TITLE: A simple but weird functional equation
QUESTION [7 upvotes]: Let $f$ be a function $f:\mathbb R\to\mathbb R$.
Find all functions $f$ that satisfy:
$$f(x^2+x+3)+2f(x^2-3x+5)=x^2-x+ \frac{18}{4} +  \frac{111}{444} +  \frac{222}{333}$$
Maybe the question is wrong; I found it on a forum on the internet without any answers.

Note: The question has no solutions, I'm sorry.  On the forum someone says he made it so probably there are no functions that satisfy this.

REPLY [2 votes]: It happens that this question is pure algebra...

No solution $f$ (whatever its regularity) can be defined on the five points $\frac{11}4$, $\frac{11}4\pm4$, $\frac{11}4\pm16$.

Consider the identity to be solved when the RHS is $x^2-x+c$, for some $c$ whose value will be irrelevant, and let $a=\tfrac34+c$. Some tedious but easy algebraic computations show that the sequence $(x_n)$ defined by
$$
x_n=f\left(4n^2+\tfrac{11}4\right)
$$ 
solves the identities
$$
x_n+2x_{n-1}=4n(n-1)+a.
$$
For example, $x_1=a-2x_0$, $x_2=8-a+4x_0$, $x_{-1}=\frac12a-\frac12x_0$ and $x_{-2}=4+\frac14a+\frac14x_0$.
By definition, $x_{-n}=x_n$ for every $n$, in particular,
$$
a-2x_0=\tfrac12a-\tfrac12x_0,\qquad 8-a+4x_0=4+\tfrac14a+\tfrac14x_0.
$$
The first identity implies that $a=3x_0$, and the second identity reads $4+\frac{15}4x_0=\frac54a$, which has no solution when the first identity holds since then, $\frac{15}4x_0=\frac54a$.