Tajland, 2014

Let $p$ be a prime and $n$ be a positive integer such that
$$2p^2 - 3p - 1 = n^3. \qquad (1)$$
Since
$$n^3 = 2p^2 - 3p - 1 < 2p^2 \le p^3,$$
it follows that $n < p$, so $n + 1 \le p$.

Consider the case $p = n + 1$. We then get by (1) that
$$n^3 - 2n^2 - n + 2 = 0$$
which implies
$$(n - 2)(n - 1)(n + 1) = 0, \quad \therefore n = 1, 2.$$
Thus $p = 2, 3$.

Consider the case $p > n + 1$. By (1) we get
$$p(2p - 3) = n^3 + 1 = (n + 1)(n^2 - n + 1). \qquad (2)$$
But $p > n + 1$ and $p$ is a prime, hence $p \mid n^2 - n + 1$. Thus there is an integer $k$ such that
$$n^2 - n + 1 = kp. \qquad (3)$$
Setting (3) into (2) we then get $2p = k(n + 1) + 3$ which implies in particular that $k$ is odd.

Setting $p = \frac{k(n+1)+3}{2}$ into (3) we have
$$2n^2 - (k^2 + 2)n - (k^2 + 3k - 2) = 0. \quad (4)$$
By (4) and that $n$ is a positive integer, we have that
$$(-k^2 + 2)^2 - 4(2)(-k^2 + 3k - 2) = k^4 + 12k^2 + 24k - 12$$
is a square.
If $k \ge 9$ then we have
$$(k^2 + 6)^2 < k^4 + 12k^2 + 24k - 12 < (k^2 + 7)^2$$
which implies that $k^4 + 12k^2 + 24k - 12$ cannot be a square, contradicting the above. Thus $1 \le k \le 8$. Now $1 \le k \le 8$ is an odd integer, hence $k \in \{1, 3, 5, 7\}$. It is directly to check that these numbers cannot make $n$ to fulfil the condition of the problem. So there are no $p > n + 1$.

We conclude that $p = 2, 3$ are the only primes making $2p^2 - 3p - 1$ a cubic. $\square$