<s>[INST] A function $f(x)$ on the real domain and real constants $a$ and $b\neq 0$ for which $f(x)-f(x-\delta)+a+bx^2=0$ for some real $\delta\neq 0$ Is there a function $f(x)$ on the real domain and real constants $a$ and $b\neq 0$ for which the following is true:
$$f(x)-f(x-\delta)+a+bx^2=0$$
for some real $\delta\neq 0$?
EDIT: I missed a very important constraint in the original posting of this question: $b\neq 0$. Sorry!
 [/INST] $f(x)-f(x-\delta)+a+bx^2=0$
$f(x)-f(x-\delta)=-bx^2-a$
In fact this belongs to a functional equation of the form http://eqworld.ipmnet.ru/en/solutions/fe/fe1108.pdf.
The general solution of this functional equation is $f(x)=\Theta(x)+f_p(x)$ , where $\Theta(x)$ is an arbitrary periodic function with period $\delta$
Luckily we can find $f_p(x)$ by method of undetermined coefficients:
Let $f_p(x)=Ax^3+Bx^2+Cx$ ,
Then $f_p(x-\delta)=A(x-\delta)^3+B(x-\delta)^2+C(x-\delta)=Ax^3-3A\delta x^2+3A\delta^2x-A\delta^3+Bx^2-2B\delta x+B\delta^2+Cx-C\delta=Ax^3+(B-3\delta A)x^2+(3\delta^2A-2\delta B+C)x-\delta^3A+\delta^2B-\delta C$
$\therefore Ax^3+Bx^2+Cx-(Ax^3+(B-3\delta A)x^2+(3\delta^2A-2\delta B+C)x-\delta^3A+\delta^2B-\delta C)\equiv-bx^2-a$
$3\delta Ax^2+(2\delta B-3\delta^2A)x+\delta^3A-\delta^2B+\delta C\equiv-bx^2-a$
$\therefore\begin{cases}3\delta A=-b\\2\delta B-3\delta^2A=0\\\delta^3A-\delta^2B+\delta C=-a\end{cases}$
$\begin{cases}A=-\dfrac{b}{3\delta}\\B=-\dfrac{b}{2}\\C=-\dfrac{b\delta}{6}-\dfrac{a}{\delta}\end{cases}$
$\therefore f(x)=\Theta(x)-\dfrac{bx^3}{3\delta}-\dfrac{bx^2}{2}-\dfrac{b\delta x}{6}-\dfrac{ax}{\delta}$ , where $\Theta(x)$ is an arbitrary periodic function with period $\delta$
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