Prime divides element with index that prime

Let $s$ be a positive integer. Let $\{a_n\}_{n=1}^{\infty}$ be a sequence given by $$a_1=0, \ a_2=2s, \ a_3=3, \ a_{n+3}=s a_{n+1}+a_n \ \ \ \text{for} \ n\ge 1.$$
Prove that for any prime number $p$, the number $a_p$ is divisible by $p$.

Solution. Let the roots of the equation $x^3-sx-1=0$ be $x_1, x_2$ and $x_3=-x_1-x_2$ (Vieta's formulas), they are algebraic integers (by definition). One directly observes (using the initial conditions) that $a_n=x_1^n+x_2^n+(-x_1-x_2)^n$. Assume $p>2$ (the other is obvious). We have
$$a_p=x_1^p+x_2^p+(-x_1-x_2)^p=-\sum_{k=1}^{p-1}{p\choose k}x_1^kx_2^{p-k}$$
${p\choose k}$ is divisible by $p$ for $1\le k \le p-1$, so we obtain that $a_p=p A$ where $A$ is an algebraic integer, since algebraic integers form a ring. However, from here we see that $A$ is also rational, hence $A$ is integer. Thus $p$ divides $a_p$.

I would like to thank Mr. Bozhilov for finding a mistake in the original proof, and his fruitful discussion, which led to a resolution.
The problem appeared at AKHIMO 2018.


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