# Irreducible Polynomials from a Reccurence

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).

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