Let me try to solve your original problem differently. First write the wanted polynomial in the form $f(x,y)=\sum_kx^kA_k(y)$ where the leading polynomial $A_0(y)$ is not identically zero (otherwise we can always replace $f(x,y)$ by $f(x,y)/x^\ell$ for a suitable $\ell$). Denote by $N$ the degree of the polynomial $A_0(y)$. For any prime $p>N!$ the numbers $0$ and $(-1)^kk!$, where $k=0,\dots,N$, k=0,\dots,N-1$, are distinct residues modulo$p$, so that$p!\equiv 0\pmod p$and$(p-k)!=(p-1)!/\prod_{j=1}^{k-1}(p-j)\equiv(-1)^k(k-1)!^{-1}\pmod p$are pairwise noncongruent modulo$p$as well. Substituting$x=(2p-2k)!\equiv0\pmod p$and$y=(p-k)!$for each$k=0,1,\dots,N$into$f(x,y)=0$and reducing modulo$p$, we obtain$N+1$different solutions of the polynomial equation$A_0(x)\equiv0\pmod p$, so that all coefficients of$A_0(x)$are divisible by$p$. Since this is true for any prime$p>N!$, the polynomial$A_0(x)$is identically zero, which contradicts our assumption. Is it elementary enough? 1 Armin, Let me try to solve your original problem differently. First write the wanted polynomial in the form$f(x,y)=\sum_kx^kA_k(y)$where the leading polynomial$A_0(y)$is not identically zero (otherwise we can always replace$f(x,y)$by$f(x,y)/x^\ell$for a suitable$\ell$). Denote by$N$the degree of the polynomial$A_0(y)$. For any prime$p>N!$the numbers$0$and$(-1)^kk!$, where$k=0,\dots,N$, are distinct residues modulo$p$, so that$p!\equiv 0\pmod p$and$(p-k)!=(p-1)!/\prod_{j=1}^{k-1}(p-j)\equiv(-1)^k(k-1)!^{-1}\pmod p$are pairwise noncongruent modulo$p$as well. Substituting$x=(2p-2k)!\equiv0\pmod p$and$y=(p-k)!$for each$k=0,1,\dots,N$into$f(x,y)=0$and reducing modulo$p$, we obtain$N+1$different solutions of the polynomial equation$A_0(x)\equiv0\pmod p$, so that all coefficients of$A_0(x)$are divisible by$p$. Since this is true for any prime$p>N!$, the polynomial$A_0(x)\$ is identically zero, which contradicts our assumption.