statistical mechanics - Can a system entirely of photons be a Bose-Einsten condensate?

Background:

In Bose-Einstein stats the quantum concentration $N_q$ (particles per volume) is proportional to the total mass M of the system:

$$ N_q = (M k T/2 \pi \hbar^2)^{3/2} $$ where k Boltzmann constant, T temperature

Questions:

A) For a B-E system "entirely of Photons" - what is the total mass of the system? (answered, see below)

B) Does an ensemble of photons have a temperature? (answered, see below)

C) Is this a Bose-Einstein condensate?

I've found a paper here (like the paper put forward by Chris Gerig below) which finds a BEC, but it is within a chamber filled with dye, and the interaction of the photons with the dye molecules makes it a dual system, as to one purely of photons. I think in this case there is a coupling between the dye molecules and the photons that is responsible for the chemical potential in the partition equation
$$N_q = \frac{g_i}{e^{\left.\left(\epsilon _i- \mu \right)\right/\text{kT} - 1}}$$

where $g_i$ is the degeneracy of state i, $\mu$ is the chemical potential, $\epsilon_i$ is the energy of the ith state.

I suspect an Ansatz along the lines of $\mu$ = 0, and $\epsilon_i$ = $\hbar \nu_i$, where $\nu_i$ is the frequency of the i photon.

another edit:

After going for a walk, I've realized the Ansatz is almost identical to Planck's Radiation Law but the degeneracy = 1 and chemical potential = 0.

So, in answer to my own questions:

A) is a nonsensical question, as photons have no mass, noting from wiki on Quantum Concentration: "Quantum effects become appreciable when the particle concentration is greater than or equal to the quantum concentration", but this shouldn't apply to non-coupling bosons.

B) yes the ensemble has a temperature, but I was too stupid to remember photons are subject to Planck's Law.

C) Is this a Bose-Einstein condensate? No, as photons have no coupling or chemical potential required for a BEC.

So, for an exotic star composed entirely photons, all the photons should sit in their lowest energy levels and the star will do nothing more than disperse.

Is this right?