3 corrected typo

Based on a small number of small cases I suspect that the majority of those of those do factor.

Let $r$ be a primitive root $\mod p$ then there is an $n$ such that $r^n=r+1 \mod p$. Then $x-p$ x-r$is a factor of$x^n-x-1$in$\mathbb{Z}_p$. On average$n$should be about$p/2$. Of course one can use$x^{n+p-1}-x-1$but that seems like cheating. That is just cases with a linear factor. For$p=19, x^k-x-1$factors for$2 \le k \le 18$except for$k=4$and$k=15$. 2 deleted 93 characters in body Based on a small number of small cases I suspect that the majority of those of those do factor. Let$r$be a primitive root$\mod p$then there is an$n$such that$r^n=r+1 \mod p$. Then$x-p$is a factor of$x^n-x-1$in$\mathbb{Z}_p$. On average$n$should be about$p/2$. Of course one can use$x^{n+p-1}-x-1$but that seems like cheating. That is just cases with a linear factor. Based on a small number of small cases I suspect that the majority of the trinomials do factor: For$p=19$p=19,$ $x^k-x-1$ factors for $2 \le k \le 18$ except for $k=4$ and $k=15$.

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Based on a small number of small cases I suspect that the majority of those of those do factor.

Let $r$ be a primitive root $\mod p$ then there is an $n$ such that $r^n=r+1 \mod p$. Then $x-p$ is a factor of $x^n-x-1$ in $\mathbb{Z}_p$. On average $n$ should be about $p/2$. Of course one can use $x^{n+p-1}-x-1$ but that seems like cheating. That is just cases with a linear factor.

Based on a small number of small cases I suspect that the majority of the trinomials do factor: For $p=19$ $x^k-x-1$ factors for $2 \le k \le 18$ except for $k=4$ and $k=15$.