Irish Mathematical Olympiad, 2014

Note first that
$$\begin{align*} a_3 &= a_1^{10} + a_2^{10} \le 2^{10} + 4^{10} < 2 \cdot 4^{10} = 2^{21} < 10^{21} < m \quad \text{and} \\ a_4 &= a_2^{10} + a_3^{10} < 4^{10} + 10^{20} < 2 \cdot 10^{20} < m. \end{align*}$$
Hence, if $m = a_n$, then $n \ge 5$.
Given the possible values for $a_1$ and $a_2$, we see that $a_3$ is congruent to either $1$ or $2$ modulo $3$, depending on whether or not we select $0$ as one of the two initial terms of the sequence. Similarly, $a_4$ is congruent to either $1$ or $2$ modulo $3$, depending on whether or not $a_2 = 0$. In any case, neither $a_3$ nor $a_4$ is divisible by $3$.
By the Little Theorem of Fermat, $a_k^{10} \equiv 1 \pmod{3}$ unless $a_k \equiv 0 \pmod{3}$.
Therefore, if neither $a_{k-1}$ nor $a_{k-2}$ is divisible by $3$, then
$$a_k = a_{k-1}^{10} + a_{k-2}^{10} \equiv 2 \pmod{3}.$$
By induction it follows now that $a_n \equiv 2 \pmod{3}$ for all $n \ge 5$.
Finally, using that any integer is congruent to its sum of digits modulo $3$, we find that
$$m \equiv 2014 \cdot (2 + 0 + 1 + 4) \equiv 2014 \equiv 1 \pmod{3}$$
and we conclude that $m \neq a_n$ for all $n \ge 1$.