You can create a tree that is meant to solve the thing. We don’t completely understand the tree yet.

The tree is meant to find n or x+n (these are VQC’s words but this could also mean finding u).

The tree has three parts.

It starts with a decision-based component in part 1.

>if e is 0, return d (because c is a square in this case)

>if(GCD(e,d)!=1) return GCD(e,d)

That’s the end of part 1.

Part 2 is a tree made with numbers. The node at the top is c.

Each time you create a child node, you divide it by 2 until it’s odd.

The descendents of c in this tree are d and e.

You create further descendents by finding d and e of both d and e each time.

There is a third part to the tree that we never seemed to learn.

This part of the tree is where the grid comes in.

It’s meant to work with patterns in either row (e,1) or in a different way for column zero.

It’s meant to also have something to do with triangle numbers.

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Just summing up the leafs (from part 2) for a completely trimmed RSA100 tree and dividing by two gets us to

40364365001756683226800431241044541563293164124405

which is pretty close to our factor

40094690950920881030683735292761468389214899724061

This doesn’t appear to be the solution though.