Cubic Reciprocity, solving $x^3 \equiv 19 \pmod{43}$ I know that there is no solution for the congruence $x^3 \equiv 19 \pmod{43}$, but I don't know how to go about showing it. I know I should use cubic reciprocity, but I'm not sure how that works. Can somebody walk me through how to show this?
 A: You can use this: Let $p$ be a prime and $p \equiv 1 \space (mod 3)$ and $a$ an integer coprime with $p$. Then $a$ is a cubic residue if and only if $a^{\frac{p-1}{3}}\equiv 1 \space(mod\space p).$
Proof:
First direction: Let $g$ be a primitive root $\space (mod\space p)$. We assume $a\equiv b^3$ for some $b$. Now $b \equiv g^s$ for some $s$. Hence $a^{\frac{p-1}{3}}\equiv g^{\frac{3s(p-1)}{3}} = g^{s(p-1)} = (g^{p-1})^s \equiv 1$ (by Fermat's litttle theorem).
Other direction: Assume $a^{\frac{p-1}{3}}\equiv 1 \space(mod\space p)$. We have $a=g^t$ for some $t$. So we have $g^{\frac{t(p-1)}{3}} \equiv 1$. Since $g$ is a primitive root, it must be that the power $\frac{t(p-1)}{3}$ is a multiple of $p-1$ and so $t = 3r$ for some $r$. Hence $a=g^{3r}$ is a cubic residue $mod\space p$.
EDIT: The case $p \equiv 2 \space (mod\space 3)$. This is the proof found on wikipedia (I add it here to make it more easy to find it). In this case every number is a cubic residue. Clearly $0 = 0^3$ is. So let $x$ be a number not divisible by $p$. Now we have $p=3n + 2$ for some $n$. Hence (by Fermat's little theorem)
$$x^{3n+1}\equiv 1.$$
$$x^{3n+2}\equiv x $$
From these we get that 
$$x=1\cdot x \equiv x^{3n+2} \cdot x^{3n+1} = x^{6n+3} = (x^{2n+1})^3$$
and $x$ is a cubic residue $\space (mod\space p)$.
A: Cubic reciprocity says your equation is solvable iff $x^3 \equiv 43 \mod 19$, or $x^3\equiv 5 \mod 19$ is solvable.
One of Euler's conjectures states that $5$ is a cubic residue of $p$ iff $15|b$, or $3|b$ and $5|a$, or $15 | (a \pm b)$, or $15|(2a \pm b)$. Here $19 = a^2 + 3b^2$, and this conjecture applies because $19$ is a prime $1 \mod 3$.  
Solving, $a = 4, b=1$. None of the conditions hold, so the equation has no solutions. 
A: Since $43$ is a prime, Fermat says that $a^{42}\equiv 1\pmod{43}$ for all $a$. Since $19^{14}\equiv36\pmod{43}$, $19$ can't be a third power modulo $43$.
