Find Functions Satisfying $f^d(x)=2015-x$ for All Divisors of 2015

[anemone]

Here is this week's POTW:

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Denote by $f^n(x)$ the result of applying the function $f$ $n$ times to $x$ (e.g. $f^1(x)=f(x),\,f^2(x)=f(f(x)),\,f^3(x)=f(f(f(x)))$ Find all functions from real numbers to real numbers which satisfy $f^d(x)=2015-x$ for all divisors $d$ of 2015, which are greater than 1, and for all real $x$.

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[anemone]

Congratulations to castor28 for his correct solution, which you can find below:

Spoiler

We note first that $5$ and $13$ are divisors of $2015$.

We have:
\begin{align*}
f^5(x) &= 2015 - x\\
f^{10}(x) &= f^5(f^5(x)) \\
&= 2015 - f^5(x) \\
&= 2015 - (2015-x) \\
&= x
\end{align*}

We can continue using the same trick:

\begin{align*}
f^{13}(x) &= f^3(f^{10}(x)) = f^3(x) = 2015 - x\\
f^5(x) &= 2015 - x \\
&= f^3(f^2(x)) \\
&= 2015 - f^2(x)
\end{align*}

Which implies $f^2(x) = x$.
We can finally conclude: $f^3(x) = 2015 - x = f(f^2(x)) = f(x)$.

Official solution:

Spoiler

Since 5 is a divisor of 2015, we have for any real $z$:
$f^{25}(z)=f^5(f^5(f^5(f^5(f^5(z)))))=2015-f^5(f^5(f^5(f^5(z))))=2015-(2015-f^5(f^5(f^5(z))))=f^5(f^5(f^5(z)))=\cdots=f^5(z)=2015-z$.

Since 13 is also a divisor of 2015, we have for any real $z$:
$f^{26}(z)=f^{13}(f^{13}(z))=f(2015-z)$.

Consequently, $z=f^{26}(z)=f(f^{25}(z))=f(2015-z)$. Any real number $x$ can be written as $2015-z$ for $z=2015-x$. Hence, $z=f(2015-z)$ implies $f(x)=2015-x$ for any real $x$.

Finally, check that the function $f(x)=2015-x$ satisfies the conditions of the problem. Let $d$ be a divisor of 2015 greater than 1. Then $d$ is odd, i.e. $d=2c+1$ for a positive integer $c$. Since $f^2(x)=2015-(2015-x)=x$, we have $f^{2c}(x)=f^2(f^2(\cdot f^2(x)\cdots))=x$, which implies $f^d(x)=f(f^{2c}(x))=f(x)=2015-x$.