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Let $p$ be a prime number, $p \equiv 2\,(\bmod~3\,), x \in \mathbb{Z}, x \neq 0\,(\bmod~p\,)$ $$a_{n} \equiv x^{3^{n}}+x^{-3^{n}}\,(\bmod~p\,)$$ with $a_{0} \equiv 5\,(\bmod~p\,)$. Show that there exists a positive integer $n$ such that $$a_{n} \equiv 5\,(\bmod~p\,).$$

Looking for hints. Been trying Fermat's little theorem to no avail.

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    $\begingroup$ Hint: the sequence $x^k \pmod p$ is periodic with period dividing $p - 1$ (application of Fermat's little theorem). Look for an $n$ such that $3^n \equiv 1 \pmod {p-1}$. (Why is there such an $n$? Look at the congruence condition on $p$). $\endgroup$
    – Tob Ernack
    Sep 23 '20 at 12:47
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By Euler's theorem, because $\gcd(3,p-1)=1$ and $\gcd(x,p)=1$ we have $$3^{\varphi(p-1)}\equiv1\pmod{p-1}\qquad\text{ and }\qquad x^{p-1}\equiv1\pmod{p}.$$ It follows that $a_{\varphi(p-1)}\equiv a_0\pmod{p}$.

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