Let $Q = \{1, 2,\dots, p −1\}$ for some prime $p$.
Is it possible to use the pigeonhole principle to prove that for each integer $x \in Q$, there is precisely one integer $y \in Q$ such that $xy \equiv 1 \pmod{p}\,$?
Edit: I am a beginner in modular arithmetic. pls reduce the use of technical terms in your response as far as possible. Thx!
Assume there are two members of $Q$ such that $xy_1\equiv 1 \bmod p$ and $xy_2\equiv 1 \bmod p$. Then $(xy_1)-(xy_2)=x(y_1-y_2)\equiv 0 \bmod p$ which means either that $x\equiv 0 \bmod p$ (which is excluded because $x \in Q$) or $(y_1-y_2)\equiv 0 \bmod p$ which is only possible if $y_1=y_2$, i.e. the "two" members of $Q$ are in fact the same member of $Q$
Since each $xy_k$ is unique, and there are $p-1$ possible such products for a given $x$, by the pigeonhole principle, the set $\{xy_1,xy_2,\dots,xy_{p-1}\}$ has the same members as $Q$, although not necessarily in the same order.
