The square of the c we want to factor will show up in (0,n), since this is the column where c is a square. It'll show up in the following ways:
a=1, b=c^2
a=a, b=ab^2
a=b, b=a^2b
a=a^2, b=b^2
So it'll show up four times. We can easily find the first example, where a=1 and b=c^2. If we find any of the others, we can factor c.
Something interesting about the cells in (0,n) where c=1c^2 and where c=a^2b^2 is that the difference between the x values in these cells is always divisible by 12.
That %12==0 thing only applies to semiprimes, by the way.
((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a
To try to explain this notation a bit more clearly since this is copied from PMA's work in a specific context in a thread
(((x from the cell where a=1 and b=cc) - (x from the cell where a=aa and b=bb)) / ((x from the cell where a=1 and b=cc) - (x from the cell where a=a and b=bc))) - 1 = the a we're looking for
(1xcc).x - (aaxbb).x + 1 == (aaxbb).a
(x from the cell where a=1 and b=cc) - (x from the cell where a=aa and b=bb) + 1 = (a from the cell where a=aa and b=bb)
This also means that the square of any prime number minus 1 is divisible by 12.
If you take a square and look for it as a b value in (0,n), the difference in n values from record to record is equal to (2x)+(4T(x)).
e.g. 9*9=81
These are the records we will get (since there can't be opposite parity between a and b as it would create an invalid n, we will only see the square of all odd numbers below 9 in a[t]):
{0:2:63:14:49:81}
{0:8:45:20:25:81}
{0:18:27:18:9:81}
{0:32:9:8:1:81}
The difference in n values can be expressed as follows:
8 - 2 = 6, 6=2+4
18 - 8 = 10, 10=2+4+4
32 - 18 = 14, 14=2+4+4+4
This is true for the product of any pair of odd squares. So, for example, 11*11=121:
(0:2:10) = {0:2:99:18:81:121} f=-199
(0:8:15) = {0:8:77:28:49:121} f=-155
(0:18:16) = {0:18:55:30:25:121} f=-111
(0:32:13) = {0:32:33:24:9:121} f=-67
(0:50:6) = {0:50:11:10:1:121} f=-23
And another example, 13*13=169
(0:2:12) = {0:2:143:22:121:169} f=-287
(0:8:19) = {0:8:117:36:81:169} f=-235
(0:18:22) = {0:18:91:42:49:169} f=-183
(0:32:21) = {0:32:65:40:25:169} f=-131
(0:50:16) = {0:50:39:30:9:169} f=-79
(0:72:7) = {0:72:13:12:1:169} f=-27
Squares only ever appear in a and b in (0,n) where n=2, 8, 18, 32, 50, 72, etc
This pattern follows the formula n = 2+(2x)+(4T(x)), where x is the number of odd squares away in a you want to be. a begins as the highest square below the square in b (e.g. in (0,2) where b=9x9=81, a will equal 7x7=49). If you want a to be a different square, the x in that n formula is equal to the number of squares away you want to be (so if you wanted the cell where b=81 and a=3x3=9, 9 is two odd squares away from 49, so n will equal 2+(2x2)+(4T(2)), which is 2+4+12=18, which is one of the examples posted near the top of this post).