Is the sum of two stationary states of different energies also a stationary state? The question title kind of speaks for itself really. I was thinking of maybe using the orthogonality relation to try to show this: $$\int_{-\infty}^{\infty}\phi_n(x)\phi_m(x)dx=\delta_{nm}.$$ Alternatively, maybe since a stationary wavefunction is given by $$\phi(x)=A\sin{\frac{n\pi x}{L}}$$ it follows that it can be a superposition of two wavefunctions, and the new one would just have a different constant $A$ at the beginning. This sounds like a reasonable explanation so may well work, but does the statement different energies have any major significance?
 A: 
Is the sum of two stationary states of different energies also a
  stationary state?

A stationary state is necessarily an energy eigenstate.  But a sum of two stationary states with different energy eigenvalues is not an a energy eigenstate:
$$H(|E_1\rangle + |E_2\rangle) = E_1|E_1\rangle + E_2|E_2\rangle \ne C(|E_1\rangle + |E_2\rangle)\,\mathrm{for}\, E_1 \ne E_2$$
A: For some Hamiltonians, with appropriate boundary conditions, the stationary states form a ladder with equally spaced eigenvalues, your energy states.  
But imagine a finite potential well with such evenly spaced energy states.  Pick two states that are more than halfway to the top: their total energy, if combined, does not fit into the potential well at all, thus there is no energy state equal to their sum.
However, your system could maintain itself in a superposition of the two states.
A: You seem to be considering the infinite square well energies.  I will first note this is not the only scenario in which stationary states arise.  But considering that scenario, the energy state in the infinite square well is defined by n, not A.  A is just the normalization constant, which ensures that the square-integral of the wavefunction is 1.  If you add two sine functions with different values of "n", you will not end up with another sine function (hopefully this makes sense to you), so the sum cannot be a stationary state (because all stationary states have the form you noted).  On the other hand, it is pointless to "add" two wavefunctions of the same energy in this case, because they would be exactly identical, and then you would just have to normalize this new wavefunction by dividing by 2, and then get the same wavefunction you began with.
Edit: I will note that in higher-dimensional cases with more than one quantum number, you could have a superposition two distinct wavefunctions which had the same energy state, but which had different eigenvalues for some other observable, say angular momentum, you would get a third distinct state which would be an eigenfunction of energy but not of angular momentum.  The eigenvalue of energy that superposition would simply be the same as the energy of the two original functions.
