Peter Scholze explains the same example in more detail in this talk https://youtu.be/cFdm0B9KLcQ?t=48m and gives even more detail here https://youtu.be/gI3Ta73yVuo?t=23m55s.

In brief, the relationship is that for every prime $p$ not dividing $183$, the polynomial has a root mod $p$ if and only if the eigenvalue of the $p$th Hecke operator on a certain 3-torsion class in that integral homology group is nonzero.

The $p$th Hecke operator comes from the subgroup $\Gamma_0(p)$ of $\Gamma$ which has two equivalent descriptions - as the subgroup of $\Gamma$ consisting of elements where $\gamma_{1,2} \equiv 0\mod p$ and the subgroup consisting of elements where $\gamma_{2,1}\equiv 0 \mod p$. The isomorphism between these two subgroups comes from multiplying $\gamma_{1,2}$ by $p$ and dividing $\gamma_{2,1}$ by $p$.

This isomorphism means there are two distinct injections $\Gamma_0 (p) \to \Gamma$, hence two distinct covering maps maps $\mathbb H^3/\Gamma_0(p) \to \mathbb H^3/\Gamma$. Pulling back a homology class along the first map (possible because the map is a finite covering) and pushing it forward along the second map produces an operator $T_p: H_1( \mathbb H^3/\Gamma,\mathbb Z) \to H_1( \mathbb H^3/\Gamma,\mathbb Z)$.

Figuiredo located a $3$-torsion class in this group that is an eigenvector of all these operators, and Scholze's result implies that the eigenvalue of $T_p$ on this class is nonzero if and only if the polynomial has a root mod $p$.

Peter Scholze explains the same example in more detail in this talk https://youtu.be/cFdm0B9KLcQ?t=48m and gives even more detail here https://youtu.be/gI3Ta73yVuo?t=23m55s.

In brief, the relationship is that for every prime $\mathfrak p$ of $\mathbb Z[i]$ not dividing $183$, the polynomial has a root mod $\mathfrak p$ if and only if the eigenvalue of the $\mathfrak p$th Hecke operator on a certain 3-torsion class in that integral homology group is nonzero.

The $p$th Hecke operator comes from the subgroup $\Gamma_0(\mathfrak p)$ of $\Gamma$ which has two equivalent descriptions - as the subgroup of $\Gamma$ consisting of elements where $\gamma_{1,2} \equiv 0\mod \mathfrak p$ and the subgroup consisting of elements where $\gamma_{2,1}\equiv 0 \mod \mathfrak p$. The isomorphism between these two subgroups comes from multiplying $\gamma_{1,2}$ by $\mathfrak p$ and dividing $\gamma_{2,1}$ by $\mathfrak p$.

This isomorphism means there are two distinct injections $\Gamma_0 (\mathfrak p) \to \Gamma$, hence two distinct covering maps maps $\mathbb H^3/\Gamma_0(\mathfrak p) \to \mathbb H^3/\Gamma$. Pulling back a homology class along the first map (possible because the map is a finite covering) and pushing it forward along the second map produces an operator $T_\mathfrak p: H_1( \mathbb H^3/\Gamma,\mathbb Z) \to H_1( \mathbb H^3/\Gamma,\mathbb Z)$.

Figuiredo located a $3$-torsion class in this group that is an eigenvector of all these operators, and Scholze's result implies that the eigenvalue of $T_\mathfrak p$ on this class is nonzero if and only if the polynomial has a root mod $\mathfrak p$.

Peter Scholze explains the same example in more detail in this talk: https://youtu.be/cFdm0B9KLcQ?t=48m

In brief, the relationship is that for every prime $p$ not dividing $183$, the polynomial has a root mod $p$ if and only if the eigenvalue of the $p$th Hecke operator on a certain 3-torsion class in that integral homology group is nonzero.

The $p$th Hecke operator comes from the subgroup $\Gamma_0(p)$ of $\Gamma$ which has two equivalent descriptions - as the subgroup of $\Gamma$ consisting of elements where $\gamma_{1,2} \equiv 0\mod p$ and the subgroup consisting of elements where $\gamma_{2,1}\equiv 0 \mod p$. The isomorphism between these two subgroups comes from multiplying $\gamma_{1,2}$ by $p$ and dividing $\gamma_{2,1}$ by $p$.

This isomorphism means there are two distinct injections $\Gamma_0 (p) \to \Gamma$, hence two distinct covering maps maps $\mathbb H^3/\Gamma_0(p) \to \mathbb H^3/\Gamma$. Pulling back a homology class along the first map (possible because the map is a finite covering) and pushing it forward along the second map produces an operator $T_p: H_1( \mathbb H^3/\Gamma,\mathbb Z) \to H_1( \mathbb H^3/\Gamma,\mathbb Z)$.

Figuiredo located a $3$-torsion class in this group that is an eigenvector of all these operators, and Scholze's result implies that the eigenvalue of $T_p$ on this class is nonzero if and only if the polynomial has a root mod $p$.

Peter Scholze explains the same example in more detail in this talk https://youtu.be/cFdm0B9KLcQ?t=48m and gives even more detail here https://youtu.be/gI3Ta73yVuo?t=23m55s.

In brief, the relationship is that for every prime $p$ not dividing $183$, the polynomial has a root mod $p$ if and only if the eigenvalue of the $p$th Hecke operator on a certain 3-torsion class in that integral homology group is nonzero.

The $p$th Hecke operator comes from the subgroup $\Gamma_0(p)$ of $\Gamma$ which has two equivalent descriptions - as the subgroup of $\Gamma$ consisting of elements where $\gamma_{1,2} \equiv 0\mod p$ and the subgroup consisting of elements where $\gamma_{2,1}\equiv 0 \mod p$. The isomorphism between these two subgroups comes from multiplying $\gamma_{1,2}$ by $p$ and dividing $\gamma_{2,1}$ by $p$.

This isomorphism means there are two distinct injections $\Gamma_0 (p) \to \Gamma$, hence two distinct covering maps maps $\mathbb H^3/\Gamma_0(p) \to \mathbb H^3/\Gamma$. Pulling back a homology class along the first map (possible because the map is a finite covering) and pushing it forward along the second map produces an operator $T_p: H_1( \mathbb H^3/\Gamma,\mathbb Z) \to H_1( \mathbb H^3/\Gamma,\mathbb Z)$.

Figuiredo located a $3$-torsion class in this group that is an eigenvector of all these operators, and Scholze's result implies that the eigenvalue of $T_p$ on this class is nonzero if and only if the polynomial has a root mod $p$.