sqrt(a/2) in the (0,1, t for ccna) row is always equal to the 1xc row x+n.
x1 = x at 1 * c
xa = x at a * b
n1 = n at 1 * c
na = n at a * b
n1 - (x1^2 - xa^2) = aa * m
Where m is a odd number.
This implies that if we knew the multiple, we'd know a.
13 - (4^2 - 0)/2 = 5 * 1
69 - (11^2 - 5^5)/2 = 7 * 3
1237 - (49^2 - 19^2)/2 = 31 * 7
369 - (27^2 - 15^2)/2 = 13 * 9
Since 1=11, the cells where a=1 and b=cc will also appear in these cells where n follows 2+2x+4T(x) (2, 8, 18, etc). The lowest possible 1cc in (0,n) is a=1 and b=3*3=9, which is at (0,2,2). t=2. As you cycle through these n values keeping a=1 and moving b up through the odd squares, t increases by 1 each time. So you find 1 and 25 in (0,8,3), you find 1 and 49 in (0,18,4), you find 1 and 81 in (0,32,5), etc.
The solution lies in (e,1), (f,1) and (0,n). He said this quite a few times.
In column zero, x is a multiple of n