the solution is : $ f(x)=cx^{2}$

... and maybe you have a proof of that?

Trying to be intimidating with 3 variables, I see. You'll notice this solution sets $ z = 0$ every time the given equation is used. Let $ x = y = z = 0$. We get $ f(0) = 0$. Then let $ y = z = 0$, and we get $ f(x) = f(-x)$. So $ f$ is even. We now prove by strong induction that for all nonnegative integers $ k$, $ f(kx) = k^{2}f(x)$. Base cases $ k = 1, k = 0$ are obvious. Suppose it's true for all nonnegative integers up to $ k$. Let $ y = kx, z = 0$. We get \[ f(x(k+1))+f(x(k-1)) = 2f(kx)+2f(x)\] \[ f(x(k+1)) = 2k^{2}f(x)+2f(x)-(k-1)^{2}f(x) = (k^{2}+2k+1)f(x) = (k+1)^{2}f(x)\] completing the induction. From this and from that $ f$ is even, we conclude that $ f(n) = n^{2}f(1)$ for all integers $ n$. Let $ p$ and $ q\neq 0$ be integers. Then \[ f\left(\frac{p}{q}\right) =\frac{f(p)}{q^{2}}=\frac{p^{2}}{q^{2}}f(1)\] We see that all rational multiples of $ x^{2}$ are indeed solutions to the equation, so we're done.

It is obviosly $ x=y=z=0$ give $ f(0)=0$, $ x=y=z$ give $ f(3x)=9f(x)$. $ y=z=0$ give $ f(x)=f(-x)$, z=0 give $ f(x+y)+f(x-y)=2f(x)+f(y)\to f(2x)=4f(x)$, by induction $ f(nx)=n^{2}f(x)$. It give solution for Q, but for R $ f(x)=cg(x)^{2}$ is solution too, were $ g(x)$ is any additive function $ g(x+y)=g(x)+g(y)$.

Can you be more specific, Rust? How do the cases $ n=2$ and $ n=3$ yield the general statement?

If $ f((n-1)x)=(n-1)^{2}f(x)$ and $ f(nx)=n^{2}f(x)$ and $ f(z+y)+f(z-y)=2f(z)+2f(y)$, then $ z=nx,y=x$ give $ f((n+1)x)=2n^{2}f(x)+2f(x)-(n-1)^{2}f(x)=(n+1)^{2}f(x)$. Therefore sufficiently for induction $ f(2x)=4f(x)$.

MellowMelon wrote: Trying to be intimidating with 3 variables, I see. You'll notice this solution sets $ z = 0$ every time the given equation is used. Let $ x = y = z = 0$. We get $ f(0) = 0$. Then let $ y = z = 0$, and we get $ f(x) = f( - x)$. So $ f$ is even. We now prove by strong induction that for all nonnegative integers $ k$, $ f(kx) = k^{2}f(x)$. Base cases $ k = 1, k = 0$ are obvious. Suppose it's true for all nonnegative integers up to $ k$. Let $ y = kx, z = 0$. We get \[ f(x(k + 1)) + f(x(k - 1)) = 2f(kx) + 2f(x) \] \[ f(x(k + 1)) = 2k^{2}f(x) + 2f(x) - (k - 1)^{2}f(x) = (k^{2} + 2k + 1)f(x) = (k + 1)^{2}f(x) \] completing the induction. From this and from that $ f$ is even, we conclude that $ f(n) = n^{2}f(1)$ for all integers $ n$. Let $ p$ and $ q\neq 0$ be integers. Then \[ f\left(\frac {p}{q}\right) = \frac {f(p)}{q^{2}} = \frac {p^{2}}{q^{2}}f(1) \] We see that all rational multiples of $ x^{2}$ are indeed solutions to the equation, so we're done. Why $ f\left(\frac {p}{q}\right) = \frac {f(p)}{q^{2}}$ : I'm very bad in Functional Equation

Take $ k=q$, $ x= \frac pq$. Then $ f(p)=f\left(q \cdot \frac {p}{q}\right) = f(qx) = q^2 f(x) = q^2 f\left(\frac {p}{q}\right)$, hence the result.

Set $z=0$ and easy to see $f(0)=0$ so denote $f(x+y)+f(x-y)+f(-x)$ $=3f(x)+2f(y)$ by $P(x,y).$ Then $P(x,0)$ gives $f(x)=f(-x).$ And further $P(x,x)$ gives $f(2x)=4f(x)$ and inducting gives $f(ax)=a^2f(x)$ i.e. $f(a)=a^2f(1)$ for all $a\in \mathbb{N}.$ Setting $a=q$ and $x=p/q$ gives $f(p/q)=p^2f(1)/q^2$ and it indeed fits.