Find all "tame" solutions of $f(x+y)+f(x-y)=2(f(x)+f(y))$ Question: Find all "tame" solutions of $f(x+y)+f(x-y)=2(f(x)+f(y))$. 

This is how I've tried to begin—
Plug in $y=0$ to get $f(0)=0$. 
$y\to x$ gives $f(2x)=2^2f(x)$. Also, I've noted that $f$ is an even function by $y\to -x$ substitution. 
Now, I somewhat think that $f(kx)=k^2f(x)$ but I'm not sure how to land onto there. 
And at the end of the day, I believe $f(x)=ax^2$ but I cannot proceed into that stage. 
 A: You have made a good start, but not sure where to go from there. My approach may help you.
Let $$e(x,y) := f(x+y)+f(x-y)-2(f(x)+f(y)). $$
We are given that $\;e(x,y)=0\;$ for all $\;x,y.\;$
Since $\;e(0,0) = -2f(0),\;$ then $\;f(0)=0.\;$ Since $\;e(0,x) = -2f(0) + f(-x) -f(x),\;$ then $\;f(-x) = f(x).\;$ Let
$$ g(x,y) := f(x+y)-f(x)-f(y). $$
Now evaluate to get
$$ 2(g(x_1,y)+g(x_2,y)-g(x_1+x_2,y)) = \\
e(x_1,x_2)+e(x_1+x_2+y,y)-e(x_1+y,x_2+y), $$
but this is $0$ by property of $\;e(x,y).\;$
Thus, $\;h(x):=g(x,y)\;$ satisfies Cauchy's functional equation, and assuming $h(x)$ is "tame", then
$\;h(x) = g(x,y) = x\, c(y)$ for some function $c(y).$ Using the same reasoning with
$\;h(y) := g(x,y)\;$ we have $\;g(x,y)=y\,d(x)\;$
for some function $d(x).$ Thus,
$\;g(x,y) = t\,xy\;$ for some constant $t.\;$ Now $\;\lim_{y\to0} g(x,y)/y = t\,x $
and, assuming $f'(0)$ exists, by definition, also $\lim_{y\to0} g(x,y)/y = f'(x)-f'(0).\;$ But $f(-x)=f(x)$ so $f'(0)=0$ and thus $f'(x)=t\,x\;$
and $f(x)=t\,x^2/2.$
