First, we derive meson condensates from the one-point function $\mathcal{M}_{x}=\bar{\chi}_{x}\chi_{x}$. The equation for the one-point function is obtained as shown in Fig. \[One-A\], $$\begin{aligned}
- \Sigma_x
 & \equiv
 - \langle \mathcal{M}_x \rangle
{\nonumber}\\
 & =
 - \langle \mathcal{M}_x \rangle_0 + 2K^2 \sum_\mu \Sigma_x \Sigma_{x+\hat{\mu}}
 - 2 \cdot \displaystyle \frac {1}{(4!)^2 \cdot 2^3} (Kr)^2 
 \sum_{\mu \neq \nu \neq \rho \neq \sigma} 
 \Sigma_x \Sigma_{x+\hat{\mu}+\hat{\nu}+\hat{\rho}+\hat{\sigma}}
\ .
\label{Eq:HPE-Adams1}\end{aligned}$$

![Feynman diagram for mesonic one-point functions in the $\mathcal{O}(K^{3})$ self-consistent equations of HPE with the Adams fermion. There is a 4-hopping fundamental diagram, which is peculiar to this fermion.[]{data-label="One-A"}](HPEwAdams.eps){height="3cm"}

We substitute $\Sigma_x=\sigma_x +i \epsilon_x \pi_x$ for $\Sigma_{x}$ in Eq. (\[Eq:HPE-Adams1\]) and obtain the self-consistent equation $$- \left( \sigma +i \epsilon_x \pi \right)=
 -1 + 2K^2 \cdot 4 \left( \sigma^2 + \pi^2 \right)
 - 2 \cdot \displaystyle \frac {1}{(4!)^2 \cdot 2^3} (Kr)^2 \cdot 4! 
 \left( \sigma +i \epsilon_x \pi \right)^2 
\ .
\label{HPE-AdamsSelf1}$$ From this, we obtain $- \sigma = -1 + 16 K^2 \pi^2$ and $- i \pi = - 8K^2 \cdot 2 i \sigma \pi$. Here we have set $r=16\sqrt{3}$ to make the equation simple. We again have two solutions: For $\pi=0$, we have a trivial solution $\sigma=1$. For $\pi \neq 0$, we have 