This post doesn't have a goal, I'm just recording some behavior I came across. Sometimes people approach me at parties and tell me how interested their children are into math; now I can point them here.
Let $p$ be prime, and let $\zeta$ be a $p$-th root of unity. This means that $\zeta^p=1$ (but $\zeta \neq 1$). For every integer $k$, defined $\epsilon_k$ to be $\zeta^k + \zeta^{-k}$.
Note that $\epsilon_0 = 2$, and $\epsilon_k = \epsilon _{-k}$. Let's see how these multiply:
$$ \begin{align} \epsilon_k \epsilon_l &= (\zeta^k + \zeta^{-k})(\zeta^l + \zeta^{-l}) \\ &= (\zeta ^ {k+l} + \zeta ^ {k-l} + \zeta ^ {-k+l} + \zeta ^ {-k-l}) \\ &= \epsilon_{k+l} + \epsilon _ {k-l} \end{align} $$
Now I will show that all the $\epsilon_k$'s can actually be expressed as polynomials in $\epsilon_1$. First let's expand powers of $\epsilon_1$ using the binomial expansion: $$ \begin{align} \epsilon_1^n &= (\zeta + \zeta^{-1})^n \\ &= \sum_{l=0}^n \binom{n}{l} \zeta^l \zeta^{-(n-l)} \\ &= \sum_{l=0}^n \binom{n}{l} \zeta^{2l-n} \end{align} $$
Let's add the terms in pairs, starting with the first and last: $$ \binom{n}{0} \zeta^{-n} + \binom {n}{n} \zeta^n. $$ This is just $\epsilon_n$.
Now add the second and second-to-last terms: $$ \begin{align} \binom{n}{1} \zeta^{2-n} &+ \binom {n}{n-1} \zeta^{2(n-1)-n} \\ \binom{n}{1} \zeta^{2-n} &+ \binom {n}{n-1} \zeta^{n-2} \end{align} $$ which is $n \epsilon_{n-2}$.
In this way, every term pairs up with another term, giving an integer times an $\epsilon_k$ (remember that binomial coefficients are always integers!). When $n$ is even, there is also the middle term $\binom{2l}{l}$, coming from $l=n/2$. This term is actually even making it an integral multiple of $\epsilon_0$.
To summarize: every $\epsilon_1 ^n$ is an integral sum of $\epsilon_k$'s for $k$ <= $n$. Even better, it is $\epsilon_n$ + an integral sum of $\epsilon_k$'s for $k$ < $n$! Which means we can turn it around to say that $\epsilon_n$ is $\epsilon_1^n$ minus an integral sum of $\epsilon_k$'s for $k$ < $n$.
Why is this good? Because continuing in this way, we can replace each $\epsilon_k$ with an $\epsilon_1^k$ plus/minus some other stuff, which we can then replace in the next step. And each time our coefficients stay integers.
So we get that $\epsilon_n$ is an integral polynomial in $\epsilon_1$. Here are the first few for $n$ <= $4$, I'm just going to write $\epsilon$ for $\epsilon_1$ now: $$ \begin{align} \epsilon_1 &= \epsilon \\ \epsilon_2 &= \epsilon^2 - 2 \\ \epsilon_3 &= \epsilon^3 - 3 \epsilon \\ \epsilon_4 &= \epsilon^4 - 4 \epsilon^2 + 2 \end{align} $$
As I find time I will study these polynomials and record my results here. Feel free to make contributions in the comments!
Let $p$ be prime, and let $\zeta$ be a $p$-th root of unity. This means that $\zeta^p=1$ (but $\zeta \neq 1$). For every integer $k$, defined $\epsilon_k$ to be $\zeta^k + \zeta^{-k}$.
Note that $\epsilon_0 = 2$, and $\epsilon_k = \epsilon _{-k}$. Let's see how these multiply:
$$ \begin{align} \epsilon_k \epsilon_l &= (\zeta^k + \zeta^{-k})(\zeta^l + \zeta^{-l}) \\ &= (\zeta ^ {k+l} + \zeta ^ {k-l} + \zeta ^ {-k+l} + \zeta ^ {-k-l}) \\ &= \epsilon_{k+l} + \epsilon _ {k-l} \end{align} $$
Now I will show that all the $\epsilon_k$'s can actually be expressed as polynomials in $\epsilon_1$. First let's expand powers of $\epsilon_1$ using the binomial expansion: $$ \begin{align} \epsilon_1^n &= (\zeta + \zeta^{-1})^n \\ &= \sum_{l=0}^n \binom{n}{l} \zeta^l \zeta^{-(n-l)} \\ &= \sum_{l=0}^n \binom{n}{l} \zeta^{2l-n} \end{align} $$
Let's add the terms in pairs, starting with the first and last: $$ \binom{n}{0} \zeta^{-n} + \binom {n}{n} \zeta^n. $$ This is just $\epsilon_n$.
Now add the second and second-to-last terms: $$ \begin{align} \binom{n}{1} \zeta^{2-n} &+ \binom {n}{n-1} \zeta^{2(n-1)-n} \\ \binom{n}{1} \zeta^{2-n} &+ \binom {n}{n-1} \zeta^{n-2} \end{align} $$ which is $n \epsilon_{n-2}$.
In this way, every term pairs up with another term, giving an integer times an $\epsilon_k$ (remember that binomial coefficients are always integers!). When $n$ is even, there is also the middle term $\binom{2l}{l}$, coming from $l=n/2$. This term is actually even making it an integral multiple of $\epsilon_0$.
To summarize: every $\epsilon_1 ^n$ is an integral sum of $\epsilon_k$'s for $k$ <= $n$. Even better, it is $\epsilon_n$ + an integral sum of $\epsilon_k$'s for $k$ < $n$! Which means we can turn it around to say that $\epsilon_n$ is $\epsilon_1^n$ minus an integral sum of $\epsilon_k$'s for $k$ < $n$.
Why is this good? Because continuing in this way, we can replace each $\epsilon_k$ with an $\epsilon_1^k$ plus/minus some other stuff, which we can then replace in the next step. And each time our coefficients stay integers.
So we get that $\epsilon_n$ is an integral polynomial in $\epsilon_1$. Here are the first few for $n$ <= $4$, I'm just going to write $\epsilon$ for $\epsilon_1$ now: $$ \begin{align} \epsilon_1 &= \epsilon \\ \epsilon_2 &= \epsilon^2 - 2 \\ \epsilon_3 &= \epsilon^3 - 3 \epsilon \\ \epsilon_4 &= \epsilon^4 - 4 \epsilon^2 + 2 \end{align} $$
As I find time I will study these polynomials and record my results here. Feel free to make contributions in the comments!