This question is inspired by a [recent one](https://mathoverflow.net/questions/146156) : 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


$$\begin{align\*}
a\_2 &= c
\\ a\_3 &={c}^{2}-1= \left( c-1 \right) \left( c+1 \right)
 \\ a\_4 &= {c}^{3}-2\,c=c \left( {c}^{2}-2 \right) 
\\ a\_5 &={c}^{4}-3\,{c}^{2}+1 = \left( {c}^{2}+c-1 \right) \left( {c}^{2}-c-1 \right)
\\ a\_6 &={c}^{5}-4\,{c}^{3}+3\,c= c \left( c-1 \right) \left( c+1 \right) \left( {c}^{2}-3 \right)
\\ a\_7 &={c}^{6}-5\,{c}^{4}+6\,{c}^{2}-1=
 \left( {c}^{3}-{c}^{2}-2\,c+1 \right) \left( {c}^{3}+{c}^{2}-2\,c-1
 \right)
\end{align\*}$$




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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.




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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 $$a\_{(m+1)+m}=a\_{m+1}^2-a\_{m}^2=(a\_{m+1}+a\_m)(a\_{m+1}-a\_m)$$



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> > **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 $\mathbb{Z}[c]$?**
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**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 $(c-2)f(c)+1.$



Irreducible Polynomials from a Reccurence
The polynomials $s\_p$ and $d\_p$ are indeed irreducible: We have $a\_p(c)=s\_p(c)d\_p(c)$ with $s\_p$ and $d\_p$ monic of degree $m=(p-1)/2$. Let $z$ be another variable. Then $z^{2m}a\_p(z+1/z)=(z^ms\_p(z+1/z))(z^md\_p(z+1/z))$. The two factors on the right hand side are monic polynomials, and if we show that they are irreducible, then $s\_p(x)$ and $d\_p(c)$ are irreducible even more.


Using the explicit expression of $a\_p(c)$ from Paolo's answer to [this related question](https://mathoverflow.net/questions/146156) shows that
\begin{equation}
z^{2m}a\_p(z+1/z)=(1+z+\dots+z^{p-2}+z^{p-1})(1-z+\dots-z^{p-2}+z^{p-1})=\Phi(z)\Phi(-z),
\end{equation}
where $\Phi(z)$ is the $p$-th cyclotomic polynomial. But $\Phi(z)$ is well known to be irreducible, and the claim follows.


