For the positive prime integer $p$, Let $\mathbb{F}_p=\{0,1,\cdots, p-1\}$ be the finite field of order $p$.

For $x\in \mathbb{F}_p$, define $f_p(x)$ to be the maximum element in the set $\{ x^n+x^{-n}\in \mathbb{F}_p | n\in {\mathbb{Z}}\}$.

For instance, when $p=17$, for $x=2$, we have $f_{17}(x)=\max\{2^0+2^0=2, 2^1+2^7=11, 2^2+2^6=0, 2^3+2^5=6, 2^4+2^4=15\}=15$.

My question is: when $p$ is relatively small, we can compute every $f_p(x)$ in the brute-force manner. However, such strategy does not work when $p$ is very large; in such cases, is there some "efficient" way to compute every $f_p(x)$?

Plus, could you suggest some materials that might be relevant to my question?

Thanks!!

For the positive prime integer $p$, Let $\mathbb{F}_p=\{0,1,\cdots, p-1\}$ be the finite field of order $p$.

For $x\in \mathbb{F}_p$, define $f_p(x)$ to be the maximum element in the set $\{ x^n+x^{-n}\in \mathbb{F}_p | n\in {\mathbb{Z}}\}$.

For instance, when $p=17$, for $x=2$, we have $f_{17}(x)=\max\{2^0+2^0=2, 2^1+2^7=11, 2^2+2^6=0, 2^3+2^5=6, 2^4+2^4=15\}=15$.

My question is: when $p$ is relatively small, we can compute every $f_p(x)$ in the brute-force manner. However, such strategy does not work when $p$ is very large; in such cases, is there some "efficient" way to compute every $f_p(x)$?

Or more broadly, could we find a non-trivial upper bound on $f_p(x)$ for some $\textit{special}$ $x$'s? For instance, when the multiplicative order of $x$ is some special factor of $(p-1)$?

Plus, could you suggest some materials that might be relevant to my question?

Thanks!!