Show that the equation $2a^3 - 7b^2 = 1$ has no solution over the integers

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Here is this week's POTW:

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Show that the equation $2a^3 - 7b^2 = 1$ has no solution over the integers.

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No one answered this week's problem. You can read my solution below.

Spoiler

Reducing the equation modulo $7$ yields $2a^3 \equiv 1 \pmod{7}$, or $a^3 \equiv 4 \pmod{7}$. Note that $\pm 1, \pm 2$, and $\pm 3$ all have cubes that are $\equiv \pm 1\pmod{7}$. Hence, the congruence $a^3 \equiv 3\pmod{7}$ has no solution. This implies the original Diophantine equation has no solution.