= Solution
Expanding $e^{iS_{\rm int}}$ gives the <Feynman vertex>
$$
\boxed{V_{ab}=-ig(\gamma^5)_{ab}.}
$$
There is one scalar leg and a continuous oriented fermion line; no extra factor $i$ is absorbed into $\gamma^5$. The <Dirac propagator> is $S_F(k)=i(\not{k}+M)/(k^2-M^2+i0)$, where $\not{k}=\gamma^\mu k_\mu$. Each closed <fermion loop> supplies a minus sign.
\Image[/past-exam-of-the-quantum-field-theory-course-of-the-university-of-oxford/2025-pseudoscalar-loops.png]
{title=Pseudoscalar Yukawa vertex and the vanishing one-loop bosonic one-point and three-point diagrams}
The <tadpole diagram> is proportional to
$$
-\int\frac{d^4k}{(2\pi)^4}\operatorname{tr}\left[(-ig\gamma^5)\frac{i(\not{k}+M)}{k^2-M^2+i0}\right]=0,
$$
because both numerator <traces> vanish by <1/a>. This is zero before the <loop momentum> integration.
For either orientation of the <triangle Feynman diagram>, let $a,b,c$ be its three internal <four-momenta>. Its numerator <trace> is
$$
\begin{aligned}
N&=\operatorname{tr}\left[\gamma^5(\not{a}+M)\gamma^5(\not{b}+M)\gamma^5(\not{c}+M)\right]\\
&=\operatorname{tr}\left[(\not{a}-M)(\not{b}+M)\gamma^5(\not{c}+M)\right]=0.
\end{aligned}
$$
After using <cyclic property of the trace>, each expanded term is a <trace> of $\gamma^5$ times at most three ordinary <gamma matrices>. Such <traces> vanish in four dimensions. For an odd number of ordinary <gamma matrices>, conjugation by $\gamma^5$ changes the sign. For two ordinary <gamma matrices>, the contracted <Clifford algebra> identity $\gamma_\rho\gamma^\mu\gamma^\nu\gamma^\rho=4\eta^{\mu\nu}I$ gives
$$
4\operatorname{tr}(\gamma^5\gamma^\mu\gamma^\nu)=-4\eta^{\mu\nu}\operatorname{tr}\gamma^5=0.
$$
The zero-matrix case is <1/a>. Hence \b[both the one-point and three-point one-loop bosonic vertices vanish], independently of their external <momenta>. No cancellation between different loop orientations or delicate shift of a divergent <integral> is needed.
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