Difference in equations from four-vector conservation

This is a question, kind of related to my previous question but is here as a stand-alone. I’ve been reading the book ‘essential relativity’ by Rindler and have a question similar to 5.20 which is bothering me. It is the problem that when a photon hits a stationary proton, a pion is created at an angle $theta$ relative to the x-axis (which is the direction of the photon) and the proton is scattered away, now we want to find the minimal required energy of the photon for the pion to be created. I solved this question as follows:
$$P^mu_gamma + P^mu_p = P^mu_{p’} + P^mu_{pi^0}$$
squaring both sides and projecting using $eta_{mumu}$:
$$0 + 2P^mu_gamma P_{mu}^{(p)} = m_{pi^0}^2c^2 + 2P^mu_{p’}P_{mu}^{(pi^0)}$$
Now solving the left in the proton rest frame (i.e the lab frame) and solving the right in the pion rest frame giving rise to a $P^0$ for the scattered proton perceived by the pion of $m_pcgamma(v_{rel})$:
$$2frac{hnu}{c}m_pc = 2m_{pi^0}cm_pcgamma(v_{rel}) + m_{pi^0}^2c^2$$

which gives
$$hnu = frac{(m_{pi^0}^2 + 2m_pm_{pi^0})c^2}{2m_p}$$
which is stated to be the correct answer, however when I try to do it the following way: I projected the first equation onto $P_{mu}^{(pi^0)}$ giving rise to (not writing the $mu$ indexes):
$$P_{gamma}P_{pi^0} + P_pP_{pi^0} = P_{p’}P_{pi^0} + m_{pi^0}^2c^2$$
with the right side then being the same as (from the second equation):
$$=P_{gamma}P_p + frac{m_{pi}^2c^2}{2}$$
and thus:
$$P_{gamma}(P_{pi} – P_p) = frac{m_{pi}^2c^2}{2} – P_pP_{pi^0}$$
which is, after filling in
$$P_{gamma} = (frac{hnu}{c},frac{hnu}{c},0,0)$$
$$P_p = (m_p c,0,0,0)$$
$$P_{pi^0} = (gamma m_{pi^0}c,gamma m_{pi^0} vec{v})$$

and setting $gamma$ to one (and v=0) gives rise to the equation

$$hnu = frac{(m_{pi}^2 – 2m_pm_{pi})c^2}{2(m_p – m_{pi})}$$

which is different from the one found before (which is stated as correct by the book), why is this different? Am I considering a different frame of reference somewhere without realising?