This question is inspired by a recent one : Let $c$ be a variable and define a sequence by $a_0=0$ $a_1=1$ and $a_{n+1}=a_{n}c-a_{n-1}$ . So

$$\begin{align*} a_2 &= c \\ a_3 &={c}^{2}-1= \left( c-1 \right) \left( c+1 \right) \\ a_4 &= {c}^{3}-2\,c=c \left( {c}^{2}-2 \right) \\ a_5 &={c}^{4}-3\,{c}^{2}+1 = \left( {c}^{2}+c-1 \right) \left( {c}^{2}-c-1 \right) \\ a_6 &={c}^{5}-4\,{c}^{3}+3\,c= c \left( c-1 \right) \left( c+1 \right) \left( {c}^{2}-3 \right) \\ a_7 &={c}^{6}-5\,{c}^{4}+6\,{c}^{2}-1= \left( {c}^{3}-{c}^{2}-2\,c+1 \right) \left( {c}^{3}+{c}^{2}-2\,c-1 \right) \end{align*}$$

Note that $$a_{n+1}=\sum_{j=0}^t (-1)^j\binom{n-j}{j}c^{n-2j}$$ where $t=\lfloor\frac{n}{2}\rfloor.$ So the array of coefficients is just Pascal's triangle shifted with alternating signs. Not that I see a connection to the questions below.

The defining recurrence $a_{n+1}=a_{n}c-a_{n-1}$ i.e. $$a_{n+1}=a_{n}a_{2}-a_{n-1}a_1$$ is the first non-trivial case (and base for an induction proof of) of the more general $$a_{n+m}=a_{n}a_{m+1}-a_{n-1}a_{m}. \tag{$\ast$}$$

From this is follows that $$\gcd(a_n,a_m)=a_{\gcd(n,m)} \tag{$\ast \ast$}$$ so the polynomial $a_n$ factors unless, possibly, $n$ is prime. But for *any* odd index $2m+1$ we have $$a_{(m+1)+m}=a_{m+1}^2-a_{m}^2=(a_{m+1}+a_m)(a_{m+1}-a_m)$$

Q: If $p=2m+1$ is prime, must the monic polynomials $s_p=a_{m+1}+a_m$ and $d_p=a_{m+1}-a_m$ be irreducible in $\mathbb{Z}[c]$?

**random remarks**

This is the case up to $p=199$ (according to Maple) so it seems likely.

Replacing $c$ by $-c$ changes $s_p$ into $\pm d_p$ so either both are irreducible or both factor (in the same way).

$a_{m+1}+a_{m}=$$c^m+c^{m-1}-\binom{m-1}{1}c^{m-2}-\binom{m-2}{1}c^{m-3}+\binom{m-3}{2}c^{m-4}+\binom{m-4}{2}c^{m-5}-\cdots$

A heuristic argument (which should perhaps not be trusted) is as follows: For a fixed integer value $c \ge 2$ the $a_n$ become integers but $(\ast)$ and $(\ast \ast)$ remain true (now as statements in $\mathbb{Z}$) so if a prime $q$ divides $a_k$ for the first time at $k=n$ then it divides $a_k$ exactly when $n \mid k$. Accordingly, when $p=2m+1$ is prime, we have the factorization $a_p=s_p \cdot d_p$ where the two factors are co-prime and any prime $q$ which divides either of them (and hence $a_p$) does not divide $a_k$ for any $k \lt p$. Of course for a given $c$ one or both of $s_p,d_p$ might be composite, but it seems likely (weak!) that each (or even just one or the other) is prime for some value of $c$ and that would require both to be irreducible as polynomials.

When $c$ is set to $c=2$ we have $a_k=k$ so $d_p=a_{m+1}-a_m=1$ and $s_p=a_{m+1}+a_m=2m+1$ is prime when $p$ is. This makes it seem somewhat more likely that the polynomial $s_p$ is irreducible, however there could be factors of the form $(c-2)f(c)+1.$