AA !LF1mmWigHQ ID: b8ea74 July 21, 2019, 7:46 p.m. No.9765   ๐Ÿ—„๏ธ.is ๐Ÿ”—kun   >>9766

>>9764

Well, the solution record is one obvious place. If we're talking squares in general, they turn up as c values in (0,n), a and b values in (0,n) where n is twice a square and f values in (e,1). Conceptually, a few things are used to create squares in different situations, such as d, i and j in any given element, so one of them from another element could equal the x+n we're looking for. There's a few ideas.

AA !LF1mmWigHQ ID: b8ea74 July 21, 2019, 7:58 p.m. No.9767   ๐Ÿ—„๏ธ.is ๐Ÿ”—kun   >>9768

>>9766

I guess x+n could maybe be between two x+n values from different elements in a different cell, if that's why you're referencing D and C. This is just an idea off the top of my head, but given the solution is recursive in some way and just directly finding i or j would be O(1), maybe we find two elements that we know for sure x+n will be between (but we don't know its exact value, just that it's between two values), and then there's something inherent to the same concept applied to d in that it's the same distance between two variables in two elements as it is for c. Maybe that's why looking at a transform of the D and D2 elements is meant to be like "zooming in". That's just an idea based on a few of the concepts I do understand though. I don't know where you're going with this.

AA !LF1mmWigHQ ID: b8ea74 July 22, 2019, 1:06 a.m. No.9792   ๐Ÿ—„๏ธ.is ๐Ÿ”—kun   >>9793

>>9790

This is another confusing and dense post. Are you saying we should look at all of the c values from the recursive d -x+n elements and compare them to our given c? I kind of already did and didn't see anything. That's the only way I can think to interpret this post.

AA !LF1mmWigHQ ID: b8ea74 July 22, 2019, 4:23 a.m. No.9795   ๐Ÿ—„๏ธ.is ๐Ÿ”—kun

>>9779

>>9782

Since these clues are kind of hard to interpret, I had a wild stab in the dark. I tried seeing if any representations of x or x+n as the difference of two squares turn up in any of the i and j values in the recursive d -x+n elements, as well as checking if any values that are off by 2d+1 turn up instead. No dice on either.

AA !LF1mmWigHQ ID: b8ea74 July 23, 2019, 4:57 a.m. No.9830   ๐Ÿ—„๏ธ.is ๐Ÿ”—kun   >>9831

>>9815

This would be so much easier to read if they just went through an example with some numbers. Maybe I recognize a lot of the words but I feel like I need a PhD to understand any of the symbols and diagrams. When this is all over I'm going to go over everything we've done and put together an explanation in the most basic English I possibly can so that anyone can understand it.

AA !LF1mmWigHQ ID: b8ea74 July 28, 2019, 8:51 p.m. No.9952   ๐Ÿ—„๏ธ.is ๐Ÿ”—kun   >>9953 >>9961

Just a thought. If we're traversing (e,1) and (f,1) inwards in O(log n) then it might be useful to see which cells contain n as a factor somewhere in each of the (sqrt,1) cells, and if not starting from e and f maybe starting from other numbers as the e and f values logarithmically inwards. If it's meant to happen for every number then you'll continually find it in the same place (parity-permitting), and maybe there'll be a pattern to the t values as well. But then again, this pattern is meant to be used to adjust the estimate i^2-j^2 values. But it's also meant to converge to n. More confusion, as usual. I might look into this later, but I felt like posting my thoughts here since there haven't been any major clues in a few days.