_{13}$ (and for the same reason, it must also contain any term of the expansion of $f$).

Let $\{e_k\}_{1\le k\le 16}$ denote the standard basis of $\mathbb{R}^{16}$, and let's consider the difference operator in the $k$-th variable, $\delta_k:\mathbb{R}[ x_1,\dots,x_{16} ]\to\mathbb{R}[ x_1,\dots,x_{16} ]$, that is $\delta_kp(x):=p(x+e_k)-p(x)$. So $$\delta_{13}\delta_9 \delta_5 \delta_1p(x)=\sum_{\epsilon } (-1)^{|\epsilon|_1}p(x+\epsilon_i ),$$ the sum being extended over all $\epsilon\in\{0,1\}^{16}$ with support in the set $S:=\{1,5,9,13\}$: it vanishes if and only if $p=\sum_{k\in S} p_k$ with $\deg_k p_k=0 $ for any $k\in S$. Now if $g\in \mathbb{R}[x]$ verifies the assumptions, $\delta_{13}\delta_9 \delta_5 \delta_1g(0)=g(e_1+e_5+e_9+e_{13})\neq0$, proving that $g$ contains a monomial of positive degree in all variables $x_1, x_5, x_9,$ and $x_{13}$ (and for the same reason, it must also contain any term of the expansion of $f$).

Let $\{e_k\}_{1\le k\le 16}$ denote the standard basis of $\mathbb{R}^{16}$, and $x:=(x_1,\dots,x_{16})$. Let's consider the difference operator in the $k$-th variable, $\delta_k:\mathbb{R}[x]\to\mathbb{R}[x ]$, that is $\delta_kp(x):=p(x+e_k)-p(x)$. So $$\delta_{13}\delta_9 \delta_5 \delta_1p(x)=\sum_{\epsilon } (-1)^{|\epsilon|_1}p(x+\epsilon_i ),$$ the sum being extended over all $\epsilon\in\{0,1\}^{16}$ with support in the set $S:=\{1,5,9,13\}$: it vanishes if and only if $p=\sum_{k\in S} p_k$ with $\deg_k p_k=0 $ for any $k\in S$. Now if $g\in \mathbb{R}[x]$ verifies the assumptions, $\delta_{13}\delta_9 \delta_5 \delta_1g(0)=g(e_1+e_5+e_9+e_{13})\neq0$, proving that $g$ contains a monomial of positive degree in all variables $x_1, x_5, x_9,$ and $x_{13}$ (and for the same reason, it must also contain any term of the expansion of $f$).

Let $\{e_k\}_{1\le k\le 16}$ denote the standard basis of $\mathbb{R}^{16}$, and let's consider the difference operator in the $k$-th variable, $\delta_k:\mathbb{R}[x]\to\mathbb{R}[x]$, that is $\delta_kp(x):=p(x+e_k)-p(x)$. So $$\delta_{13}\delta_9 \delta_5 \delta_1p(x)=\sum_{\epsilon } (-1)^{|\epsilon|_1}p(x+\epsilon_i ),$$ the sum being extended over all $\epsilon\in\{0,1\}^{16}$ with support in the set $S:=\{1,5,9,13\}$: it vanishes if and only if $p=\sum_{k\in S} p_k$ with $\deg_k p_k=0 $ for any $k\in S$. Now if $g\in \mathbb{R}[x]$ verifies the assumptions, $\delta_{13}\delta_9 \delta_5 \delta_1g(0)=g(e_1+e_5+e_9+e_{13})\neq0$, proving that $g$ contains a monomial of positive degree in all $x_1 x_5 x_9$ and $x_{13}$ (and for the same reason, it must also contain any term of the expansion of $f$).

Let $\{e_k\}_{1\le k\le 16}$ denote the standard basis of $\mathbb{R}^{16}$, and let's consider the difference operator in the $k$-th variable, $\delta_k:\mathbb{R}[ x_1,\dots,x_{16} ]\to\mathbb{R}[ x_1,\dots,x_{16} ]$, that is $\delta_kp(x):=p(x+e_k)-p(x)$. So $$\delta_{13}\delta_9 \delta_5 \delta_1p(x)=\sum_{\epsilon } (-1)^{|\epsilon|_1}p(x+\epsilon_i ),$$ the sum being extended over all $\epsilon\in\{0,1\}^{16}$ with support in the set $S:=\{1,5,9,13\}$: it vanishes if and only if $p=\sum_{k\in S} p_k$ with $\deg_k p_k=0 $ for any $k\in S$. Now if $g\in \mathbb{R}[x]$ verifies the assumptions, $\delta_{13}\delta_9 \delta_5 \delta_1g(0)=g(e_1+e_5+e_9+e_{13})\neq0$, proving that $g$ contains a monomial of positive degree in all variables $x_1, x_5, x_9,$ and $x_{13}$ (and for the same reason, it must also contain any term of the expansion of $f$).