Find the set of integers $a$ such that the equation $2 a^{2}=7 k+2$ has an integer solution $k$.

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The problem can be restated as follows: find the integers $a$ such that $2 a^{2} \equiv 2$ (mod. 7). Such a relation depends only on the congruence of $a$ modulo 7. We
then verify by hand that $2 \times 0^{2} \equiv 0(\bmod .7), 2 \times 1^{2} \equiv 2 \times 6^{2} \equiv 2(\bmod .7), 2 \times 2^{2} \equiv 2 \times 5^{2} \equiv$ $1(\bmod .7)$ and $2 \times 3^{2} \equiv 2 \times 4^{2} \equiv 4(\bmod .7)$. It follows that $\frac{2 a^{2}-2}{7}$ is an integer if and only if $a$ is of the form $7 \ell+1$ or $7 \ell-1$.