<s>[INST] Solve $x^2+x+1\equiv0\pmod5$ 
Solve:
$$x^2+x+1\equiv0\pmod5$$

My attempt:
Our proffesor told us that if we have $ax^2+bx+c\equiv\pmod p$ we need to multiply by $4a$, to get form of $(\text{ something})^2\equiv D\pmod p$.
$p$ is a prime and $\gcd(4a,p)=1$, so I tried to do that here:
$$4x^2+4x+4\equiv 0 \pmod 5$$
But now
$$(2x+1)^2+3\equiv 0 \pmod 5$$
$$(2x+1)^2\equiv -3 \pmod 5$$
 [/INST] In this specific case, there is yet another way:
Since $(x-1)(x^2+x+1)=x^3-1,$ we find that a solution to the equation must also satisfy $x^3-1\equiv0\pmod5.$
But if $x^3\equiv1\pmod5,$ then clearly $x^4\equiv x\pmod5.$ Also, Fermat's little theorem says that $x^4\equiv1\pmod5$ if $5\not\mid x.$ Thus the solution is either $x\equiv0\pmod5$ or $x\equiv x^4\equiv1\pmod5,$ while these two cases can easily be seen not to satisfy the equation in question. Therefore there are no solutions.  
Hope this helps.
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