Let $\mathrm{p}$ be a prime number at least three and let $\mathrm{k}$ be a positive integer smaller than $\mathrm{p}$. Given $\mathrm{a}_1, \ldots, \mathrm{a}_{\mathrm{k}} \in \mathbb{F}_{\mathrm{p}}$ and distinct elements $\mathrm{b}_1, \ldots, \mathrm{b}_{\mathrm{k}} \in \mathbb{F}_{\mathrm{p}}$, prove that there exists a permutation $\sigma$ of $[\mathrm{k}]$ such that the values of $\mathrm{a}_{\mathrm{i}}+\mathrm{b}_{\sigma(\mathrm{i})}$ are distinct modulo $\mathrm{p}$.
I am not sure if the below solution logic contains fallacy or not, but this is my progress so far:
We will prove this by induction on $k$.
Base Case:
Let's start with the case where $k=1$.
We only have one possible permutation here, $\sigma(1) = 1$, and since $a_1, b_1 \in \mathbb{F}_p$, $a_1 + b_1 \mod p$ is unique. So, the claim holds for $k=1$.
Induction Step:
Assume that the claim holds for some $k < p$. We will show that it also holds for $k+1$.
Consider $k+1$ values of $a$ and $b$, denoted as $a_1, a_2, \ldots, a_{k+1}$ and $b_1, b_2, \ldots, b_{k+1}$.
For every permutation $\sigma$ of $[k]$, $a_i + b_{\sigma(i)}$ are distinct modulo $p$ for $i = 1, 2, \ldots, k$ by our induction hypothesis.
Now, we consider the value $a_{k+1} + b_i$ for each $i \in [k+1]$. If for every $i$, $a_{k+1} + b_i \mod p$ is different from all $a_j + b_{\sigma(j)} \mod p$ where $j \in [k]$, then we can simply append $i$ to the permutation $\sigma$ to form a permutation of $[k+1]$, and we are done.
Otherwise, suppose that there exists an $i$ such that $a_{k+1} + b_i \equiv a_j + b_{\sigma(j)} \mod p$ for some $j \in [k]$. We can write this as $a_{k+1} - a_j \equiv b_{\sigma(j)} - b_i \mod p$.
Because $a_{k+1} \neq a_j$ and $b_i$ is distinct from $b_{\sigma(j)}$, the quantity $a_{k+1} - a_j$ is distinct from $b_{\sigma(j)} - b_i$ in the field $\mathbb{F}_p$, thus they cannot be equivalent. This contradiction shows that such an $i$ cannot exist, and that there always exists a permutation $\sigma$ of $[k+1]$ such that $a_i + b_{\sigma(i)}$ are distinct modulo $p$.
Therefore, by the principle of mathematical induction, the claim holds for all $k < p$.