Find $f$ such that $f(a-b)+f(c-d)=f(a)+f(b+c)+f(d)$ Denote the set of non-negative real numbers by $\mathbb R^+_0$. Find all functions $f:\mathbb R \rightarrow \mathbb R_0^+$ s.t. $\forall a,b,c,d\in\mathbb R$ satisfying $ab+bc+cd=0$ we have $$f(a-b)+f(c-d)=f(a)+f(b+c)+f(d)$$
My attempt: 
set $b=d=0$ and get $f(0)=0$. 
Now set $a=b=c=0$ to get $f(-d)=f(d)$.
Now we pretty much reduced this fe's domain to $\mathbb R^+_0\rightarrow\mathbb R^+_0$.
That's where I got stuck. I tried a few things but none of them gave me any progress. Any help/hints appreciated.
 A: Let $c \neq 0$. To remove our condition, we write $d$ in terms of the rest of the variables :
$$d=\frac{-ab-bc}{c}=-b-\frac{ab}{c}$$
Now, we can substitute this in our functional equation :
$$f(a-b)+f(b+c+\frac{ab}{c})=f(a)+f(b+c)+f(-b-\frac{ab}{c})$$
For $c=-b$, we have :
$$f(a-b)+f(-a)=f(a)+f(0)+f(a-b) \implies f(0)=f(-a)-f(a)$$
Replacing $a$ by $-a$ :
$$f(0)=f(a)-f(-a) \implies f(0)=0 \implies f(a)=f(-a) \space \forall \space a \in \mathbb{R}$$
Next, for $a=b=c$ :
$$f(0)+f(3a)=f(2a)+f(2a)+f(a) \implies f(3a)=2f(2a)+f(a) \space \forall \space a \in \mathbb{R}$$
For $b=-nc=-na$ :
$$f((n+1)a)+f((2n-1)a)=f(a)+f((n-1)a)+f(2na)$$
At $n=2$, this yields: 
$$f(4a)=2f(3a)-2f(a)=4f(2a) \implies f(2a)=4f(a) \space; f(3a)=9f(a)\space; f(4a)=16f(a)$$
For all naturals $n \leqslant 4$, we see that $f(na)=n^2 \cdot f(a)$. We continue with induction hypothesis. Let $f(na)=n^2 \cdot f(a) \space \forall \space n<2n-1$. We can see $f(2na)=4f(na)$. Observe the equation:
$$f((n+1)a)+f((2n-1)a)=f(a)+f((n-1)a)+4f(na)$$
All coefficients of $a$ inside the function are less than or equal to $2n-1$, and we can solve this equation for $f((2n-1)a)$ to yield one solution in terms of $f(a)$. We can see that in our functional equation, $f(ta)=t^2 \cdot f(a)$ satisfies our condition. Since there is only one solution in terms of $f(a)$, it must be $f(2n-1)=(2n-1)^2 \cdot f(a)$. Thus, induction hypothesis yields us the equation $f(na)=n^2 \cdot f(a) \space \forall \space n \in \mathbb{N}$
If $f$ is continuous, we conclude that the only solution to this functional equation is $f(x)=kx^2$ for some real number $x \geqslant 0$. If not, $k$ may vary (or not) for linearly independent $n$. I am unaware of how to tackle this part of the problem.
