$f \colon \mathbb{R} \rightarrow \mathbb{R}$, $f$ is continuous , and $f(x+1)+f(x)=x^2$ I would appreciate if somebody could help me with the following problem:

Find $f(x)$ ($f(x)$ is not Polynomial function), given that:
$f \colon \mathbb{R} \rightarrow  \mathbb{R}$, $f$ is continuous, and $f(x+1)+f(x)=x^2$

I tried  but couldn't get it that way.

It might be helpful to first find $f:\Bbb N \to \Bbb R$, then extend the result.
We can solve the recurrence relation $f(n) + f(n+1) = n^2$ via generating functions: define $F(x) = \sum_{n=0}^\infty f(n) x^n$.  Then
$$
f(n) + f(n+1) = n^2\\
\sum_{n=0}^\infty [f(n) + f(n+1)]x^n = \sum_{n=0}^\infty n^2 x^n \\
\sum_{n=0}^\infty f(n)x^n +
\sum_{n=0}^\infty f(n+1)x^n = 
\sum_{n=0}^\infty n^2 x^n\\ 
\sum_{n=0}^\infty f(n)x^n +
\sum_{n=1}^\infty f(n)x^{n-1} = 
\sum_{n=0}^\infty n^2 x^n\\ 
F(x) + 
\frac 1x \left(F(x) - f(0)\right) = 
\frac{x(x+1)}{(x-1)^3}\\
xF(x) + 
F(x) - f(0) = 
\frac{x^2(x+1)}{(x-1)^3}\\
(x+1)F(x) = 
\frac{x^2(x+1)}{(x-1)^3} + f(0)\\
F(x) = 
\frac{x^2}{(x-1)^3} + \frac{f(0)}{x+1}\\
\sum_{n=0}^\infty f(n) x^n = 
\sum_{n=0}^\infty \frac{n(n+1)}{2} x^n + 
f(0)\sum_{n=0}^\infty (-1)^n x^n
$$
So setting $f(0) = a$ for any $a$, we find
$$
f(n) = \frac{n(n+1)}{2} + a(-1)^n = 
\frac{n(n+1)}{2} + a\cos(\pi n)
$$
if a solution to the problem over $\Bbb R$ exists, it must be an extension of this function for some value of $a$.
In fact, $f(x) = \frac{x(x+1)}{2} + a\cos(\pi x)$ seems to work for any $a \in \Bbb R$.

A more general solution (taken from Salomo's work) is
$$
f(x) = \frac{x(x+1)}{2} + g(x)
$$
where $g(x)$ satisfies $g(x+1) + g(x) = 0$.
