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

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The discussion focuses on identifying functions \( f \) that satisfy the equation \( f^d(x) = 2015 - x \) for all divisors \( d \) of 2015 greater than 1 and for all real numbers \( x \). Participants explore the implications of this functional equation and its recursive nature. A correct solution was provided by user castor28, which is acknowledged in the thread. The problem emphasizes the relationship between the function and its iterations, leading to insights about the structure of potential solutions. The thread concludes with the official solution shared below.
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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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Congratulations to castor28 for his correct solution, which you can find below:
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:
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 $f^d(x)=f(f^{2c}(x))=f(x)=2015-x$.
 
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