Looking for a proof relativistic momentum is conserved Using first principals Could someone help show that in special relativity, conservation of momentum is independent of inertial frame by applying Lorentz transform.Or better, can you derive the formula for relativistic momentum under the requirement of conservation of momentum for inertial 
Here’s what I’m hoping you can help me with. Your naive to SR, so you try and define momentum just as in high school. You suppose conservation of momentum in frame S and apply Lorentz transform And realize an observer in frame S’ does not observed conservation of momentum. So you need to adjust the definition of momentum so that conservation of momentum is in all inertial frames. As a mathematician, how do you go about finding the new formula for momentum? This should not require high power tools
 A: Conservation of energy-momentum is a fundamental principle in relativity; it is "built in" to Einstein's equation 
$$G^{ab}=8\pi G T^{ab}.$$
It can be proven for interactions in quantum field theory, but that is a heavy duty proof. Otherwise, it is best to take it as a fundamental principle (it can also be proven from Noether's theorem, but that depends on an equivalent reformulation of Newton's laws and the argument can be seen as circular). 
As for the other part of your question, $4$-momentum, or energy-momentum $(E,\mathbf p)$, is a $4$-vector. So there is nothing to prove. $4$-vectors are same in all frames. Perhaps it may help if you define the velocity $4$-vector for a body at rest $v=(1,0,0,0)$, and obtain its form after Lorentz transformation. Then you can define $4$-momentum in the usual way $$p = mv.$$
A: Since the OP asks to find the, and I quote here, 

formula for relativistic momentum under the requirement of conservation of momentum for inertial frames

(the last word being my guess which makes most sense), we do the following.
We first define the particle orbits as functions $x^\mu(\tau)$ in spacetime, where $\tau$ is an arbitrary Lorentz-invariant parameter. The action is $$A=\int d\tau\ L(x^\mu(\tau),\dot x^\mu(\tau),\tau)$$ where $\dot x^\mu(\tau)$ denotes the derivative with respect to the parameter $\tau$. If the Lagrangian depends only on invariant scalar products of the form $x^\mu x_\mu,x^\mu\dot x_\mu,\dot x^\mu \dot x_\mu$,  then it is
invariant under Lorentz transformations $$x^\mu\to \dot x^\mu=\Lambda^\mu_\nu x^\nu$$ where $\Lambda$ satisfies $\Lambda g\Lambda^T=g$ with $g_{\mu\nu}=(1,-1,-1,-1)$.
For a free massive point particle in spacetime, the Lagrangian is $$L=-mc\sqrt{g_{\mu\nu}\dot x^\mu\dot x^\nu}.$$ It is invariant under $\tau\to f(\tau)$ for arbitrary and sufficiently smooth $f$. Under translations like $$\delta_sx^\mu(\tau)=x^\mu(\tau)-\epsilon^\mu(tau)$$ the Lagrangian is invariant, satisfying $\delta_sL=0$. Thus applying Euler-Lagrange to calculate the variance, we get
$$0=\int_{\tau_\mu}^{\tau_\nu}d\tau\left(\frac{\partial L}{\partial x^\mu}\delta_sx^\mu+\frac{\partial L}{\partial \dot x^\mu}\delta_s\dot x^\mu\right)=-\epsilon^\mu\int_{\tau_\mu}^{\tau_\nu}d\tau\frac{d}{d\tau}\left(\frac{\partial L}{\partial \dot x^\mu}\right).$$
Thus the Noether charges are $$-\frac{\partial L}{\partial\dot x^\mu}=mc\frac{\dot x^\mu(\tau)}{\sqrt{g_{\mu\nu}\dot x^\mu\dot x^\nu}}=mcu^\mu\equiv p^\mu$$
and satisfies $$ \frac{d}{d\tau}p^\mu(\tau)=0$$
This is the conservation of 4-momentum, once we note that $p^\mu$ is indeed the 4-momentum which can be noted by defining $\tau$ to be the physical time $t=x^0/c$. Also note that $u^\mu$ is the dimensionless 4-velocity of the particle, and hence the 4-momentum retains it's look from Newtonian mechanics. 
Thus if one can agree on the Lagrangian, then the definition of 4-momentum as the conserved Noether charge of it falls from the definitions and Euler-Lagrange.
A: Conservation of momentum is mingled with the conservation of energy when you go to relativistic speed. 
$$E^{'} = \gamma (E - v p)$$
$$p^{'} = \gamma (p - \frac{v E}{c^2})$$
Now if you have conservation $E_1+E_2 = E_3+E_4$ and $p_1 + p_2 = p_3 + p_4$ then because Lorentz transformation is linear, they will just transform into $E^{'}_1 + E^{'}_2 = E^{'}_3 + E^{'}_4$ and $p^{'}_1 + p^{'}_2 = p^{'}_3 + p^{'}_4$ in the new frame. 
A: In a collision, the conservation of 4-momentum can be described by a polygon (just like a free-body diagram of forces on an object in static equilibrium):
$$\sum_i \tilde P_{i,\rm before} - \sum_j \tilde P_{j,\rm after}=\tilde 0.$$
Then, as @stackoverblown says, Lorentz Transformations are linear transformations (just like Euclidean rotations and Galilean Transformations are). So, this polygon transforms into another polygon (as determined by the Lorentz Transformation).
A: Momentum is only conserved if there is no external force on your system. Since force is the time derivative of momentum, momentum is conserved if the external force on a particle is zero. Let me make this more mathematical: $0 = \vec f = d\vec p / dt$. This holds for the non-relativistic and the relativistic case.
Of course one can write down a lagrangian and then apply Noether's theorem, as this is more math. But basically the answer is as simple as I put it.
