Question 34: Let prime $p \geq 5$, for a positive integer $k$, define $R(k)(0 \leq R(k) \leq p-1)$ as the remainder when $k$ is divided by $p$. Try to find all positive integers $a$, such that for every $m \in\{1,2, \ldots, p-1\}$, we have $m+R(m a)>a$.


Question 34:
Solution: When $a \geq p$, take $m=1$, then $m+R(m a) \leq 1+(p-1)=p \leq a$, which does not satisfy the condition. When $a=1$, for any $m \in\{1,2, \ldots, p-1\}, p \nmid m a$, hence $R(m a) \geq 1$. Therefore, $m+R(m a) \geq 1+1 > a$, which satisfies the condition.
Now assume $2 \leq a \leq p-1$.
Let $p=a s+r\left(s=\left[\frac{p}{a}\right] \in \mathbb{N}, 0 \leq r \leq a-1\right)$, then $r \neq 0$.
Notice that $R(m a)=m a-p\left[\frac{m a}{p}\right]$, so
$$
m+R(m a)>a
$$
$$
\begin{array}{l}
\Leftrightarrow m+m a-p\left[\frac{m a}{p}\right]>a \\
\Leftrightarrow m+(m-1) a>p\left[\frac{m a}{p}\right]
\end{array}
$$

First, we take $m=s+1$, then $\left[\frac{m a}{p}\right]=\left[\frac{(s+1) a}{p}\right] \geq 1$. Substituting into (1) we get
$$
(s+1)+s a>p \Rightarrow s+1>r \Rightarrow s \geq r
$$

We now prove: If $r \leq s$, then the corresponding $a$ satisfies the condition.
In fact, for any given $m \in\{1,2, \ldots, p-1\}$, let $\left[\frac{m a}{p}\right]=t$. Since $p \nmid m a$, we have $\frac{m a}{p} > \left[\frac{m a}{p}\right]=t$. Therefore,
$$
\begin{array}{l}
m a-1 \geq p t=(a s+r) t \\
\Rightarrow m \geq s t+\left\lceil\frac{r t+1}{a}\right\rceil \geq s t+1
\end{array}
$$

Thus,
$$
m+(m-1) a \geq s t+1+s t a > r t+s t a=(s a+r) t=p\left[\frac{m a}{p}\right]
$$

Therefore, (1) holds.
According to the above analysis, $a(2 \leq a \leq p-1)$ satisfies the condition if and only if $s \geq r \Leftrightarrow\left[\frac{p}{a}\right] \geq p-a\left[\frac{p}{a}\right] \Leftrightarrow s=\left\lfloor\frac{p}{a}\right\rfloor \geq \frac{p}{a+1}$.
When $s=1$, $\frac{p}{a+1} \leq 1 \Rightarrow a=p-1$, which satisfies the condition.
When $2 \leq s \leq p-1$, $\frac{p}{a+1} \leq s \Leftrightarrow a s<p<(a+1) s \Leftrightarrow a=\left[\frac{p}{s}\right]$.
In summary, the required $a=p-1$ or $\left[\frac{p}{s}\right](2 \leq s \leq p-1)$, which is $a=\left[\frac{p-1}{s}\right]$ for $1 \leq s \leq p-1$.