Let p>=3 be a prime number. Show that whatever the distict numbers x_1,x_2,...,x_(p-1) beloning to the set {1,2,...,p-1}, there exists a pair Let $p\geq3$ be a prime number. Show that whatever the distinct numbers $x_1, x_2, \ldots , x_{p−1}$ belonging to the set $\{1, 2, \ldots , p−1\}$, there exists a pair $(i, j)$ with $1 \leq i <\ j \leq p−1$ such that $p\vert ix_i − jx_j$.
I've observed that from Bezout's Theorem it is clear that for any number $h\in\{1, 2, \ldots, p−1\}$ there exists a $h'\in\{1, 2, \ldots , p−1\}$ such that $h'h\equiv1 \pmod{p}$.
I have also observed that if $h\in\{1, 2, \ldots , p−1\}$ then $h,2h,\ldots(p-1)h$ have all different ressidues when divided with p (all non-zero). This can be easily deduced if we assume the contrary.
I've just found this this question that uses Wilson's theorem that would also solve my problem. But I do not seem to be able to understand the way in which it is used.

By Wilson's theorem:
$$1\cdot 2 \dotsb (p-1) \equiv \underbrace{x_1 \cdot x_2 \dotsb x_{p-1}}_{\text{any permutation of } \{1,2, \ldots ,p-1\}} \equiv -1 \pmod{p}.$$
If $\{ix_i\}_{i=1}^{p-1}$ was a permutation of $1,2, \ldots ,p-1$, then By Wilson's theorem, it should have also been equivalent to $-1 \pmod{p}$. However,
$$(1x_1)(2x_2) \dotsb ((p-1)x_{p-1}) \equiv (1\cdot 2 \dotsb (p-1))\,\,(x_1 \cdot x_2 \dotsb x_{p-1}) \equiv (-1)^2 \pmod{p}.$$
Thus, $\{ix_i\}_{i=1}^{p-1}$ is not a permutation of $1,2, \ldots, p-1$. It is clear that for all $1 \leq i \leq p-1$, $ix_i \not\equiv 0 \pmod{p}$. Thus, the number of residues mod $p$ in $\{ix_i\}_{i=1}^{p-1}$ is no more than $p-2$.
