L.Lawliet03 wrote: The polynomial $P(x,y)$ is such that for every integer $n\geq 0$ each of the polynomials $P(x,n)$ and $P(n,y)$ either is a constant zero or has a degree not greater than n. Is it possible that $P(x,x)$ has an odd degree? We claim that this is not possible. Let $m$ and $n$ be the highest powers of $x$ and $y$, respectively in $P(x,y)$ . Claim : $m=n$ Proof : Write $P(x,y) = \sum_{i=0}^{n} A_i(x)\cdot y^i$ Where $A_i(x)$ is a polynomial in $x$ $\forall 0\leq i \leq n$ . Next note that since the degree of $P(k,y)$ must be less than $n$ $\forall 0\leq k < n$ ,hence we have This implies that $A_n(k)=0$ $\forall 0\leq k < n$ Hence degree of $A_n(X)$ is $\geq n$ . But it is also $\leq m$ . Hence $m=n$ Since , $P(x,x)$ ,the term $A_n(x)x^n$ contains a term of the form $Cx^{2n}$ , which cant appear in any other of the terms ,hence $\operatorname{deg} P(x,x) =2n$ . Done $\blacksquare$

Answer is NO. Claim: Given polynomial $Q(x,y)$, if both $Q(x,0),Q(0,y)$ are constant (or $0$), then $P(x,y) = a + xy \cdot R(x,y)$ for some constant $a$ and polynomial $R(x,y)$. Proof: Basically any power of $x$ in $Q(x,y)$ cannot be isolated. Similar for $y$. Our claim actually just follows. To be more formal, both $Q(x,0),Q(0,y)$ must equal the constant $Q(0,0)$. Then $Q(x,y) - Q(0,0)$ is divisible by both $x,y$. $\square$ Define $P_0(x,y) = P(x,y)$. We have $$ P(x,y) = xy \cdot P_1(x,y) + c_1$$for some constant $c_1$. But then $P_1(x,1),P_1(1,y)$ are constant. Applying Claim to polynomial $Q_1(x,y) = P_1(x+1,y+1)$ gives $P_1(x,y) = (x-1)(y-1) \cdot P_2(x,y) + c_2$. Continuing this way we get $P(x,x)$ has even degree, as desired. $\blacksquare$

My solution is same as @above,anyway I will post for storage. We notice that by subsi-ing $P(x,0);P(0,y)$ we obtain $P(x,y)=xyQ(x,y)+c$ for some constant $c$ and some 2-variate polynomial $Q(x,y)$. Then by $P(1,y);P(x,1)$ we get $Q(x,y)=(x-1)(y-1)R(x,y)+d$ repeating this arguement we get that the degree of $P$ is even.