Is total four-momentum square invariant

Question

When a particle with mass m collides with a fixed target with the same mass, are the followings true?

1. $p_{total}^{2}=\frac{\left( E_{1}+E_{2} \right)^{2}}{c^{2}} = (2m)^{2}c^{2}$ (If so, that means $E_{1}$ and $E_{2}$ vary between reference frames, but the $p_{total}^{2}=4m^{2}c^{2}$ always hold. )

2. The $p_{total}^{2}$ is conserved after the collision. (Is the $p_{total}^{2}$ not only invariant, but also conserved?)

Answer 1

In Minkowski the square of the four-momentum $P^\mu$ is
$P^2 = \eta_{\mu \nu} P^\mu P^\nu = P^\mu P_\mu = -E^2/c^2 +p^2 = -m^2 c^2$
where:
$\mu = 0, 1, 2, 3$
$\eta_{\mu \nu} = diag(-1, 1, 1, 1)$ metric tensor
$p^i$ relativistic three-momentum
$p^2 = p^i p_i$
$i = 1, 2, 3$
The square of the four-momentum is an invariant, while the total four-momentum of a physical system is a conserved quantity in a specific reference frame.
Note: Point 1. in the question is not applicable to a generic reference frame, as it neglects the three-momentum.