Is the Bose-Einstein condensation a single particle phenomenon? BEC occurs for noninteracting Bosons. Can we conclude that it can be described with a single particle? What is the significance of the number of the particles?
 A: What @Árpád Szendrei said is correct. I will add some miscellaneous points. 


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BEC occurs for non-interacting bosons

BEC occurs for interacting bosons as well, and non-interacting BEC is actually a pathological example. It has an infinite compressibility. The speed of sound is zero, and any infinitesimal drag will create excitations. A weakly interacting BEC has a non-zero speed of sound, and acts like a superfluid. It IS possible to make a non-interacting BEC, by modifying the scattering length between atoms to zero, using external fields (see Feshbach resonance). 


*The "wavefunction" that people usually discuss ($\psi(r) = \sqrt{n(r)}e^{i\phi(r)}$) is technically not the actual many-body wavefunction, but an order parameter of the condensate. This "wavefunction" obeys a non-linear Schrodinger-like equation called the Gross-Pitaevskii equation. 

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What is the significance of the number of the particles?

It would help if the question is more precise, but usually a common question is whether the form of the order parameter mentioned above conserves the number of particles. The fact is, it doesn't, because it has a well-defined phase. It has a well-defined average of numbers, though. There is fluctuation in the number of particles, but (fluctuation)/(average) quickly goes to zero in the thermodynamic limit. [to find fluctuation in numbers, you need to look at the full Hamiltonian in second quantization form to get answers quick, so what I said is not really rigorous but just a sketch].
A: The Bose Einstein condensate is a QM effect of collective quantum state in which a macroscopic number of particles occupy the lowest energy state and thus is described by a single wavefunction.
All the bosons will be described by the same wavefunction.
So it is not a single particle, but all the particles (their probability distribution) are described by the same wavefunction.
A: 
BEC occurs for noninteracting Bosons. Can we conclude that it can be described with a single particle?

No, generally not. Most BEC systems in literature are described by a Hamiltonian 
\begin{equation}
H = \sum_{i=1}^N h({\bf r_i}) + \lambda_0\sum_{i < j}W({\vert \bf r_i - \bf r_j\vert})),
\end{equation}
where $h({\bf r})$ is a single-particle Hamiltonian and $W({\bf r})$ is an interaction potential. 
Now write $\phi_0({\bf r})$ for the ground state of $h({\bf r})$.
For $\lambda_0=0$, the ground state of $H$ is $\Psi_0({\bf r}_1,\dots,{\bf r}_N)=\phi_0({\bf r}_1)\times\dots\times \phi_0({\bf r}_N)$. So $\Psi_0$ is just a single particle wave function, $\phi_0({\bf r})$,  multiplied $N$-times.
But for most physical systems $\lambda_0\neq 0$. The many-body ground state $\Psi_0$ then no longer factorizes into a simple product as above. In fact, it can be very far away from that and consists of a superposition of $N$-boson states. So a single particle wave function no longer suffices to describe your system. 
For that reason, BEC is nowadays usually defined in terms of the eigenvalues and eigenfunctions of the first order reduced density matrix, see Penrose and Onsager and the answer to this question. This criterion extends the concept of "${\mathcal O}(N)$ particles in the same state" to interacting systems. The book by Leggett is a more accessible reference on this topic than the original paper and also discusses the issues that arise in other attempts to define BEC. 
