A false conjecture by Goldbach

Observe that $5777\equiv 2\pmod 3$ and $2a^2\equiv 2\pmod 3$ unless $3\mid a$. Hence once you check that $5777-3$ is not twice a square, you need only check $a$ with $3\mid a$ (cutting down the effort by two thirds).

Likewise, $5777\equiv 2\pmod 5$, which allows you to drop all cases where $a\equiv \pm1\pmod 5$ (after checking that $5777-5$ is not twice a square).

Say $p = 5777 - 2a^2$.

Now, if $a \equiv 1,2 \mod 3$ then $p$ is divisible by $3$, which is impossible as we can check that $p \neq 3$.

If $a \equiv 0 \mod 3$ then $p \equiv 17 \mod 18$, so it is sufficient to only check a few values for $p$.