why does $P(X) = X^3+X+1$ has at most 1 root in $F_p$ ?
I could fact check this on Sage for small values of $p$.
For example $p=5$ or $7$ or $19$; there is no root.
If $p = 11, 2$ is the only root.
If $p = 13, 7$ is the only root.
If $p = 17, 11$ is the only root.
I also realize that there can't be only 2 roots a1 and a2 because a3 = -(a1+a2) will be a root as well.
On
The first prime for which $x^3+x+1$ has $3$ distinct roots is $47$, where the roots are $24$, $34$ and $35$. Hmm... it looks like the primes for which $x^3+x+1$ splits are OEIS sequence A033221, the primes of the form $x^2 + 31 y^2$.
On
The primes other for which $x^3 + x + 1$ has three distinct roots are all those that can be represented by $$ x^2 + xy + 8 y^2 $$ other than 31 itself
31, 47, 67, 131, 149, 173, 227, 283, 293, 349,
379, 431, 521, 577, 607, 617, 653, 811, 839, 853,
857, 919, 937, 971, 1031, 1063, 1117, 1187, 1213,
The primes for which $x^3 + x + 1$ has no roots are all those that can be represented by $$ 2x^2 + xy + 4 y^2 $$ other than 31 itself
2, 5, 7, 19, 41, 59, 71, 97, 101, 103,
107, 109, 113, 157, 163, 191, 193, 211, 233, 257,
281, 307, 311, 317, 359, 373, 397, 419, 421, 439,
443, 467, 479, 503, 541, 547, 563, 593, 599, 659,
661, 683, 691, 701, 727, 733, 751, 769, 877, 887,
907, 977, 997,
The primes for which the cubic has one root are all odd primes $p$ such that Legendre symbol $$ (-31|p) = -1 $$ Note that $31 \equiv 3 \pmod 4,$ so for any odd prime other than $31$ itself, $$ (-31|p) = (p|31) $$
On
Meanwhile, the numbers that can not be expressed with integers as $$ 2x^2 + xy + 4 y^2 - z^3 - z $$ are $\pm C_n,$ where the sequence of $C_n$ begins $$ 1, 869, 25171, 21118439, 611705641, 513220303709, 14865670462411, $$ and obeys $$ C_{n+4} = 24302 C_{n+2} - C_n. $$ Put another way, the numbers $$ 27 C_n^2 + 4 $$ are of the form $$ 31 D_n^2 $$
It's not true.
$3$ and $14$ are roots in $F_{31}$.