$p(x,y)$ is a symmetric polynomial. $(x-y)|p(x,y) \implies (x-y)^2$ is a factor of $p(x,y)$ I'm confused on how to approach this problem. So using symmetric property, I was trying this:
$ p(x,y)^2 = (x-y)^2 (q(x,y))^2$
How to I go about it from here?
 A: Based on what we've been given
$$\text{Let } p(x,y) = (x-y)g(x,y)$$
Now, using the symmetry of the polynomial, we have
$$p(y,x) = (y-x)g(y,x) \implies g(y,x) = -g(x,y)$$
Hence, $g$ is anti-symmetric. Hence, for $x=y$, we have
$$2g(x,x) = 0 \implies g(x,x) = 0$$
Hence, $(x-y)$ is a factor of $g(x,y)$
A: Hint: Since we can write $x=\frac{x+y}{2}+\frac{x-y}{2}$ and $y=\frac{x+y}{2}-\frac{x-y}{2}$, there is a polynomial $q$ of two variables such that $p(x,y)=q(x+y,x-y)$.  Now what does $p(x,y)=p(y,x)$ say about $q$?  Can you see why the conclusion follows?
A: No don't square both sides.  Instead look at the equation
$$ p(x,y) = (x-y) (q(x,y)),$$
and apply the symmetry property of $p$.  You get $$q(x,y)=-q(y,x).$$
Examining the coefficients on $q$ you get that it is a linear combination of terms of the form $x^iy^j-x^jy^i$.
Without loss of generality suppose $i\leq j$, so you can factor these terms as:$$x^iy^i(y^{j-i}-x^{j-i}).$$
Now you just need the identity: $$y^r-x^r=(y-x)(y^{r-1}+y^{r-2}x+\cdots+x^{r-1}).$$
