Problem 1692

Prove that dividing the given $p+1$ numbers by their greatest common divisor obviously does not affect the conclusion of this problem, so we can assume that these $p+1$ numbers are coprime. In particular, there must be a number that is not divisible by $p$. Let these $p+1$ numbers be
$$x_{1}, \cdots, x_{k}, x_{k+1}=p^{l_{k+1}} y_{k+1}, \cdots, x_{p+1}=p^{l_{p+1}} y_{p+1}$$

Here, $\dot{x}_{1}, \cdots, x_{k}$ are distinct and coprime with $p$ $(k \geqslant 1)$, $l_{k+1}, \cdots, l_{p+1}$ are positive integers, and $y_{k+1}, \cdots, y_{p+1}$ are positive integers not divisible by $p$.

Among the $p+1$ numbers
$$x_{1}, \cdots, x_{k}, y_{k+1}, \cdots, y_{p+1}$$

there must be two that are congruent modulo $p$. We discuss three cases.

(1) At least three numbers in (1) are equal. In this case, the conclusion is easy to prove. If $y_{r}=y_{s}=y_{t}$, then $p^{l_{r}}, p^{l_{s}}, p^{l_{t}}$ are distinct, and the largest number is at least $p^{2}$ times the smallest. Without loss of generality, assume $p^{l_{r}} \geqslant p^{2} \cdot p^{l_{t}}$, then $x_{r}$ and $x_{t}$ satisfy the condition; if $y_{r}=y_{s}=x_{t}(1 \leqslant t \leqslant k)$, without loss of generality, assume $l_{r}>l_{s}$, then $l_{r} \geqslant 2$, so $x_{r}$ and $x_{t}$ satisfy the condition.

(2) Two pairs of numbers in (1) are equal. If $y_{i}=y_{j}, y_{r}=y_{s}$, then when $\left|l_{i}-l_{j}\right| \geqslant 2$ or $\left|l_{r}-l_{s}\right| \geqslant 2$, the conclusion holds as above; when $\left|l_{i}-l_{j}\right| \leqslant 1$ and $\left|l_{r}-l_{s}\right| \leqslant 1$, we can rename $x_{i}, x_{j}, x_{r}, x_{s}$ as $a, a p, b, b p$, and $ap$

Thus, the integer $\frac{b p}{(a, b p)} \geqslant p+1$.
If $x_{i}=y_{r}, x_{j}=y_{s}(1 \leqslant i, j \leqslant k)$, the conclusion can also be proven similarly.

(3) Exactly two numbers in (1) are equal. This can only be $y_{r}=y_{s}$, or $x_{i}=y_{r}(1 \leqslant i \leqslant k)$. In this case, we can remove $y_{r}$ from (1), leaving $p$ numbers that are distinct but still have two that are congruent modulo $p$. Now there are three possibilities:
(i) Suppose $y_{r} \equiv y_{s}(\bmod p)$. Without loss of generality, assume $y_{r}>y_{s}$. If $l_{r}>l_{s}$, the conclusion is obvious. If $l_{r} \leqslant l_{s}$, let $y_{r}=y_{s}+n$, then $n>0$, and $p \mid n$. Let $\left(y_{r}, y_{s}\right)=d$, then $p \nmid d$, so $\left(x_{r}, x_{s}\right)=p^{l} d$. We have (note $d|n, p| n$, and $p \nmid d$)
$$\frac{x_{r}}{\left(x_{r}, x_{s}\right)}=\frac{y_{r}}{d}=\frac{y_{s}}{d}+\frac{n}{d} \geqslant 1+p$$

Thus, the larger of $x_{r}$ and $x_{s}$, when divided by their greatest common divisor, yields a quotient of at least $p+1$.
(ii) Suppose $x_{r} \equiv x_{s}(\bmod p)(1 \leqslant r < s \leqslant k)$. Without loss of generality, assume $x_{r}>x_{s}$. If $y_{s} < y_{r}$, the conclusion is obvious. If $y_{s} > y_{r}$, let $y_{s}=y_{r}+n$, then $n>0$, and $p \mid n$. Let $\left(x_{r}, y_{s}\right)=d$, then $p \nmid d$, so $\left(x_{r}, x_{s}\right)=\left(x_{r}, p^{l_{s}} y_{s}\right)=d$, thus
$$\frac{x_{r}}{\left(x_{r}, x_{s}\right)}=\frac{y_{s}}{d}+\frac{n}{d} \geqslant 1+p$$

Thus, the larger of $x_{r}$ and $x_{s}$, when divided by their greatest common divisor, yields a quotient of at least $p+1$. This completes the proof of the problem.

We note that if the $p+1$ integers in Example 9 are replaced by $p$ integers, the conclusion may not hold. For example, the $p$ numbers $1,2, \cdots, p$ clearly do not have two numbers that satisfy the condition.