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This question already has an answer here:

Find all primes $p$ such that $$x^3+x+1\equiv0\pmod p$$ has $3$ incongruent solutions.

By a standard result in number theory, it has $3$ incongruent solutions iff there exists $q(x),r(x)\in\mathbb{Z}[x]$ with $\deg r(x)<3$ such that $$x^p-x=(x^3+x+1)\,q(x)+p\cdot r(x).$$ I am unable to proceed from here. I tried to write out the coefficients of $q(x)$ and multiply out, but it was in vain. How do I solve this problem? Thanks in advance!

I am unclear about what tags to add, so please fix this.

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marked as duplicate by Watson, Namaste, Nosrati, Claude Leibovici, The Phenotype Feb 18 '18 at 9:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ The related problem $x^3 - 2 \equiv 0 \pmod p$ can be done with cubic reciprocity, it is done completely in the book by Ireland and Rosen. The problem you quote is much, much harder. $\endgroup$ – Will Jagy Apr 14 '17 at 17:49
  • $\begingroup$ Closely related: math.stackexchange.com/questions/378331 $\endgroup$ – Watson Feb 17 '18 at 22:39
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Such questions lead rapidly to problems in algebraic number theory. Here we need to know at which primes the number field generated by a zero of $x^3+x+1$ splits completely. As this number field does not have an Abelian Galois group, the set of such primes cannot be described simply in terms of congruence conditions.

The discriminant of $f(x)=x^3+x+1$ is $-31$. This means that the splitting field of $f$ is the Hilbert class field of $\mathbb{Q}(\sqrt{-31})$. The primes $p$ that split completely in this field, that is those for which $f(x)\equiv0$ (mod $p$) has three solutions, are those represented by the quadratic form $a^2+ab+8b^2$ (the principal binary quadratic form of discriminant $-31$).

For proofs and much more discussion see David Cox's book Primes of the Form $x^2+ny^2$.

PDF of the entire article, page shown below

enter image description here

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  • $\begingroup$ So is this problem solvable using elementary number-theoretic methods? And what should the answer be? $\endgroup$ – Colescu Apr 14 '17 at 7:53
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    $\begingroup$ @YuxiaoXie, no, nothing elementary about this. The "answer" is that the two sets of primes are the same, your set and the set of primes $p$ that can be expressed $p = u^2 + uv + 8v^2$ with integers $u,v.$ I have edited in, without permission, a page from an article that finishes all these problems. The main topic is "class field theory" $\endgroup$ – Will Jagy Apr 14 '17 at 17:39
  • $\begingroup$ Shark, apologies. The Hudson and Williams article is not widely known, at least on this site. I once sent a problem to the M.A.A. Monthly problem section, confirm that $4 x^2 + 2 xy + 7 y^2 - z^3$ fails to integrally represent integers that factor certain ways. Problem December 2010, the single correct answer December 2012. $\endgroup$ – Will Jagy Apr 14 '17 at 17:44
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    $\begingroup$ The current problem leads to the difficult Show that certain numbers given by a Pell type equation are not integrally represented by $2 x^2 + xy + 4 y^2 - z^3 - z.$ Kevin Buzzard got the first of this set of harder problems, see mathoverflow.net/questions/12486/… $\endgroup$ – Will Jagy Apr 14 '17 at 17:56
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CW answer. To fill in Shark's answer, these are the primes up to 2100 that you want. Well, almost. Most people do not call $1$ a prime, but it is my computer program. Also things "ramify" at $31,$ there is a double root, which they don't want. A Thought: maybe this came up because $2017$ is one of these primes, and this is year $2017.$ Still not a reasonable question for any contest anywhere.

GP-PARI:

parisize = 4000000, primelimit = 500509
? factormod( x^3 + x + 1, 31)
%1 = 
[Mod(1, 31)*x + Mod(17, 31) 2]

[Mod(1, 31)*x + Mod(28, 31) 1]

? factormod( x^3 + x + 1, 47)
%2 = 
[Mod(1, 47)*x + Mod(12, 47) 1]

[Mod(1, 47)*x + Mod(13, 47) 1]

[Mod(1, 47)*x + Mod(22, 47) 1]

? factormod( x^3 + x + 1, 67)
%3 = 
[Mod(1, 67)*x + Mod(4, 67) 1]

[Mod(1, 67)*x + Mod(9, 67) 1]

[Mod(1, 67)*x + Mod(54, 67) 1]

=======================================================

? factormod( x^3 + x + 1, 2017)
%4 = 
[Mod(1, 2017)*x + Mod(176, 2017) 1]

[Mod(1, 2017)*x + Mod(267, 2017) 1]

[Mod(1, 2017)*x + Mod(1574, 2017) 1]

? 

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jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./primego
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2 
1 1 8
Discriminant  -31

Modulus for arithmetic progressions? 
31
Maximum number represented? 
2100

      1,     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,
   1237,   1259,   1303,   1327,   1451,   1493,   1523,   1559,   1583,   1619,
   1663,   1721,   1723,   1741,   1879,   1931,   1973,   1993,   2003,   2017,


    0    1    2    4    5    7    8    9   10   14   16   18   19   20   25   28

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ 
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ date
Fri Apr 14 11:04:55 PDT 2017
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ 
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