<s>[INST] $f(x+1) + f(x-1) = x^2$ ; $f(x+4) + f(x-4) = 2\sin x$ , then $f(x) =$? Given $f$ is a complex valued function satisfying $$f(x+1) + f(x-1) = x^2 \\ 
f(x+4) + f(x-4) = 2\sin x$$
what is $f(x)$ ?
Here,  only for the first part MathWolfram alpha is showing $f(x)$ to be of type $c_1(i)^x + c_2(-i)^x + \dfrac{x^2 - 1}{2}$ but how are they introducing "$i$"? I used a generating function but was still unable to get those terms containing $i$ . Please help me with this problem.  
 [/INST] Let's do two change of variables. First, $u = x-1$:
$$f(u+2) = (u+1)^2 - f(u)$$
Let's apply that multiple times:
$$\begin{align}
f(u+4) &= (u+3)^2 - f(u+2) = (u+3)^2 - (u+1)^2 + f(u)\\
f(u+6) &= (u+5)^2 - f(u+4) = (u+5)^2 - (u+3)^2 + (u+1)^2 - f(u)\\
f(u+8) &= (u+7)^2 - f(u+6) = (u+7)^2 - (u+5)^2 + (u+3)^2 - (u+1)^2 + f(u)
\end{align}$$
Which can be simplified to:
$$f(u+8) = 8u + 32 + f(u)$$
However we can also do the variable substitution $u = x - 4$ in our second identity:
$$f(u+8) = 2\sin (u+4) - f(u)$$
Giving equation:
$$8x + 32 + f(x) = 2\sin (x+4) - f(x)$$
$$2f(x) = 2\sin (x+4) - 8x - 32$$
$$f(x) = \sin (x+4) - 4x - 16$$
</s>