Tintarn wrote: Let $x_0=2024^{2024}$ and $x_{n+1}=|x_n-\pi|$ for $n \ge 0$. Show that there exists a value of $n$ such that $x_{n+2}=x_n$. Let $n$ be number such a $\pi>x_n>0$ (obviously, such $n$ exist). Then $x_{n+1}=\pi - x_n < \pi$, so $x_{n+2}=\pi-x_{n+1}=\pi-(\pi-x_n)=x_n$.

Since the first couple of terms of the sequence are decreasing, we must have a some $n$ such that $\pi\geq x_n$. So $$x_{n+2}=|x_{n+1}-\pi|=||x_n-\pi|-\pi|=|\pi-x_n-\pi|=x_n$$thus we are done.

Note that until $x_n>\pi$, it decreases by $\pi$, so it eventually becomes less than $\pi$. But then $x_{n+1}=\pi-x_n$ and $x_{n+2}=\pi-x_{n+1}=x_n$

Choose the $x_i$ with $0 < x_i < \pi$ (which exists due to $2024^{2024}$ not being an integer multiple of $\pi$). Then $x_{i+1} = \pi - x_i$, so $0 < x_{i+1} < \pi$. Therefore, $x_{i+2} = \pi - x_{i+1} = x_i$, yielding the result. $\square$

Let $x_n$ be the first term less in magnitude than $\pi$, then $x_{n+1}= \pi -x_n$ so $x_{n+2}= |-x_n|$ . Note that $x_n$ must be positive so $x_n=x_{n+2}$.