The bigrade  

The magic bigrade is a line of m numbers out of the numbers [1 .. m^{n}] with _{k=0}∑^{m1}B_{k} = m (m^{n} + 1) / 2 and _{k=0}∑^{m1}B_{k}^{2} = m (2 m^{2n} + 3 m^{n} + 1) / 6 

centralized sum 
changing numbers B_{k} > (m^{n} + 1) / 2 + b_{k} it follows: _{k=0}∑^{m1}b_{k} = 0 and _{k=0}∑^{m1}b_{k}^{2} = m (m^{2n}  1) / 12 

_{k=0}∑^{m1}B_{k} =
_{k=0}∑^{m1}(m^{n} + 1) / 2 + b_{k} =
m (m^{n} + 1) / 2 + _{k=0}∑^{m1}b_{k}
m (m^{n} + 1) / 2 ==> _{k=0}∑^{m1}b_{k} = 0 _{k=0}∑^{m1}[(m^{n} + 1) / 2 + b_{k}]^{2} = _{k=0}∑^{m1}[(m^{n} + 1)^{2} / 4 + (m^{n} + 1) * b_{k} + b_{k}^{2}] = m (m^{n} + 1)^{2} / 4 + _{k=0}∑^{m1} b_{k}^{2} = m (2 m^{2n} + 3 m^{n} + 1) / 6 ==> _{k=0}∑^{m1} b_{k}^{2} = m (2 m^{2n} + 3 m^{n} + 1) / 6  m (m^{n} + 1)^{2} / 4 = m (m^{2n}  1) / 12 

bigrade move 
the move from one bigrade to another is given by B2_{k} > B1_{k} + d_{k} with 2 _{k=0}∑^{m1}B1_{k}d_{k} =  2 _{k=0}∑^{m1}B2_{k}d_{k} =  _{k=0}∑^{m1}d_{k}^{2} 

_{k=0}∑^{m1}B2_{k}^{2} =
_{k=0}∑^{m1}(B1_{k} + d_{k})^{2} =
_{k=0}∑^{m1}B1_{k}^{2} +
2 _{k=0}∑^{m1}B1_{k}d_{k} +
_{k=0}∑^{m1}d_{k}^{2} ==> 2 _{k=0}∑^{m1}B1_{k}d_{k} =  _{k=0}∑^{m1}d_{k}^{2} =  2 _{k=0}∑^{m1}B2_{k}d_{k} 

The bigrade move make it possible to "hang" the square onto the first line with a set of differences the twisted condition these difffernces is curiously bound to the first lines numbers, the validity of these difference squares remain to be seen 