Show that there exists a polynomial $p(X) ∈ F[X]$ satisfying $c^n + c^{−n} = p(c + c^{−1})$. Let $c$ be a nonzero element of a field $F$ and let $n > 1$ be
an integer. Show that there exists a polynomial $p(X) ∈ F[X]$ satisfying
$c^n + c^{−n} = p(c + c^{−1})$.
I made some particular cases, but not seems to have a form so I need some help, Thank you.
This is an exercise from the book: The Linear Algebra a Beginning Graduate Student Should Know by Golan.
 A: Complete solution:
We can prove this by induction, starting with a base case of $n=0$.  If $n=0$, then taking $p(X)=2$ is a solution to the problem.  Suppose $n>0$ and that we have polynomials $p_k(X)\in F[X]$ such that $p_k(c+c^{-1})=c^k+c^{-k}$ for all $k<n$.  Then notice that $(c+c^{-1})^n=c^n+nc^{n-1}c^{-1}+\left(\!\begin{array}{c}n\\2\end{array}\!\right)c^{n-2}c^{-2}+\cdots \left(\!\begin{array}{c}n\\n-2\end{array}\!\right)c^2c^{-(n-2)}+ncc^{-(n-1)}+c^{-n}.$  By noting that$\left(\!\begin{array}{c}n\\i\end{array}\!\right)=\left(\!\begin{array}{c}n\\n-i\end{array}\!\right)$, and by simplifying and gathering terms, we have that\begin{align*}(c+c^{-1})^n&=c^n+c^{-n}+n(c^{n-2}+c^{-(n-2)})+\cdots +\left(\!\begin{array}{c}n\\\lfloor n/2\rfloor\end{array}\!\right)(c^{n-2\lfloor n/2\rfloor}+c^{-(n-2\lfloor n/2\rfloor)})\\&=c^n+c^{-n}+np_{n-2}(c+c^{-1})+\cdots+\left(\!\begin{array}{c}n\\\lfloor n/2\rfloor\end{array}\!\right)p_{n-2\lfloor n/2 \rfloor}(c+c^{-1})\end{align*}if $n$ is odd and\begin{align*}(c+c^{-1})^n&=c^n+c^{-n}+n(c^{n-2}+c^{-(n-2)})+\cdots
 +\left(\!\begin{array}{c}n\\n/2\end{array}\!\right)\\&=c^n+c^{-n}+np_{n-2}(c+c^{-1})+\cdots+\left(\!\begin{array}{c}n\\ n/2\end{array}\!\right)\end{align*}if $n$ is even.  Therefore, taking $p_n(X)=X^n-np_{n-2}(X)-\left(\!\begin{array}{c}n\\2\end{array}\!\right)p_{n-4}(X)-\cdots -\left(\!\begin{array}{c}n\\\lfloor n/2\rfloor\end{array}\!\right)p_{1}(X)$ if $n$ is odd and $p_n(X)=X^n-np_{n-2}(X)-\left(\!\begin{array}{c}n\\2\end{array}\!\right)p_{n-4}(X)-\cdots -\left(\!\begin{array}{c}n\\ n/2\end{array}\!\right)$ if $n$ is even we have our desired polynomial for $n$.  So we have completed the inductive step and proved that such a polynomial exists for all $n\geq 0$.
A: Hint.  For example, if $n=5$ we have
$$c^5+c^{-5}=(c+c^{-1})^5-(5c^3+10c+10c^{-1}+5c^{-3})\ .$$
Note that the coefficients in the final bracket are "symmetric".  See if you can prove that this is always the case (not only for $n=5$) and then use induction.
A: Outline. Use strong induction.
Base case: $n=1$ and $n=2$: $c^1 +  c^{-1} = p(c^1 + c^{-1})$, in particular with $p = x$. $(c^2 + c^{-2}) = p(c^1 + c^{-1})$ with $p = x^2 - 2$.
Inductive step: Assume that there exists $p_m(x)$ with $c^{m} + c^{-m} = p(c^1 + c^{-1})$ for $m \leq n$.
$$c^{n+1} + c^{-(n+1)} = (c^1 + c^{-1})^{n+1} - \left(\displaystyle \sum_{k=1}^{n-1} {n+1 \choose k} c^{n+1-2k}\right)$$
And the right sum can be expressed as some $p(x)$ by induction (note that there may be some constant term if $n+1$ is even, in which case you may just add $c$ to the polynomial), so $p_{n+1} (x) = p(x) + x^{n+1}$
A: Hint: Use complete induction on $n$, multiplying your polynomial $p(c + 1/c) = c^n + c^{-n}$ by $(c + 1/c)$ to initiate the inductive step. You'll need the cases for $n$ and $n-1$ to prove the case for $n+1$.
