Functional equation in $\displaystyle \mathbb{Z} \to \mathbb{Z}$: $2014f(f(x)) + 2013f(x) = x$

Question

Solve in $\mathbb{Z} \to \mathbb{Z}$ the following functional equation: $$2014f(f(x)) + 2013f(x) = x$$

Modulo $2014$, we have that: $$f(x) \equiv -x \pmod {2014}$$

Modulo $2013$ we have that: $$f(f(x)) \equiv x \pmod {2013}$$

By easy computations, we may prove that: $$f(x) \equiv -x \ \pmod {2013}$$

So, $f \equiv -\text{Id}$ on $\{1, 2, \cdots, 2013 \cdot 2014\}$. However, I am unable to extend on $\mathbb{Z}$. (Another easy observation is that $f$ is injective)

Answer 1

If we rewrite your equation as $$ 2014 ( f(f(x)) + f(x)) = f(x) + x$$ we see that $f(x) + x$ is divisible by $2014$. In fact repeating with $f(x)$ in place $x$ we find that $f(x) + x$ is divisible by $2014^2$. By simple induction we see that $f(x) + x$ is divisible by any power of $2014$, hence it must be zero since $f(x)+x$ was assumed to be an integer.