Functional Equation for $f(x-y)+f(y-z)+f(z-x)=2f(x+y+z)$ The following functional equation proved quite difficult.
$1.$ $f(x)$ is a polynominal with real coeffecients. 
$2.$ $f(1)=2,f(2)=20$. 
$3.$ When for real $x,y,z$ satisfies the condition  $xy+yz+zx=0$, $$f(x-y)+f(y-z)+f(z-x)=2f(x+y+z)$$ 
Find $f(x)$ that satisfies all conditions given above. 
While it is easy enough to guess that $f(x)=x^4+x^2$, proving it proved difficult. 
First, putting in $(x,0,0)$ gives us that $f(-x)=f(x)$. 
Second, putting in $(0,0,0)$ gives us $f(0)=0$
I also substituted $f(x)=g(x^2)$. 
However, I was able to progress no further from here. Any help would be appreciated. 
 A: After some comments from Michael, and upon further examination, I was able to formulate a solution.
$f(x-y)+f(y-z)+f(z-x)=2f(x+y+z)$. If $f(x)=g(x^2)$, this gives us that 
$g(x^2-2xy+y^2)+g(y^2-2yz+z^2)+g(z^2-2zx+x^2)=2g(x^2+y^2+z^2)$. 
From the fact that $xy+yz+zx=0$. If $x-y=s,y-z=t$, then $x-z=s+t$. This gives us that $xy+yz+zx=(y+s)y+y(y-t)+(y+s)(y-t)=3y^2+2(s-t)y-st=0$. $b^2-4ac=4(s^2+st+t^2)>0$. Thus such a $y$ always exists. 
Notice that $s^2+st+t^2=x^2-2xy+y^2+y^2-2yz+z^2-zx-y^2+xy+yz=x^2+y^2+z^2$
Then, $g(s^2)+g(t^2)+g(s^2+2st+t^2)=2g(s^2+st+t^2)$ gives us that $2g(s^2)+g(4s^2)=2g(3s^2)$. If the degree of $g(x)$ is $n$, comparing the coeffecients give us that $2 \times 1 +4^n=2 \times 3^n$. The only solution for this is $n=1,2$. It is easy to proceed from here.  
A: Here There is no need to assume $f(x)=g(x^2)$, just assume $f(x)$ to be a real polynomial of degree 'n'. Eliminate z by taking $x-y = r,y-z = s$. Then invoke the polynomial equation and apply it to each term in both sides of given equation and by comparing you can arrive that $$f(x) = ax^2 +bx^4$$
