TITLE: $(x^2 + 3)(x^3 - q) = 0$
QUESTION [6 upvotes]: I'm trying to show that $(x^2 + 3)(x^3 - q) = 0$ is a counter-example to the Hasse Principle when $q \equiv 1 $ (mod $27$) is prime.
That is the above equation has solutions in $\Bbb{R}$, $\Bbb{Q}_p$ for all primes $p$ (p-adic integers) but none in $\Bbb{Q}$.
It is clear that the equation has solutions in $\Bbb{R}$ and it's also clear it doesn't have any in $\Bbb{Q}$.
For the $\Bbb{Q}_p$ case, I have shown that it has solutions when $p \equiv 1$ modulo 3 as when $p \equiv 1$ (mod 3) then -3 is a quadratic residue, and so the equation has a solution in $\Bbb{F}_p$ which lifts to one in $\Bbb{Q}_p$ by Hensel's Lemma.
I have also show than the equation has a solution in $\Bbb{Q}_p$ when $p = 3$.
Thus I'm left to showing that it has solutions in $\Bbb{Q}_p$ when $p \equiv 2$ modulo 3.
I'm stuck on this part! I believe it will come down to showing there is some $a$ such that $a^3 \equiv q$ (mod p) for all such primes $p$, and then such a solution can be lifted by Hensel's lemma. However I've no idea how to do this.
Any help with showing this much appreciated.

REPLY [5 votes]: I have no idea how to show there is some $a$ such that $a^3 \equiv q$ (mod p) for all primes $p\equiv2\pmod3$

When $p\equiv2\pmod3$, we can say that $2p-1\equiv0\pmod3$, and then the solution
to  $a^3\equiv q\pmod p$   is $a\equiv q^{(2p-1)/3}\pmod p$, because, by Fermat's little theorem,
then $a^3\equiv(q^{(2p-1)/3})^3\equiv q^{2p-1}\equiv q^pq^{p-1}\equiv q\cdot q^{p-1}=q^p\equiv q\pmod p$.