Really nice functional equation with second partial deratives. Show that $f:\mathbb{R}^2\to\mathbb{R}$, $f \in C^{2}$ satisfies the equation
$$\frac{\partial^2 f}{\partial x^2} - \frac{\partial^2 f}{\partial y^2} = 0$$
for all points $(x,y) \in \mathbb{R}^2$ if and only if for all $(x,y)\in \mathbb{R}^2$ and $t \in \mathbb{R}$ we have:
$$f(x, y + 2t) + f(x, y) = f(x + t,y + t) + f(x - t, y +t).$$
Note. In such case, $f$ is said to satisfy the parallelogram's law.
 A: Try setting $u=x+y,v=x-y$, and notice that the inverse equations are $x=(u+v)/2,y=(u-v)/2$. This changes your equation to $f_{uv}=0$, which makes the problem much easier.
Edit: Apparently we have some critics. Do what I said to do; the solution of this equation is any function of the form $f=g(u)+h(v)$. Then $f(x,y+2t)=g(u+2t)+h(v-2t)$, while $f(x+t,y+t)=g(u+2t)+h(v)$ and $f(x-t,y+t)=g(u)+h(v-2t)$.
For the other direction, take the second derivative with respect to $t$ twice in the equation and plug in 0 for $t$. 
A: The PDE $f_{xx}-f_{yy}=0$ is the wave equation with general solution
$$
f(x,y)=F(x-y)+G(x+y)
$$
where $F$ and $G$ are any functions.
Forall $t\in \Bbb R$ the function $f(x,y)=F(x-y)+G(x+y)$ satisfies $$f(x, y + 2t) + f(x, y) = f(x + t,y + t) + f(x - t, y +t)$$
because
$$\small
\underbrace{F(x-y-2t)+G(x+y+2t)}_{f(x, y + 2t)}+\underbrace{F(x-y)+G(x+y)}_{f(x,y)}=\underbrace{F(x+t-y-t)+G(x+t+y+t)}_{f(x + t,y + t)}+\underbrace{F(x-t-y-t)+G(x-t+y+t)}_{f(x - t, y +t)}
$$
that is
$$\small
\underbrace{\color{red}{F(x-y-2t)}+\color{blue}{G(x+y+2t)}}_{f(x, y + 2t)}+\underbrace{\color{green}{F(x-y)}+G(x+y)}_{f(x,y)}=\underbrace{\color{green}{F(x-y)}+\color{blue}{G(x+y+2t)}}_{f(x + t,y + t)}+\underbrace{\color{red}{F(x-y-2t)}+G(x+y)}_{f(x - t, y +t)}
$$
A: First assume that (replacing $y$ by $y-t$ in the assumption)
$$
f(x, y + t) + f(x, y) = f(x + t,y) + f(x - t, y -t),
$$
for all $t,x,y\in\mathbb R$. This implies that
$$
\frac{f(x, y + t) -2f(x,y)+ f(x, y-t)}{t^2} = \frac{f(x + t,y)-2f(x,y) + f(x - t, y)}{t^2},
$$
for $x,y\in\mathbb R$ and $t\ne 0$. We shall now explain why, the left hand side tends to $f_{yy}(x,y)$, while the right hand side tends to $f_{xx}(x,y)$, as $t\to 0$. In general, if $g$ is $C^2$, then
\begin{align}
\frac{g(x+h)-2g(x)+g(x-h)}{h^2}&=\frac{\int_0^h g'(x+t)\,dt-\int_0^h g'(x-h+t)\,dt}{h^2}\\
&=\frac{\int_0^h\int_0^h g''(x-h+t+s)\,ds\,dt}{h^2}\to g''(x),
\end{align}
when $h\to 0$, as $g''$ is continuous. 
Assume now that $f_{xx}(x,y)=f_{yy}(x,y)$. We shall use the following fact:
Fact. If $g_x=g_y$, then $g$ is of the form $g(x,y)=h(x+y)$.
Let now $F(x,y)=f_x(x,y)+f_y(x,y)$. Clearly,
$$
F_x-F_y=(f_x+f_y)_x-(f_x+f_y)_y=f_{xx}+f_{yx}-f_{xy}-f_{yy}=f_{xx}-f_{yy}=0,
$$
as $f_{xy}=f_{yx}$ since $f$ is $C^2$. Thus, using the above fact, $F(x,y)=H(x+y)$, where $H\in C^1$. Thus
$$
f_x(x,y)+f_y(x,y)=H(x+y).
$$
Now, let see how $f$ behaves along lines of the form $x-y=s$, parametrised as $x=t+s, y=t$. Then
$$
\frac{d}{dt}f\big(x(t),y(t)\big)=\frac{d}{dt}f(t+s,t)=f_x(t+s,t)+f_y(t+s,t)=H(2t+s),
$$
and thus
$$
f(t+s,t)-f(s,0)=\int_0^t H(2\tau+s)\,d\tau,
$$
or, setting $x=s+t$ and $y=t$ 
$$
f(x,y)=f(x-y,0)+\int_0^{y} H(2\tau+x-y)\,d\tau.
$$
If $g'=H$, then
\begin{align}
f(x,y)&=f(x-y,0)+\int_0^{y} H(2\tau+x-y)\,d\tau =f(x-y,0)+\left.\frac{1}{2}g(2\tau+x-y)\right|_0^y\\ &= f(x-y,0)+\frac{1}{2}g(2y+x-y)-\frac{1}{2}g(x-y)=f_1(x-y)+f_2(x+y),
\end{align}
where
$$
f_1(x)=f(x,0)-\frac{1}{2}g(x) \quad\text{and}\quad f_2(x)=\frac{1}{2}g(x).
$$
Proof of the Fact. Fix $s\in\mathbb R$ and Let $\big(x(t),y(t)\big)=(t,s-t)$, $t\in\mathbb R$. Then
$$
\frac{d}{dt}g\big(x(t),y(t)\big)=\frac{d}{dt}g(t,s-t)=g_x-g_y=0.
$$
Thus $g(t,s-t)=g(0,s)$. Now $t=x$ and $s=x+y$, and therefore
$$
g(x,y)=g(t,s-t)=g(0,s)=g(0,x+y)=H(x+y).
$$
