= Solution
{c}
There is one four-gluon <tadpole diagram>. Temporarily replace its internal <gluon propagator> by $-ig_{\rho\sigma}\delta^{cd}/(l^2-m_g^2+i0)$. Contract the supplied four-gluon <Feynman vertex> with $g_{\rho\sigma}\delta^{cd}$. Its first <colour factor> vanishes because $f^{cce}=0$. Each of the other two terms contributes $C_A\delta^{ab}(D-1)g^{\mu\nu}$, so
$$
\delta^{cd}g_{\rho\sigma}V_4^{\mu\nu\rho\sigma,abcd}=-2ig_D^2C_A\delta^{ab}(D-1)g^{\mu\nu}.
$$
Include the <symmetry factor> $1/2$ and the factor $-i$ from the internal <gluon propagator>:
$$
i\Pi_{4g}^{\mu\nu,ab}=-g_D^2C_A\delta^{ab}(D-1)g^{\mu\nu}J_1(m_g^2).
$$
The exact regulated <tadpole integral> is
$$
J_1(m_g^2)=-\frac{i}{(4\pi)^{D/2}}\Gamma(1-D/2)(m_g^2)^{D/2-1}.
$$
In particular,
$$
i\Pi_{4g}^{\mu\nu,ab}\big|_{\mathrm{UV}}=-\frac{i g^2C_A\delta^{ab}}{8\pi^2\epsilon}(D-1)m_g^2g^{\mu\nu}.
$$
The entire expression is proportional to $(m_g^2)^{1-\epsilon/2}$ and vanishes as the auxiliary <mass> tends to zero near four dimensions. Equivalently, its expanded terms are proportional to $m_g^2/\epsilon$, $m_g^2\log m_g^2$ and $m_g^2$, all of which disappear when the regulator is removed. Hence
$$
\boxed{\Pi_{4g}^{\mu\nu,ab}=0\qquad(m_g\to0).}
$$
This is the massless <scaleless integral> result obtained directly in <dimensional regularization>. The artificial <mass> is only an intermediate device for this graph; retaining it in the full theory would modify the <gauge symmetry> and cannot be used when applying the massless transversality identity in part (iv).
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