= Solution
The <Maxwell field> has the <gauge symmetry> $A_\mu\mapsto A_\mu+\partial_\mu\alpha(x)$ for an arbitrary smooth scalar function $\alpha$. Commuting the two derivatives shows that $F_{\mu\nu}$ is unchanged, so the <Maxwell Lagrangian> is invariant. Vary the action, use antisymmetry of $F$ and integrate by parts with variations vanishing on the boundary:
$$
\delta S=-\frac12\int d^4x\,F^{\mu\nu}\delta F_{\mu\nu}
=-\int d^4x\,F^{\mu\nu}\partial_\mu\delta A_\nu
=\int d^4x\,(\partial_\mu F^{\mu\nu})\delta A_\nu.
$$
The <Euler-Lagrange field equations> are therefore $\partial_\mu F^{\mu\nu}=0$. Expanding $F$ gives
$$
\Box A^\nu-\partial^\nu(\partial_\mu A^\mu)=0,
\qquad \Box=\partial_\mu\partial^\mu.
$$
In <Lorenz gauge> the second term vanishes, leaving
$$
\boxed{\Box A_\nu=0.}
$$
The PDF calls this “Lorentz gauge”; the standard name is <Lorenz gauge>. A residual <gauge symmetry> preserves this condition when $\Box\alpha=0$.
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