Let $$ f_{a}(x)=x^3+(u-2-a)x^2+ax+1, $$ where $u\in\mathbb{Z}_p^*$ is fixed. Let $S$ be the set consisting of all $a\in\mathbb{Z}_p$ such that $f_{a}(x)$ factor linearly. Then what is the cardinality of $S$? Can we get an exact formula?
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$\begingroup$ I'm reminded of what Shanks called the simplest cubics, $x^3=ax^2+(a+3)x+1$. $\endgroup$– Gerry MyersonCommented Jun 1, 2023 at 11:58
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$\begingroup$ See also mathoverflow.net/questions/447972/… from same OP. $\endgroup$– Gerry MyersonCommented Jun 1, 2023 at 11:59
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1$\begingroup$ Probably not - this is a genus 1 curve. $\endgroup$– Will SawinCommented Jun 1, 2023 at 13:21
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