similar logic to fibonacci seq periodic modulo $m$ so we can apply the same logic to get periodic modulo $10^4$

augustin_p wrote: Consider the sequence $(x_n)_{n\in\mathbb{N^*}}$ such that $$x_0=0,\quad x_1=2024,\quad x_n=x_{n-1}+x_{n-2}, \forall n\geq2.$$Prove that there is an infinity of terms in this sequence that end with $2024.$ Could you please provide me with the full question? (An image file is also acceptable.) Alternatively, could you let me know which question number this is and the date, so I can cite the source accurately? Thank you!

We have the sequence $(x_n)_{n\in\mathbb{N^*}}$ such that $$x_0=0,\quad x_1=2024,\quad x_n=x_{n-1}+x_{n-2}, \forall n\geq2.$$ Consider the pairs $\left(x_k, x_{k+1}\right) \pmod{10^4}$, yielding ordered pairs of residues. If we were to pick $10^8+1$ distinct pairs, then by Pigeonhole principle, we are guaranteed to find $m\neq n$ such that $x_m\equiv x_n \pmod{10^4}$ and $x_{m+1}\equiv x_{n+1} \pmod{10^4}$. WLOG, say, $m>n$. We can prove by induction that $x_{m-k}\equiv x_{n-k} \pmod{10^4}$ for $k\leq n$. Thus, $x_{m-n+1}\equiv x_1\equiv 2024 \pmod{10^4}$. We can safely repeat this process infinitely many times by choosing $10^8+1$ different pairs to vary $m-n$ and find an infinite subsequence of numbers ending with $2024$. $\blacksquare$

I got another solution. Starting with assuming otherwise. Like the number of n is not infinitive ($ \equiv 2024 $mod$ 10^4$). I assumed that $a_k$ is the last number which pays this condition. (Giving $2024$ mod $10^4$) Then say $x_{k-1}=a_k$ , $x_{k-2}=b_k$ and find $x_{k+1}$, $x_{k+2}$, for $ \geq k+1$ , it can not give $2024$ for $mod$ $10^4$. After some equation solvings, we will get a contradiction. Which finishes the proof.