In setting up a translationally invariant Hamiltonian the following expressions are helpful. The center-of-mass (CoM) momentum is $$ P=\sum_{i=1}^A\boldsymbol{p}_i, $$ and we have that $$ \sum_{i=1}^A\boldsymbol{p}_i^2 = \frac{1}{A}\left[\boldsymbol{P}^2+\sum_{i < j}(\boldsymbol{p}_i-\boldsymbol{p}_j)^2\right] $$ meaning that $$ \left[\sum_{i=1}^A\frac{\boldsymbol{p}_i^2}{2m} -\frac{\boldsymbol{P}^2}{2mA}\right] =\frac{1}{2mA}\sum_{i < j}(\boldsymbol{p}_i-\boldsymbol{p}_j)^2. $$