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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