Question 10
For any prime $p$ and any integer $n$, let $d_p(n)\in\{0,1,\ldots,p-1\}$ denote the remainder when $n$ is divided by $p$. A sequence $(a_0,a_1,a_2,\ldots)$ of positive integers is called a $p$-sequence if $a_0$ is coprime to $p$ and
$$
a_{n+1}=a_n+d_p(a_n)\qquad(n\ge0).
$$
Part (a). Are there infinitely many primes $p$ for which there are two $p$-sequences $(a_n)$ and $(b_n)$ such that $a_n\gt b_n$ for infinitely many $n$, and $b_n\gt a_n$ for infinitely many $n$?
Part (b). Are there infinitely many primes $p$ for which there are two $p$-sequences $(a_n)$ and $(b_n)$ such that $a_0\lt b_0$, but $a_n\gt b_n$ for every $n\ge1$?
Source-language clarification: the Chinese statement allows any prime, whereas its English version says “odd prime”. Both readings are addressed in the solution; including $p=2$ does not change either answer.
$$
a_{n+1}=a_n+d_p(a_n)\qquad(n\ge0).
$$
Part (a). Are there infinitely many primes $p$ for which there are two $p$-sequences $(a_n)$ and $(b_n)$ such that $a_n\gt b_n$ for infinitely many $n$, and $b_n\gt a_n$ for infinitely many $n$?
Part (b). Are there infinitely many primes $p$ for which there are two $p$-sequences $(a_n)$ and $(b_n)$ such that $a_0\lt b_0$, but $a_n\gt b_n$ for every $n\ge1$?
Source-language clarification: the Chinese statement allows any prime, whereas its English version says “odd prime”. Both readings are addressed in the solution; including $p=2$ does not change either answer.