The difference between pure and mixed states is the difference in their density matrix structure.

For density matrix $\rho$ of mixed state the trace of $\rho^{2}$ should be less than 1.  For pure state corresponding trace $Tr(\rho^{2}) = 1$.

But when i tried to check the Bell two-qubit state, i got:
$$ \rho = \frac{1}{2}\begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1\end{pmatrix}$$
$$ \rho^{2} = \frac{1}{4}\begin{pmatrix} 2 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 2 & 0 & 0 & 2 \end{pmatrix}$$
Trace of which is equal to 1.
I think i've got incomplete understanding of this check.