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Problem: I am trying to prove that $$ x^2 + x + 1 \equiv 0 \mod p $$ has a solution where $p$ is a prime such that $p \equiv 1 \mod 3$, without using quadratic reciprocity. I am also suspecting that it has no solution if $p \equiv -1 \mod 3$.

Attempt: I checked for some small primes and tried rewriting the equation, like $$ x(x + 1) \equiv -1 \mod p, $$ and substituting $x$, e.g. $x = y -1$ gives $$ (y-1)^2 + y-1 + 1 = y^2 - y + 1 \equiv 0 \mod p. $$ (Using that $\mathbb{F}_p$ is a field this question is equivalent.)

Bart Michels
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Math Student 020
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2 Answers2

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The multiplicative group modulo $p$ has order $p-1$ and is cyclic with some generator $a$. If $p-1=3k$, then $1$, $a^k$, $a^{2k}$ are three distinct roots of $X^3-1$. Divide out $(X-1)$ to obtain that $a^k$, $a^{2k}$ are roots of $X^2+X+1$.

On the other hand, if $p-1$ is not a multiple of $3$, then the cyclic group of order $p-1$ has no element of order $3$, hence $1$ is the only root of $X^3-1$, hence the only candidate as a root fo $X^2+X+1$, which works out only if $p=3$.

Hagen von Eitzen
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If $p=2$, the equation has no solution.

Otherwise you can use the old quadratic formula: $$x=\frac{-1\pm\sqrt{1-4}}{2}$$ having in mind, of course, that these 'numbers' and operations are in $\Bbb F_p$.

So the equation has solutions if and only if $-3$ is a square mod $p$.

ajotatxe
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