= Solution
Use dimensionless interaction couplings in the exponent of the <Boltzmann weight>. In the following formulas $K,L$ denote the physical interaction energies divided by $k_BT$ if that factor has not already been absorbed into the displayed Hamiltonian. This convention is required for temperature-independent numerical fixed-point coordinates.
Split the original <square lattice> into retained sites $E$ with even $x+y$ and eliminated sites $O$ with odd $x+y$, where coordinates are in units of $a$. Every nearest-neighbour bond joins $E$ to $O$, whereas a diagonal next-nearest-neighbour bond stays within one sublattice. Denote retained <Ising spins> by $s_e$ and eliminated <Ising spins> by $\eta_o$. For fixed retained <Ising spins> define
$$
S_o=\sum_{e:\,e\text{ nearest to }o}s_e.
$$
The blocked <Boltzmann weight> in the <checkerboard decimation of the square-lattice Ising model> is
$$
W(s)=\exp\left(L\sum_{\langle ee'\rangle_{\mathrm{diag}}}s_es_{e'}\right)
\sum_{\{\eta\}}\exp\left[K\sum_{o\in O}\eta_oS_o
+L\sum_{\langle oo'\rangle_{\mathrm{diag}}}\eta_o\eta_{o'}\right].
$$
Use the independent uniform <product measure> on <Ising spins> on $O$ to perform a <cumulant expansion>. Its one-spin mean vanishes, and $\langle\eta_o\eta_{o'}\rangle_0=\delta_{oo'}$. Consequently the first-order $K$ term, the first-order eliminated-sublattice $L$ term, and the mixed $KL$ term vanish. To the requested order,
$$
\log W(s)=\frac N2\log2+L\sum_{\langle ee'\rangle_{\mathrm{diag}}}s_es_{e'}
+\frac{K^2}{2}\sum_{o\in O}S_o^2
+O(K^4,K^2L,L^2).
$$
Here the truncation keeps $K^2$ and the linear term in $L$; equivalently count $L$ as order $K^2$. At higher orders the interaction family will generally not close.
Each eliminated site has four retained neighbours, so
$$
S_o^2=4+2\sum_{\{e,e'\}\subset\mathrm{NN}(o)}s_es_{e'}.
$$
Its six distinct neighbour pairs comprise four corner pairs at distance $\sqrt2a$ and two opposite pairs at distance $2a$. A given retained diagonal pair shares two eliminated centres; a given opposite pair shares only one. Thus the generated pair couplings are $2K^2$ for diagonal pairs and $K^2$ for axial pairs separated by two original spacings. The original $L$ interaction directly contributes to each retained diagonal pair. No single-spin term survives <spin inversion symmetry>, and no three-spin or four-spin term is present in this second cumulant. This proves closure on exactly two nonconstant interactions at the retained order.
The primitive vectors of the retained <square lattice> are $a(1,1)$ and $a(1,-1)$. Rotate the axes by $45$ degrees and rescale lengths by $1/\sqrt2$; retained diagonal pairs become nearest neighbours, and the axial distance-$2a$ pairs become next-nearest neighbours. With $b=\sqrt2$, the <leading checkerboard Ising decimation recursion> is therefore
$$
\boxed{K'=L+2K^2,\qquad L'=K^2.}
$$
The constant part of the weight is also determined: $4(K^2/2)(N/2)=NK^2$, so the original <partition function> equals
$$
Z(K,L)=2^{N/2}e^{NK^2}\,Z_{N/2}(K',L')
$$
to the same order in the logarithm of the weights. That normalization changes <free energy> but not the retained-spin interaction operators. If physical energy couplings are kept instead, at fixed $\beta_T$ the relations read $K'_{\mathrm{energy}}=L_{\mathrm{energy}}+2\beta_TK_{\mathrm{energy}}^2$ and $L'_{\mathrm{energy}}=\beta_TK_{\mathrm{energy}}^2$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-303-checkerboard-decimation.png]
{title=Two shared eliminated neighbours generate a nearest-neighbour bond; one generates a next-nearest-neighbour bond after checkerboard decimation}
The fixed-point equations are $L_*=K_*^2$ and $K_*=3K_*^2$. Their only finite real solutions are
$$
\boxed{(K_*,L_*)=(0,0)\quad\text{and}\quad(K_*,L_*)=\left(\frac13,\frac19\right).}
$$
The origin is the high-temperature fixed point; its <Jacobian matrix> is $\begin{pmatrix}0&1\\0&0\end{pmatrix}$, with both multipliers zero. The nontrivial candidate critical point is $(1/3,1/9)$. Linearize the discrete <renormalization-group transformation> there:
$$
\begin{pmatrix}\delta K'\\\delta L'\end{pmatrix}
=\begin{pmatrix}4/3&1\\2/3&0\end{pmatrix}
\begin{pmatrix}\delta K\\\delta L\end{pmatrix}+O(\delta K^2).
$$
The characteristic polynomial is $\lambda^2-(4/3)\lambda-2/3$, giving
$$
\boxed{\lambda_+=\frac{2+\sqrt{10}}3\approx1.72076,\qquad
\lambda_-=\frac{2-\sqrt{10}}3\approx-0.387426.}
$$
<Eigenvectors> may be chosen as $(1,2/(3\lambda_\pm))^T$. The positive <eigenvalue> greater than one is the <relevant direction of a fixed point>; the other is irrelevant because its magnitude is less than one. Its negative sign only alternates the sign of the perturbation between steps. For a discrete map the magnitude relative to one decides relevance, not the sign rule used for continuous beta-function <eigenvalues>.
Let $t$ be the relevant thermal scaling coordinate, so $t'=\lambda_+t$ to first order. A <correlation length> measured in restored lattice units satisfies $\xi(t')=\xi(t)/b$. If $\xi(t)\sim|t|^{-\nu}$, then $\lambda_+^{-\nu}=b^{-1}$. Hence the <thermal exponent from a discrete renormalization map> is
$$
\boxed{\nu=\frac{\log b}{\log\lambda_+}
=\frac{\log\sqrt2}{\log[(2+\sqrt{10})/3]}\approx0.6385.}
$$
This is the value predicted by the stated low-order <real-space renormalization group>, whose nontrivial fixed point is evaluated beyond the strictly infinitesimal-coupling limit. It is an approximation, not an exact solution of the full square-lattice <Ising model>.
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