>>8841

You missed an important detail.

>Its got to have exact values of (n'-1) and we have created and controlled many of these values.

Semiprime c has two (a,b) pairs, giving it two n values including BigN, only one of which we know to begin with. Multiplying c by one prime increases the number of n values to four, of which we can directly calculate two. Multiplying c by two primes increases the number of n values to eight, of which we can directly calculate four. That's what he meant about controlling n'.

Hey Chris…

>>>/newsplus/228556

"Brexit: MPs vote by a majority of 211 to seek delay to EU departure"

https:// www.bbc.com/news/uk-politics-47576813

Does this affect your release timeline at all? It sounds like it might be necessary to make the VQC public whether Brexit happens or not, considering, like you said >>8689 here, it'll bring down the UK government and end all of that traitorous nonsense.

March 29th is exactly 17 months to the day since Q's first post.

>>8881

>>8880

Well this certainly looks interesting, doesn't it?

>>8883

Since every number (regardless of it being prime) appears as a factor of a[t] in (-1,1) at t=itself and t=itself+1, then again at t=2itself and t=2itself+1, and so on infinitely, c (and c's prime factors) are a factor at t=c and t=c+1, t=2c and t=2c+1, etc. As the example in this image shows, a[c] = c(2c-2) and a[c+1] = c(2c+2).

Pee is stored in the balls

>>8887

Looked at some more t values (same c as the previous example) from (e,1) cells that we've paid attention to in the past. There doesn't seem to be any significance in relation to (-1,1), as far as I can tell (in this example and the other examples I tried). I think aside from Chris eventually just telling us what the significance of (-1,1) is, it would be useful to look at the factors of the a[t] values in (-4,1), (-9,1), (-16,1) etc too, because I remember when I was looking into (-1,n) I found that it had basically the same patterns as the other negative squares (for example, all those negative square (f,n) columns also have valid elements in every cell).

>>8890

Same example numbers, 3173=2263. I checked the prime factors of each of the a[c] and a[c+1] values for (-1,1), (-4,1), (-9,1) etc up to (-100,1). It would appear that c (in this example 3173) is always a factor in a[c+1] and that 2(c-1) (in this example 2231329) is always a factor at a[c]. Next I'll make diagrams like this >>8883 for these other (-f,1) cells.

>>8893

If anyone reads this image, disregard all of those t values in the cell descriptions. They're all actually 2263. I was printing their index in an ArrayList rather than the actual t value. The rest of the information in this post still seems to be valid.

>>8883

Alright, same thing as here, but for (-4,1), (-9,1), (-16,1), (-25,1) and (-36,1). All the a[t] values for each of these contain every factor as a prime also, but the gaps between their appearances change compared to (-1,1).

>>8918

(c % something) * c = a or b? Am I reading that right?

>>8918

(c * a_number) % certain_value == a or b seems like it makes far more sense (thanks 5D). I put together a program that uses qc and tries to find any of the variables from cells we've studied as the mod value to find a. Every time it seems to be a different variable. I'm thinking q probably isn't the multiplier for this, but I don't know. I might adapt this program so that it tries replacing q with each of the other cell variables. If that doesn't work, then either a_number or certain_value are calculated from the other variables (like b-BigN+d or something unknown like that) or they aren't in my list of cells.

>>8921

Here are the equations that seem to work for all semiprimes (or at least the semiprimes I tested it with) for which (c * something) % something_else = a or b where something and something_else are variables we've studied. It appears that this only works when the number you multiply c by is either a or b, where the answer is also either a or b, so there are still too many unknowns for there to be a solution here.

>>8935

Yip, a(b%d)=~=c%d would appear to only work for some specific cases. e%d =~= c%d definitely appears to hold. Pics related. I have a feeling e%d=~=c%d would have already been proven somewhere but I wouldn't have any idea who did it or what they called it. Just seems like a concept someone would have studied (given it's all about squares and such).

>>8939

Why did you type a(b%d) = c%d if you meant (a(b%d))%d = c%d then? You have to be specific.

>>9033

We did see that (that was me you were talking to - I was also posting other stuff like those graphs right after Defango mentioned graphs in the stream in an attempt to fuck with him). I was meaning to look into that. It still isn't directly calculable, right? You'd need to know t.

>>9028

>whatever the hell TempleOS is

Oh boy do I have a rabbit hole for you. Terry was the one who coined the term "CIA n*s". Here's a good introductory video.

>>9040

I remember there being someone on YouTube who went through TempleOS and discovered it was extremely insecure, so I don't know about that.

>>9045

>>9033

I had a look into this and you're missing a lot of details. Firstly, a and b in (0,n) are only ever both squares when n is a square multiplied by 2 ((0,2), (0,8), (0,18), etc - the a[t] values in (0,1)). dd/2 will always be a square multiplied by two, but only when d is even. When d is odd, (dd+1)/2 gives us the sum of two consecutive squares (so the a values in (1,1)). The a and b values in (0,n) where n is the sum of two consecutive squares aren't squares themselves, so this doesn't fit. Here's another thing: a and x can be different parities, but a and b need to be the same parity, so if a=13 and x=10, you can't have a cell where a[t]=169 and b[t]=100. So this doesn't work for odd ds and a/x pairs with alternate parities. I haven't been able to figure these parts out myself. I also haven't been able to actually successfully find a negative cell where a[t]=aa and b[t]=xx but I think I'm just too tired to code at the moment. Have you looked into this?

>>9053

>would we be able to shorten the BigN - ? calculation, that is calculating 2BigN - 5xi and then testing the resulting numbers?

Your example is based on using a=q b=c, so it would give us starting points for which (e,1) a[t] is divisible by q, but would it give us any extra information about the same for a=a b=qb? I wouldn't think so. With RSA-sized numbers, we would maybe have a starting point for which a[t] is divisible by q, but the search space would still have grown exponentially compared to the smaller numbers we've all been testing with. That's if I understand what you're getting at.

>Also is it Chris who's doing the posts in templeos as VQC or someone else fucking around?

They're copied word-for-word from posts Chris made over a year ago with a trip we now know the password to. It's almost definitely not Chris, and if it was it would be incredibly strange.

>>9078

>Suppose you split the amount of possibilties in two each division, and you know that it must be between 2 and d

Too bad we don't have any patterns that fulfill this property. Otherwise binary search would have worked.

>>9103

I've spent some time analyzing your code. Is this what it's meant to do? Correct me if I'm wrong about anything.

>take a value c

>find the elements in (e,1) and (f,1) where d falls between d[t]

>check if any of the d+n values in any of these four elements or the two elements one t value further create a square remainder with our c (you're using i but it's the same concept of the difference of two squares and i=d+n)

>if they don't, try the d+n values from further (e,1) and (f,1) cells by treating a and b from each (e,1) element from t=1 upwards as c recursively and doing the same thing

I'm not really sure where you'd get the idea to try that, so if I did misunderstand anything do let me know. It's an interesting concept. I haven't seen a lot of study into the d[t] element pairs (unless I just wasn't paying attention) so it's good to see some. I don't mean this is a negative way, but the way it checks each i value for a square remainder with c seems a bit haphazard in that we don't seem to necessarily know that the right d+n value is going to show up anywhere in particular. If it does show up somewhere specific then that would be a good point for further study. Also what's the offset value for? I noticed you've calculated this number based on the d[t] values but it doesn't seem to get used for anything after it's calculated.