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For which prime numbers $p$ does the congruence $x^2+x+1\equiv0\pmod p$ have solutions?

For $p\ne2$ we have $$x^2+x+1\equiv0\pmod p\quad\Leftrightarrow\quad (2x+1)^2\equiv-3\pmod p\ .$$

Does this mean the above congruence has a solution$\pmod p$ for any $p>2$ such that the quadratic residue modulo $p$ is 1?

Vladimir Nabokov
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  • See http://math.stackexchange.com/questions/398376/show-that-7-is-a-quadratic-residue-for-any-prime-p-of-the-form-28k-1-and-28k and http://math.stackexchange.com/questions/262188/quadratic-reciprocity-determine-if-11-is-a-quadratic-residue-mod-p-for-prime – lab bhattacharjee Feb 04 '15 at 11:32

1 Answers1

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Yes, the congruence $x^2+x+1\equiv 0\pmod{p}$ has a solution for the prime $p\gt 2$ if and only if $-3$ is a quadratic residue of $p$. For by your argument, if there is a solution of $x^2+x+1\equiv 0\pmod{p}$, then $(2x+1)^2\equiv -3\pmod{p}$. In the other direction, if $w^2\equiv -3\pmod{p}$, then we can solve the original congruence by solving $2x++1\equiv w\pmod{p}$, which is always possible.

Now to finish characterizing the primes such that $x^2+x+1\equiv 0\pmod{p}$ is solvable, note that either

(i) $p$ is of the form $4k+1$, and $3$ is a quadratic residue of $p$ or

(ii) $p$ is of the form $4k+3$, and $3$ is a non-residue of $p$.

To compute the Legendre symbol $(p/3)$ in cases (i) and (ii), use Quadratic Reciprocity. It is easy to describe the odd $p$ such that $(3/p)=1$.

André Nicolas
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