Problem 1651

1. **Uniqueness of $q_i$:**
- Given that $q_i$ are the inverses modulo $p$ of $p_i$, it is clear that $q_i$ are distinct. This is because if $q_i \equiv q_j \pmod{p}$ for $i \neq j$, then $p_i \equiv p_j \pmod{p}$, which contradicts the fact that $p_i$ are distinct primes.

2. Condition for common elements:
- For a given $a$ and $b$, there exists at most one index $i$ such that $q_i \equiv aq_i + b \pmod{p}$. This can be seen by solving the equation $q_i \equiv aq_i + b \pmod{p}$, which simplifies to $q_i (a - 1) \equiv b \pmod{p}$. If $a \neq 1$, this has a unique solution $q_i \equiv b(a-1)^{-1} \pmod{p}$.

3. Assumption of at least 4 common elements:
- Suppose the sets $\{r(q_1), r(q_2), \ldots, r(q_m)\}$ and $\{r(aq_1 + b), r(aq_2 + b), \ldots, r(aq_m + b)\}$ have at least 4 common elements. Then there exist indices $i_1, i_2, i_3, i_4$ and $j_1, j_2, j_3, j_4$ such that $q_{i_k} \equiv aq_{j_k} + b \pmod{p}$ for $k = 1, 2, 3, 4$.

4. Deriving a contradiction:
- From the above, we have:
$$a \equiv \frac{q_{i_1} - q_{i_2}}{q_{j_1} - q_{j_2}} \equiv \frac{q_{i_1} - q_{i_3}}{q_{j_1} - q_{j_3}} \pmod{p}$$
This implies:
$$(q_{i_1} - q_{i_2})(q_{j_1} - q_{j_3}) \equiv (q_{i_1} - q_{i_3})(q_{j_1} - q_{j_2}) \pmod{p}$$

5. Multiplying both sides:
- Multiplying both sides by $p_{i_1} p_{i_2} p_{i_3} p_{j_1} p_{j_2} p_{j_3}$:
$$p_{i_3} p_{j_2} (p_{i_2} - p_{i_1})(p_{j_3} - p_{j_1}) \equiv p_{i_2} p_{j_3} (p_{i_3} - p_{i_1})(p_{j_2} - p_{j_1}) \pmod{p}$$

6. Absolute value consideration:
- Both numbers in absolute value are smaller than $\frac{1}{2}p$, because $p_i < \sqrt[4]{\frac{1}{2}p}$ for all these primes. This means we have equality between those numbers:
$$p_{i_3} p_{j_2} (p_{i_2} - p_{i_1})(p_{j_3} - p_{j_1}) = p_{i_2} p_{j_3} (p_{i_3} - p_{i_1})(p_{j_2} - p_{j_1})$$

7. Contradiction:
- Since $p_{i_3}$ does not divide any of these numbers: $p_{j_2}$, $p_{j_3}$, and $p_{i_3} - p_{i_1}$, it follows that $p_{i_3} \mid p_{j_2} - p_{j_1}$. Thus:
$$p_{i_3} \leq |p_{j_2} - p_{j_1}| < \max\{p_{j_2}, p_{j_1}\} \leq \max \{p_{i_1}, p_{i_2}, p_{i_3}, p_{j_1}, p_{j_2}, p_{j_3}\}$$
- This leads to a contradiction because there is some $p$ which is the maximum.

Therefore, the sets $\{r(q_1), r(q_2), \ldots, r(q_m)\}$ and $\{r(aq_1 + b), r(aq_2 + b), \ldots, r(aq_m + b)\}$ can have at most 3 common elements.

$\blacksquare$