= Solution
Use signed one-dimensional external momenta $k_1+k_2+k_3=0$, with nonzero $k_i$. Let $P_L(k)=P_L(-k)$ be the linear <cosmological density power spectrum> at the time of evaluation, normalized by $\langle\delta^{(1)}(k)\delta^{(1)}(k')\rangle=2\pi\delta^{(D)}(k+k')P_L(k)$. Using $P_L$ absorbs the factor $a^6$ multiplying a product of three initial <power spectra>. All kernels below are symmetrized <one-dimensional cosmological density kernels>.
The connected <B222 contribution to the one-loop matter bispectrum> has a triangular <Feynman diagram>: each second-order external field has two Gaussian inputs, joined pairwise to the other two vertices. The connected <B411 contribution to the one-loop matter bispectrum> has a fourth-order vertex joined to both linear external fields, with its remaining two inputs contracted into a loop at that vertex. External stubs indicate the measured momenta; blue internal lines represent linear <power spectra>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-312-loop-diagrams.png]
{title=One-loop triangle and fourth-order-vertex diagrams}
{description=The two connected one-loop topologies. Vertex labels are perturbative orders, not numbers of external measured fields. The $B_{411}$ diagram must be summed over the three choices of the fourth-order field.}
There are $2^3=8$ connected <Wick contractions> in the triangle, giving
$$
\boxed{B_{222}=8\int\frac{dq}{2\pi}\,F_2(q,k_1-q)F_2(-q,k_2+q)F_2(q-k_1,-k_2-q)
P_L(q)P_L(k_1-q)P_L(k_2+q)}.
$$
For a fixed fourth-order external field, attaching the two labelled linear fields gives $4\times3=12$ choices; the remaining pair contracts uniquely. Therefore
$$
\boxed{B_{411}=12P_L(k_1)P_L(k_2)\int\frac{dq}{2\pi}F_4(-k_1,-k_2,q,-q)P_L(q)+\text{two cyclic terms}}.
$$
These are contributions to the reduced connected <one-loop matter bispectrum>, with its external momentum-conserving <Dirac delta function> removed. Disconnected contractions have a zero external momentum and are excluded.
For the ultraviolet behaviour, the three $F_2$ denominators multiply to a negative square. In fact
$$
8F_2(q,k_1-q)F_2(-q,k_2+q)F_2(q-k_1,-k_2-q)
=-\frac{k_1^2k_2^2k_3^2}{q^2(k_1-q)^2(k_2+q)^2}.
$$
Thus for $|q|\gg |k_i|$ and a smoothly varying high-momentum spectrum,
$$
B_{222}^{\rm UV}\simeq-k_1^2k_2^2k_3^2\int_{\rm hard}\frac{dq}{2\pi}\frac{P_L(q)^3}{q^6}.
$$
Each second-order vertex exhibits the <ultraviolet softness of the second-order density kernel>, so this term contains six powers of external momentum.
The fourth-order kernel is even simpler:
$$
F_4(-k_1,-k_2,q,-q)=-\frac{k_3^4}{24k_1k_2q^2}.
$$
With a hard-loop cutoff $q_h<|q|<\Lambda$, define
$$
I_\Lambda=\int_{q_h<|q|<\Lambda}\frac{dq}{2\pi}\frac{P_L(q)}{q^2}.
$$
The <B411 contribution to the one-loop matter bispectrum> from this range is exactly
$$
B_{411}^{\rm hard}=-\frac12\sum_{\rm cyclic}\frac{k_3^4}{k_1k_2}P_L(k_1)P_L(k_2)I_\Lambda
=-\sum_{\rm cyclic}k_3^2F_2(k_1,k_2)P_L(k_1)P_L(k_2)I_\Lambda.
$$
For fixed momentum ratios, it has two external derivative powers multiplying two long-mode <power spectra>, rather than the six-derivative stochastic structure of $B_{222}$. It is the leading deterministic UV-sensitive contribution for external modes below the <nonlinear wavenumber>. This is a comparison of derivative orders and generic cutoff sensitivity; an arbitrary specially chosen spectrum or vanishing shape can change their numerical ranking. A convergent integral can still depend on nonperturbative short scales. For a high-$q$ power law $P_L(q)\propto |q|^n$, the $B_{411}$ integral diverges for $n\geq1$, whereas the $B_{222}$ integral diverges only for $n\geq5/3$, with logarithmic divergences at the thresholds.
At the field level, there are six ways to contract a hard pair within $\delta^{(4)}$. Comparing $6F_4$ with $F_2$ gives
$$
\delta^{(4)}_{\rm hard}(k)=-\frac{I_\Lambda}{2}k^2\delta^{(2)}(k).
$$
The appropriate <second-order density Laplacian counterterm> is therefore
$$
\boxed{\delta_{\rm ct}^{(2)}(x)=C(\Lambda)\partial_x^2\delta^{(2)}(x),\qquad
C(\Lambda)=-\frac12 I_\Lambda+C_{\rm finite}}.
$$
Indeed its contribution with two linear fields is $-2C(\Lambda)\sum_{\rm cyclic}k_3^2F_2P_L(k_1)P_L(k_2)$, which cancels the displayed hard-loop term. The <counterterm> sign follows from $\partial_x^2\leftrightarrow-k^2$. Its finite coefficient must be matched to short-scale dynamics, as in the <effective field theory of large-scale structure>. A term proportional to $\partial_x^2\delta^{(1)}$ similarly renormalizes a linear response, but a purely linear insertion cannot by itself cancel this $\langle\delta^{(4)}\delta^{(1)}\delta^{(1)}\rangle$ shape, since an odd Gaussian three-point function vanishes.
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