= Solution
{c}
The third panel of the figure in part <2/a> is a representative four-photon <box diagram>. Fix the external photon labelled 1 and orient the <fermion loop>. The six inequivalent cyclic orders of the labelled <interaction vertices> are
$$
\boxed{(1234),\ (1243),\ (1324),\ (1342),\ (1423),\ (1432).}
$$
They are obtained by permuting labels $2,3,4$ on the displayed box, keeping label 1 fixed. Equivalently there are three undirected boxes, each with two opposite <fermion flows>. The reversal pairs are
$$
(1234)\leftrightarrow(1432),\qquad
(1243)\leftrightarrow(1342),\qquad
(1324)\leftrightarrow(1423).
$$
For four photon <interaction vertices>, the sign in part <2/d> is positive, so the members of each pair add. They can be represented by three <box diagrams> multiplied by two. There is no tree photon self-coupling in <QED>, and therefore no triangle or bubble with a photon self-vertex to add at this <loop order>. This is the one-loop <light-by-light scattering> contribution, of order $e^4$.
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