Consider $ 3n^2 + 3n + 7$ $ \pmod 9$. If $ n\equiv 0\pmod 3$ or $ n\equiv 2\pmod 3$ then $ n^2 + n = n(n + 1)\equiv 0\pmod 3$; if $ n\equiv 1 \pmod 3$ then $ n^2 + n\equiv 2\pmod 3$. Therefore, $ n^2 + n\equiv 0$ or $ 2 \pmod 9$ hence $ 3n^3 + 3n\equiv 0$ or $ 6\pmod 9$, so $ 3n^2 + 3n + 7\equiv 7$ or $ 4\pmod 9$. However, because any cube of an integer is congruent to $ - 1,0,$ or $ 1$ $ \pmod 9$ then it is impossible for a cube of an ineteger to be congruent to $ 4$ or $ 7\pmod 9$, so no cube of an integer can be equal to $ 3n^3 + 3n + 7$ where $ n$ is an integer, QED.

@rem $ n^2 + n\equiv 0$ or $ 2 \pmod 9$ This conclusion is wrong , such as $ n \equiv 2 \pmod{9} $ then $ n^2 + n\equiv 6 \pmod{9}$

The elliptic curve $y^2 + 9y = x^3 - 189$ has Mordell-Weil group $E(\mathbf{Q}) = 0$ (no non-trivial torsion and trivial $2$-Selmer group).

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