Question 973728
Draw a model if it helps.  Top and bottom rows will have a and a pieces each.  The remaining left and right columns have a-2 spaces each for containing pieces.  The number of board pieces to fill the boundary spaces is then {{{a+a+(a-2)+(a-2)}}}.


Simplified, {{{4a-4}}} boundary spaces for the checkerboard and you have  {{{4a-4=b}}}.   You can find MANY possible values for b.  One value which would not mean much is 0, which you'd have if a=1.  No purpose in having such a checkerboard.  Also, you cannot have a=0, because you cannot have meaning for b=-4.  This value is nonsense.  You must have {{{a>1}}}; in fact, that is also impractical.  {{{a=2}}}?  No.  A board with only four spaces allows for insufficient movement of playing pieces.  


The question is open-ended.  Pick any value that makes sense for your purposes.