elementary number theory - A question on the remainders of integer division

This is a question on the remainders of integer division from my student.

Notations.

Let $p$ be a positive odd prime integer.
We write $r_{i,j}$ for the remainder of $i \times j \div p$.
Now for an integer $u \in \{1, 2, \dots, p-2\}$, we define
\begin{align*}
K_1 =&\{ i : 1 \le i \le p-2,\ r_{i,u}+uK_2 =&\{ i : 1 \le i \le p-2,\ r_{i,u}+uu+1\}, \\
K_3 =&\{ i : 1 \le i \le p-2,\ r_{i,u}+u>p,\ r_{i,p-u-1} K_4 =&\{ i : 1 \le i \le p-2,\ r_{i,u}+u>p,\ r_{i,p-u-1} >u+1\}.
\end{align*}
And for another $u' \in \{1, 2, \dots, p-2\}$, we can define $K_\nu'$ for $\nu=1,2,3,4$.

Question.
If $K_1 =K_1'$ and $K_4=K_4'$, then either $K_2=K_2'$ and $K_3=K_3'$, or $K_2=K_3'$ and $K_3=K_2'$.

What I have done.

1.
With the help of computer, I checked the correctness of this question for $p<100$.

2.

Still with the help of computer, I found that, at least for $p<100$, if $K_1 =K_1'$ and $K_4=K_4'$, then either $u \equiv u' \pmod{p}$ or $u+u' \equiv p-1 \pmod{p}$.
The first case corresponds to the result $K_2=K_2'$ and $K_3=K_3'$, and the second one corresponds to the result $K_2=K_3'$ and $K_3=K_2'$.

3.
I observed the definitions of the $K_\nu$'s, and found some equivalent conditions.
Firstly, we have
\begin{align*}
r_{i,u}+u

r_{i,u}+u>p &\quad\Leftrightarrow\quad r_{i+1,u} =r_{i,u} +u-p,
\end{align*}

and similarly,
\begin{align*}
r_{i,p-u-1} r_{i,p-u-1} >u+1 &\quad\Leftrightarrow\quad r_{i+1,p-u-1} =r_{i,p-u-1} +(p-u-1) -p.
\end{align*}

4.
Using the discussion in (3), we obtain
\begin{align*}
i \in K_1

&\;\Leftrightarrow\; r_{i,u}+r_{i,p-u-1}

p, \\
i \in K_4
&\;\Leftrightarrow\; r_{i,u}+r_{i,p-u-1}>p \text{ but } r_{i+1,u}+r_{i+1,p-u-1}i \in K_2 \cup K_3
&\;\Leftrightarrow\; \text{both $r_{i,u}+r_{i,p-u-1}$ and $r_{i+1,u}+r_{i+1,p-u-1}$ are $p$}.
\end{align*}

5.
Now by the result in (4), we can see that $K_1 =K_1'$ and $K_4=K_4'$ if and only if for every $i \in \{1, 2, \dots, p-1\}$, the signatures of $r_{i,u}+r_{i,p-u-1}-p$ and $r_{i,u'}+r_{i,p-u'-1}-p$ are the same (i.e., both positive or both negative).

However, I really do not know how to solve this question, so if you have any idea, please help me.
Thank you very much for your attention.

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Any ideas or hints are welcome. Please.