= Solution
{c}
At large <loop momentum>, each <Dirac propagator> scales as $\ell^{-1}$. One four-dimensional loop and four internal <Dirac propagators> therefore give the <superficial degree of divergence>
$$
D=4-4=0.
$$
Thus a possible <ultraviolet divergence> in the sum is logarithmic. By locality its divergent coefficient is a degree-zero polynomial in the external <four-momenta>, namely a constant Lorentz tensor,
$$
\mathcal M_{\rm div}^{\mu\nu\rho\sigma}
=\frac1\epsilon\left(a\,g^{\mu\nu}g^{\rho\sigma}
+b\,g^{\mu\rho}g^{\nu\sigma}
+c\,g^{\mu\sigma}g^{\nu\rho}\right).
$$
The photon <Ward identities> apply to the complete sum for arbitrary external <four-momenta>, including <off shell> ones: $k_{1\mu}\mathcal M^{\mu\nu\rho\sigma}=0$, and likewise for the other legs. A constant tensor with $k_{1\mu}\mathcal M_{\rm div}^{\mu\nu\rho\sigma}=0$ for every $k_1$ must be zero. Hence $a=b=c=0$ and
$$
\boxed{\mathcal M_{\rm one\ loop}^{\mu\nu\rho\sigma}\ \text{is ultraviolet finite}.}
$$
Equivalently, a logarithmic divergence would require a local dimension-four $A^4$ <counterterm>, forbidden by Abelian <gauge invariance>. A gauge-invariant four-photon <local operator> starts with four <gauge field strengths>, for example $(F_{\mu\nu}F^{\mu\nu})^2$, of <mass dimension> eight; it cannot be this <power counting> divergence. Individual boxes need not be finite; their forbidden divergent parts cancel in the sum. There are no divergent proper loop subgraphs at this <loop order>. With the massive <Dirac field> in the source, there is also no soft or collinear <infrared divergence> at generic external kinematics. The argument establishes ultraviolet finiteness, and does not claim the absence of physical threshold singularities.
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