Many people have tried and failed to extend Apery's Irrationality proof of $\zeta(3)$ to Catalan's constant, by looking for a fast converging series for Catalan's constant analogous to the one for $\zeta(3)$ that Apery utilized:
$${\displaystyle {\begin{aligned}\zeta (3)&={\frac {5}{2}}\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{{\binom {2k}{k}}k^{3}}}\end{aligned}}}.$$
see this question: Why can we not establish the irrationality of Catalan's constant the same way as $\zeta(3)$?
which unfortunately didn't receive an answer.
My question is different and specifically relates to Beukers-like irrationality proof for $\zeta(2)$/$\zeta(3)$ as most clearly articulated recently by F. M. S. Lima in Beukers-like irrationality proofs for $\zeta(2)$ and $\zeta(3)$
For analogous Beukers-like irrationality proof applied in the case Catalan's Constant, does a proof fail in the initial lemma's or in evaluating and applying the unit square integral?
Lets rewrite the Lemma's applying to $\zeta(2)$ given by F. M. S. Lima and apply them to an analogous unit square integral for Catalan's Constant, $G$, that is "Lemma N" in the paper becomes "Lemma N_G" here:
Lemma 1G A unit square integral for Catalan's Constant $$\int_0^1 \int_0^1 \frac{1}{1+(x y)^2} \,dx\,dy= G$$
Lemma 2G ($I_{2r,2r}$) For all odd integers $r>0$ $$\int_0^1 \int_0^1 \frac{x^{2r}y^{2r}}{1+(x y)^2} \,dx\,dy= G-\sum_{m=1}^{2r}\frac{(-1)^{m-1}}{(2m-1)^2}$$
Lemma 3G ($I_{2r,2s}$) Let r and s be non-negative odd integers, $r\ne s$. Then $$\int_0^1 \int_0^1 \frac{x^{2r}y^{2s}}{1+(x y)^2} \,dx\,dy=\frac{\widetilde{h_s}-\widetilde{h_r}}{2(r-s)}$$ where $\widetilde{h_n}=\sum_{m=1}^n \frac{(-1)^{m-1}}{(2m-1)}$, an alternating analog of the Harmonic Number.
Lemma 4G ($I_{2r,2r}$ as a linear form). For all odd integers $r>0$ $$I_{2r,2r}=G-\frac{z_{2r}}{(d_{2r})^2} $$ for some $z_{2r} \in {\mathbb{N}}^*$. Where $d_{r}=lcm(1^2,3^2,5^2,...,r^2)$
Lemma 5G ($I_{2r,2s}$ is a positive rational). For all odd $r,s \in {\mathbb{N}},\, r \ne s,$ $$I_{2r,2s}=\frac{z_{2r,2s}}{(d_{2r})^2} $$ for some $z_{2r,2s}\in {\mathbb{N}^*}$
Lemma 6G and Lemma 7G These are written in terms of $n$ and therefore both Lemma's can be written in terms of $2n$
If the analogous lemma's can be all proved correct for odd integer $r,s>0$ then presumably a proof in regards to $G$ must fail in the main part of the proof, what Lima labels Theorem 1, i.e. in the process of evaluating the new unit square integral and applying the result; which for me is the hardest part of this proof.
Added One difficulty is defining the even powered polynomials to multiply together - the resultant terms will have more than one factor of 2 so the lemmas above need to be modified.
