Show that specific polynomial with $p(a)=p(b)=0$ must be constantly $0$. 
Say polynomial $p\in \mathbb{R}[x]$ is such that for all
reals $x,y,z$ such that $xy+yz+zx = 0$, we have
$$ p(x-y) + p(y-z) + p(z-x) = 2p(x+y+z). $$
Prove that $p(x)=0$ for all $x$ if $p(a) = p(b)=0$ for some
different positive $a$ and $b$, without explicitly solving the
equation.


To make it easier, you can choose $a,b$ on your own explicitly.

All I can get is
\begin{align}
 x=y=z=0: &\;\; 3p(0)=2p(0) \implies \boxed{p(0) =0}\\
 x=y=0 : &\;\; p(-z)+p(z) =2p(z) \implies \boxed{p(-z) =p(z),\;\; \forall z}\\
\end{align}
so $p$ is an even polynomial with constant term $0$. From $p(a)=p(b)=0$ we can say that $\deg (p)\geq 6$.
I can't find anything usefull from here.
 A: Set $x=6t$, $y=3t$, and $z=-2t$. Then
$$xy+yz+zx=18t^2-6t^2-12t^2=0$$
and
$$p(x-y)+p(y-z)+p(z-x)=2p(x+y+z)$$
$$p(3t)+p(5t)+p(-8t)=2p(7t)$$
Then with your observation that $p(t)=p(-t)$, this simplifies to
$$p(3t)+p(5t)+p(8t)=2p(7t)$$
Now, we know that if the polynomial is not identically zero, then for very large $t$ (past all the zeros)
$$1=\frac{2p(7t)}{p(3t)+p(5t)+p(8t)}$$
If we denote the leading coefficient of $p(t)$ by $a$, then
$$1=\frac{2p(7t)}{p(3t)+p(5t)+p(8t)}=\lim_{t\to\infty}\frac{2p(7t)}{p(3t)+p(5t)+p(8t)}=\lim_{t\to\infty}\frac{2a7^nt^n}{a3^nt^n+a5^nt^n+a8^nt^n}$$
(where $n$ is the degree of $p(t)$). This simplifies to
$$=\lim_{t\to\infty}\frac{2a7^nt^n}{a3^nt^n+a5^nt^n+a8^nt^n}=\lim_{t\to\infty}\frac{2\cdot 7^n}{3^n+5^n+8^n}=\frac{2\cdot 7^n}{3^n+5^n+8^n}$$
Now, is this equation ever satisfied? Yes, at $n=2$ and $n=4$. However, you showed in your question that $n\geq 6$ (this is where we use the fact that $p(a)=p(b)=0$ for some $a,b>0$). But for these $n$ we have
$$\frac{2\cdot 7^n}{3^n+5^n+8^n}<2\left(\frac{7}{8}\right)^n\leq 2\left(\frac{7}{8}\right)^6=\frac{117649}{131072}<1$$
Thus, the polynomial must be identically zero.
