Czech-Slovak-Polish Match

All the congruences and remainder classes below are meant mod $m$. We obtain the desired congruence relation $a_k^4 - a_k - 2 \equiv 0$ as a consequence of the simpler relation $a_k \equiv -1$.

The sequence of remainder classes of the numbers $a_k$ has the following property: the remainder classes of any two consecutive elements $a_k$, $a_{k+1}$ determine uniquely the remainder classes of all subsequent elements $a_i$ ($i > k + 1$), as well as of all elements $a_i$ ($i < k$) preceding them. By the standard argument, based on the fact that the number of ordered pairs of remainder classes is $m^2$, hence finite, it follows that the sequence of remainder classes of the elements $a_i$ is periodic, starting already from its first member. Thus there exists a number $p > 0$ (depending on the given modulus $m$) such that $a_i \equiv a_{i+p}$ for any index $i$. Unless $m = 1$ (then the problem is trivial), clearly $p > 1$. Since $a_1 \equiv a_2 \equiv 1$, we also have $a_{p+1} \equiv a_{p+2} \equiv 1$, whence $a_p \equiv 0$ and $a_{p-1} \equiv -1$, so we can take $k = p-1$ and the proof is finished.