For c6107, interesting that our u1 and u2 values are 41 and 42, since our f derived polite numbers = 3+4=7

The total x+n dimensions are 83 * 83 = 6889.

4 rectangles of 41 * 42 +1 = 6889

It’s the (n-1) pattern playing out at 2 levels simultaneously.

Maybe a fluke, just brainstorming over here.

c6107 f= 134

f div 8 = 16

Sqrt(16) = 4

Sqrt(16)- 1 = 3

4+3=7. These are our two polite triangle number bases added together.

7 * unknown + 1 = correct (x+n)^2

For this example it turns out to be 984.

>>8420

This should be blazing fast at computer calc speed! since the two polite triangle bases come from f, it should scale upwards with increasing c values.

All we have to do is check each iteration to see if it's a perfect square.

This is the two equations merging together or running alongside each other. (i think)

2d(n-1)+f-1= 2xn+ x^2

F and d limit possible n and (n-1) values.

Let’s work on this!

Is it a graph line where only certain integers work?

More about eliminating values than solving?

The algorithm is ancient.

According to VQC

How to combine multiple variable equations to run side by side.

a new form of algebraic solution

Underlying fractal patterns

Triangles and squares

And how they combine

Limiting possibilities

Factors contained within f.

(D+n)^2 - (x+n)^2=c

Simple and clean.

So much complexity and simplicity combined.

Mods and patterns

(Of mods)

Factoring each square

Multiple possibilities for perfect squares in (x+n)^2

Geometric patterns in the Grid

Geometric patterns in the 8Tu+1

Leading to a Grid shortcut based on understanding of the geometric patterns.

How do we factor the two squares (d+n)^2 and (x+n)^2 ?

Using only c d e f ?

I'm reworking equations over here.

I have too much time invested here to give up.

And I'll keep going till we solve this shit.