We have the RSA function: $c = m^e (mod n)$. I would like to know the proof that there is not an $m_1$ and an $m_2$ message that produce the same $c$.
My thoughts:
We know that $m \le n$, so $m_1 \ncong m_2 (mod n)$. We also know that if $a \cong b (mod n)$, then $a^k \cong b^k (mod n)$. So if $m_1 \ncong m_2 (mod n)$ then $m_1^e \ncong m_2^e (mod n)$?
