Prove that $x^3 \equiv a \pmod{p}$ has a solution where $p \equiv 2 \pmod{3}$?
How can I prove a congruence equation has a solution? I tried to link Fermat's little theorem with this problem, but I couldn't find a way to solve it.
My attempt was: $$x^3 \equiv 1 \pmod{2}$$ $$x^3 \equiv a \pmod{p}$$
If $p \equiv 2 \pmod{3}$, I have $p = 3k + 2$, for some integers $k$. But I was stuck here :(. Any idea?
Another question is, is there are infinitely many primes of the form 3k + 2? A hint would be sufficient.
Thanks, Thanks,
has a solution
, and at most. $\endgroup$