The functional equation $x(x+1)+C(x)=(x+1)(x+2)+C(x+1)=(x+2)(x+3)+C(x+2)=...$

Question

Consider the functional equation $$x(x+1)+C(x)=(x+1)(x+2)+C(x+1)=(x+2)(x+3)+C(x+2)=...$$ The equality continues to infinity.

Is there $C(x)$ that satisfies all the equality? If there is, what is it?

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I missed the simple solution $C(x)=c-x(x+1)$. My main question though is whether there are other solutions.

Answer 1

We describe the general solution. For $0\le t\lt 1$, let $g(t)$ be arbitrary. For any real number $x$, let $f(x)=g(\{x\})$, where $\{x\}$ is the fractional part of $x$. Let $C(x)=f(x)-x(x+1)$.

Answer 2

Yes, there is.

$$C(x)=-x(x+1)$$

Note that this makes $0$ every term of the "infinite" equation.

Note also that this solution is far from being unique.