Divide e by 4

The remainder tells you whether (x+n) is odd or even.

0,1 = even

2,3 = odd

From the grid, it can be seen that columns follow a pattern mod 4. 0,1,2,3;0,1,2,3;..

(x+n): even,even,odd,odd;..

This pattern in respect of the difference of two squares where c is odd.

The factor tree is used to factor d and e.

Factoring these (and down the tree) allow for the factoring of c.

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The tree solution finds x+n or x. If we know x+n is odd and are to construct x+n from triangles, then that must mean if we know x+n is even we are to construct x (I don't know if this last point is necessarily true. It was Jan's speculation. VQC has said "it does this or this" in relation to other things before and it's turned out to only make sense with one of them (like the cell in (e,1) in which x=f-1). Thought I'd post it here anyway in case I missed another justification).

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You can create any negative x record by substituting a and b.

(1,5,4) = {1:5:12:7:5:29} = 145

(1,5,-8) = {1:5:12:-17:29:5} = 145

(1,61,6) = {1:61:12:11:1:145} = 145

(1,61,-66) = {1:61:12:-133:145:1} = 145

looks like x = d - a is the correct x.

and the formula for any negative x in (e,n) could be:

x = -( x + 2*n )

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While an will be divisible by a and n and bn will be divisible by b and n, cN will be divisible by a, b, c and N.

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Between the starting prime solution cell and the a[t]=na cell in (e,1), the difference between the a value from the first to the second is a(n-1). This is obvious since na-a = (n-1)a. To add to this, though, the difference between the d values in these cells is also a(n-1).