I am not an expert, but if we assume that n can be only 0, 1, 2... then for j=16 we have n(n+2) + m = 2j= 32 n^2 + 2n + m - 32 = 0

Now we can observe all possible cases for m. In the first case, for example, we assume m=n, and soon we can see that equation n^2 + 3n - 32 = 0 does not have integer solutions. Next case is m = n - 2, then m = n - 4. It is easy to see that in this (last) case we have n = 4 as a solution of equation n^2 + 2n + m - 32 = 0.

Hi,

the idea is as follows: the value 2j=n^2+2n+m varies from n^2+n to n^2+3n.

We solve the auxiliary equation x^2+x=2j, whence (Floor=integer part) n=Floor[x]=Floor[(-1+sqrt(1+8j))/2] . Next, m=2j-n(n+2), for just found n.

For instance, j=16, n=Floor[(-1+sqrt(1+128))/2]=Floor[(-1+11.35...)/2]=5,

m=32-5(5+2)=-3=-5+2.

I concur with Anton Vrdoljak