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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 $x-p$ 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 $k=15$. |
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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 $p=19$ 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 $k=15$. |
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