(i) If $n = 0$ : Clearly no solution (ii) If $n = 1$ : $P(x, y) = ax+by \implies$ the identification yields directly $P(x,y) = x-2y$ (iii) If $n > 1$, $a = 1, \ b = 1, \ c =-2 \implies P(2,-2)+P(-1, 1)+P(-1, 1) = 0$ $\implies ((-2)^{n}+2) P(-1, 1) = 0 \implies P(-1, 1) = 0 \implies P(-y, y) = 0 \implies$ $P(x,y)$ is divisible by $(x+y)$ It is then easy to see that $\frac{P(x, y)}{(x+y)}$, of degree $n-1$ verifies all the equations. The only solutions are thus $P(x, y) = (x-2y)(x+y)^{n-1}$

$F(a,a,a) \implies P(2a,a)=0 \implies (x-2y)$ is a factor of $P(x,y)$. We may write $P(x,y)=(x-2y)Q(x,y)$ $F(a,b,b) \implies 2P(a+b,b)+P(2b,a)=0 \implies 2(a+b-2b)Q(a+b,b)+2(b-a)Q(2b,a) \implies (a-b)(Q(a+b,b)-Q(2b,a))=0$ Thus $Q(a+b,b)=Q(2b,a) \forall a \neq b$ We may rewrite it as $Q(x,y)=Q(2y,x-y)=Q(2x-2y,3y-x=\cdots$ $Q(x+d,y-d)-Q(x,y)$ is a polynomial in $d$ of degree $n-1$ for any two fixed $x,y$,which has infinitely many zeroes,i.e,$0,2y-x,x-2y,6y-3x,\cdots$.Thus $Q(x+d,y-d)=Q(x,y)$ holds for all $d$.In particular it holds for $d=y$,i.e, $Q(x+y,0)=Q(x,y)$.Now consider the polynomial $R(x,y)=Q(x+y,0)-Q(x,y)$.Suppose that its not the zero polynomial.Then its degree $d$ is defined.With $t=\frac{x}{y}$ it can be wriiten as $y^dS(t)=y^d(A(t)-B(t))$.But $S(t)$ has infinitely many zeroes and this forces $A(t)=B(t)$,forcing $R(x,y)$ to be a zero polynomial.Contradiction!.Thus $Q(x+y,0)$ and $Q(x,y)$ are identical polynomials.This forces $Q(x,y)=c(x+y)^{n-1}$.With $Q(1,0)=1$ we get $c=1$.Thus $P(x,y)=(x-2y)(x+y)^{n-1}$

200th post on HSO!

first of all we denote the assertion as $\Omega(x,y,z)$ we have $\Omega(a,a,a,)\implies P(2a,a)=0\implies P(x,y)+(x-2y)J(x,y)$ for some polynomial $J(x,y)$ of degree $n-1$ now we have $P(1,0)=Q(1,0)=1$ $\Omega(a,b,b)$ gives $P(2b,a)+2P(a+b,b)=0\implies (2b-2a)J(2b,a)+2(a-b)J(b+a,b)\implies J(a+b)=J(2b,a)$ $\forall$ $a\neq b$ now set $\alpha=a+b,\beta=b, a=\alpha-\beta$ we have $J(\alpha,\beta)=J(2\beta,\alpha-\beta)=J(2\alpha-2\beta,3\beta-\alpha)=\cdots$ this gives $J(\alpha-\zeta,\beta+\zeta)-J(x,y)=0$ $\forall$ $\zeta$, but this a polynomial of $n-1$ degree hence we have $J(x,y)=c(x+y)^{n-1}$ , plugging $J(1,0)=1$ we have $\boxed{P(x,y)=(x-2y)(x+y)^{n-1}}$ $\forall$ $n \in \mathbb{Z^{+}}$ $\blacksquare$