Showing that the sequence $a_{n+3}=5a^6_{n+2}+3a^3_{n+1}+a^2_n$, with $a_1=2019$, $a_2=2020$, $a_3=2021$, contains no numbers of the form $m^6$

Here's a problem that I couldn't solve in a recent olympiad in which I took part. The problem says

The sequence of real numbers $a_1$, $a_2$, $a_3,\cdots$ is defined as follows: $a_1 = 2019,$ $a_2 = 2020,$ $a_3 = 2021$ and for any $ n \ge 1$ holds that: $$ a_{n+3} = 5a^{6}_{n+2} + 3a^{3}_{n+1} + a^{2}_{n} $$ Prove that this sequence does not contain numbers of the form $m^{6}$ for $m \in \mathbb{N}.$

This was the only problem I didn't even could make process, I asked some peers about it but they gave me ideas of calculus and I didn't undertand very well. I would like to get some ideas of how to solve it I don't really want a solution, instead I wanna try to solve it by myself.

Edit 1. Proving important lemma of the problem and general fact (See @TonyK comment). For proving that we'll consider the following. As it is well-know there exists 7 clases $ \equiv$ (mod 7) {0,1,2,3,4,5,6}. So a single number can only belongs to one of the 7 clases. Let $m$ be our number and $m \in \mathbb{Z} $, is clear that if $m$ belongs to the $0$ class, $m^6 \equiv 0$ (mod 7). Now supose that $m \equiv 1\pmod 7$. Then: \begin{align} m^2 &\equiv 1\pmod{7}\\ m^3 &\equiv 1\pmod{7}\\ &\; \; \vdots \notag \\ m^6 &\equiv 1\pmod{7} \end{align} Supose that $ m \equiv 2\pmod{7}$. Then: \begin{align} m^2 &\equiv 4\pmod{7}\\ m^3 &\equiv 8 \equiv 1\pmod{7} \\ m^4 & \equiv 2 \pmod{7}\\ m^5 & \equiv 4\pmod{7}\\ m^6 &\equiv 8 \equiv 1\pmod{7} \end{align} Now you can consider from $3$ up to $6\bmod 7$ the same way we just did with $m \equiv 2\pmod{7}$, for explain this consider the numbers $n_i$ such that $n_i \equiv i\pmod{7}$ and if you follow the same way as in $m \equiv 2\pmod{7}$ you'll get: \begin{align} n^{6}_1 &\equiv 1\pmod{7}\\ n^{6}_2 &\equiv 1\pmod{7}\\ &\; \; \vdots \notag \\ n^{6}_6 &\equiv 1\pmod{7} \blacksquare \end{align} Which means that $\forall m \in \mathbb{N} $ when $ m \ne 7k $ for $ k \in \mathbb{Z}$, holds that $ m^6 \equiv 1\pmod{7}$.

Second ultra easier proof. By FLTh $a^{7-1} \equiv 1 \pmod{7}$ $\blacksquare$

Solution 1:

Hint: Compute a few values modulo $7$. You will see a pattern. Can any of these values be a sixth power modulo $7$?

Updated to add: My initial values were out, so I thought this problem was easier than it is. You actually have to go as fas as $a_{15}$ to see the repeating pattern $\;4,3,3,4,\;\;4,3,3,4,\ldots$

But if $a_n$ and $a_{n+1}$ are both equal to $3$ or $4\bmod 7$, then so is $a_{n+3}$. So you only need to go as far as $a_7\equiv 3, a_8\equiv 4, a_9\equiv 3\bmod 7$ to show that $a_n$ is never equal to $0$ or $1$ $\bmod 7$ (and hence that $a_n$ is never a sixth power).