Proof of a series of congruences

Question

Prove that this is impossible: $$ \begin{cases} a_2 a_1 \equiv a_1 \pmod{n}\\ a_3 a_2 \equiv a_2 \pmod{n} \\ a_4a_3 \equiv a_3 \pmod{n} \\ \ldots\\ a_1a_k \equiv a_k \pmod{n} \end{cases}$$

For any $n>1$, any $k \le\ n$ and any $a_1,a_2,a_3...a_k \le\ n$ without repetitions (if $a_1 = 5$, there can not be another $a$ that has value 5).

Answer 1

Hint: If $\gcd(a_i, n) \ne 1$ then $\gcd(a_i - 1, n) = 1$ so $a_{i-1}$ is divisible by $n$.

Answer 2

Here is another tack you could try: Start with $a_1\equiv a_1a_2$ and substitute $a_2\equiv a_2a_3$ on the right, getting $a_1\equiv a_1a_2a_3$. Do it again with $a_3\equiv a_3a_4$, and so on. Oh, and also notice the nice, cyclical nature of the problem: There is nothing special about $a_1$ or $a_k$ or any of the $a_j$ in between.