$f: \Bbb{R} \to \Bbb{R_0^+}, f(x+2y)+f(x)=2(f(x+y)+f(y)).$ 
$f: \Bbb{R} \to \Bbb{R_0^+}, f(x+2y)+f(x)=2(f(x+y)+f(y)).$

\begin{align}
P(x, 0): \; & 2f(x)=2f(x)+2f(0), f(0)=0. \\
P(0, y): \; & f(2y)=4f(y). \\
P(x, -x): \; & f(-x)+f(x)=2f(-x), f(x)=f(-x). \\
P(x, x): \; & f(3x)+f(x)=2f(2x)+2f(x). \\
\Rightarrow \; & f(3x)=8f(x)+f(x)=9f(x). \\
P(2x, x): \; & f(4x)+f(2x)=2f(3x)+2f(x). \\
\Rightarrow \; & f(4x)=18f(x)+2f(x)-4f(x)=16f(x). \\
P(3x, x): \; & f(5x)+f(3x)=2f(4x)+2f(x). \\
\Rightarrow \; & f(5x)=25f(x). \\
& \vdots \\
P((n-2)x, x): \; &f(nx)=n^2f(x).
\end{align}
How can we proceed from here?
 A: I know this is a little "old", but just so there's an answer: Ivan Neretin from the comments above is right I believe. Now, I won't claim to know anything about Hamel bases and such myself, so I'll link this post for a reference, however what I do know is that the $f$ we're looking for here is quite special.
If you take $x - y$ instead of $x$ in the functional equation, and let $g : \mathbb{R} \to \mathbb{R}$ be a function such that $g^2 = f$, then our equation is equivalent to the familiar parallelogram identity:
$$g(x+y)^2 + g(x - y)^2 = 2\big(g(x)^2 + g(y)^2\big)$$
Thus (unless I'm wrong which I could very well be!) we should be able to create such $g$ (and therefore such $f = g^2$) by choosing the "norm" induced by any inner product on $\mathbb{R}$ seen as a $\mathbb{Q}$-vector space -- for example, any dot product using the coordinates of reals in a Hamel basis.
(And I'm assuming that if $g$ is a "$\mathbb{Q}$-norm" on $\mathbb{R}$ then $g$ is induced by an inner product since the parallelogram law should still go both ways even in $\mathbb{Q}$-spaces, but that's probably less useful?).
And these might not even be the only non-continuous solutions, since that's just for $g$ $\mathbb{Q}$-norms...
(Feel free to correct me!)
