> So if we don't know it, then a closed solution may not even exist. If we don't know whether a closed form solution exists then how can we even know if it's de

> So if we don't know it, then a closed solution may not even exist. If we don't know whether a closed form solution exists then how can we even know if it's deterministic. We can prove that something has some properties even if we don’t know everything about it. Newton’s second law in a many-body system is (in the case of Newtonian gravity) basically a system of independent second-order ODEs. We know a thing or two about these beasts. Anyway, let me try something (without LaTeX, so I apologise in advance for the notation). Let’s think about a many-body system (following Newtonian mechanics, and interacting via Newtonian gravity) from the point of view of its positions-velocities phase space. The solution is not deterministic if there are points (i.e., sets of positions and velocities for all the particles) that are part of at least two trajectories. Let’s assume we are looking at such a point. The forces F and all their derivatives dF/dr are the same along all trajectories, because they depend only on the positions (which are identical). The velocities and the positions are the same because it is the same point in the phase space. The third derivatives of the positions w.r.t time are also identical (d^3r/dt^3 = 1/m dF/dt, and dF/dt = dF/dr dr/dt, and the dr/dt are identical). You can go on to the n-th derivative, n being arbitrary large (d^4 r/dt^4 is a function of only dF/dr, dr/dt, d^2 r/dt^2 and d^2 F/dr^2, all of which are identical along both trajectories, etc). So, it means that both trajectories have the same position in the phase space, and all their derivatives are identical, i.e., both trajectories are identical. So it is deterministic (there is no point in the phase space that belongs to more than one trajectory). It is somewhat related to Liouville’s theorem.