Prime pair equation

The remainder on dividing any cube number by $7$ is always one of $\{-1,0,1\}$ . The technical way of saying this is that these are the cubic residues modulo $7$ ; we have $0^3 \equiv 0,\;\;\;1^3\equiv 2^3 \equiv 4^3 \equiv 1,\;\;\; 3^3\equiv 5^3 \equiv 6^3 \equiv -1 \pmod{7}$

The given equation becomes $-a^3-b^3 \equiv 1 \pmod{7}$

The only way this can happen is if one of $a$ or $b$ is a multiple of $7$ ; and since they're prime, would have to be $7$ itself.

$20a^3-7^3=1 \implies 20a^3=344$ has no integer solution.

$20\cdot 7^3-b^3=1 \implies b=19$ is a valid solution; so there is exactly $\boxed1$ solution pair.

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An alternative approach is to rearrange the equation as $20a^3=b^3+1=(b+1)\left(b^2-b+1\right)$

Comparing factors of $b+1$ , $b^2-b+1$ and $20a^3$ leads to a solution but involves more case analysis than the above approach.