I still check daily, but I haven't had any spare time to work on it.

>>4213

Appears to check out.

I'm probably stating something we already know, but have you noticed that the n in (0, n) for a^2, b^2 appears to be equal to 2*(x+n)^2?

>>4221

First, welcome back!

I'm very excited for today!

So I wrote a small python script to compute the remainder of e for the unsolved challenges and I believe RSA 490 is the largest unsolved number with (x + n) being odd.

>>4235

When computing the remainder of 4 for e RSA 490 we end up with 3, which means it has an odd (x + n).

>>4246

Hmm, I never doubled checked, but it appears you are correct regarding (e, 1).

Take (3, n). 3 % 4 = 3, so we should expect (x+n) to be odd. But then we have (3, 2) and (3, 3). Both of which have even and odd (x+n). So we can't deduce parity of (x + n) based on e then?

Maybe VQC can clarify.

>>4254

But take a look at (3, 2) and (3, 3).

Compute the (x + n) and look at the parity. For (3, 2) you'll have (x + n) as odd and for (3, 3) you'll have (x + n) as even.

This should hold for any (e, n) where e also has (e, n-1) as a valid infinite set of records.

>>4252

That in an of itself would be a great result.

>>4255

Unless this applies only to a=1, b=c.

Could you double check that PMA?

Okay, bear with me. I'm trying to think out loud and using images of squares on an example here to try and understand this thing better.

If I've done any mistakes or misunderstood anything, call me a faggot and point it out.

So let's try and wrap our heads around these triangles and how the 8Tu thing works. It's not a complete guide or anything. Just the ramblings of an anon.

I decided to try this on a smaller number, specifically 7 * 37 (As step 1 doesn't solve this, so we will have to build a tree and then solve it using the method VQC is outlining), which has the record: (3, 6, 16, 9, 7, 37).

The photos I've attached are of:

• (x+n)(x+n) square for the record of a=7, b=37.

• The triangle we are looking for (all blueish)

• The triangle with colors representing (n - 1) (redish), d (yellow/gold) and ((x + n)-1)/2 (base of triangle) as purpleish.

• The same triangle again, but this time marked with

So we start with c = 259.

We then compute d = floor(sqrt(c)) and e = c - d*d.

c = 259,

e = 3,

d = 16

We now want to compute f =f = (2d + 1) - e => (216 + 1) - 3 => 30.

So we have f = 30 and this gives us f - 2 = 28.

Now that we have f - 2, I'm a bit unsure about the way to go about of finding the base. So I don't know if I'm supposed to select a number to multiply 8 with when dividing, but I've opted to not do it.

This gives us (f - 2) mod 8 = 4 and (f - 2)/8 = 3.

So we compute our n0 with 3, using the GetNFromOddTriangleBase function which takes (base, c, d) as parameters.

This will give us n0 = 1.

Now I didn't include it in the list of attachment, I'll post, though. But I tried to create a new triangle and fill it with n0 squared (which is still 1) and 2*d. However, here I am a bit lost. That triangle is bigger than the triangle we are after.

As pr:

> We will add a method to calculate what would be missing from the triangles if we could only fill them with n0 squared and multiples of 2d.

So I think either I screwed something up, or maybe I didn't read close enough.

What we see from our (f-2)/8 is that it overlaps with (n - 1) and d, which I don't quite understand how to interpret. Anyhow, I'm still staring at this thing, but I'm trying to make triangles and I'm trying to follow VQC to see if I understand anything.

>>4500

Yup, that's the n we are looking for.

Well since we are supposed to be able to solve it in less than log n, where n is the length of c in bits, we should be able to solve 7 * 37 in like 2 something. So maybe it's not unreasonable.

I can keep trying with a few other numbers, just to see what we get.

>>4503

Yeah and I'm sure it only works for smaller numbers too.

>>4504

I use spreadsheets and just colour the squares.

I've also been thinking about that, but I haven't had time to play with different styles of triangles. I used the that pattern because of VQC's "this" comment, but I'm open for it being inaccurate.

>>4500

Except this doesn't work for the next record in (3, 6), specifically {3:6:34:15:19:61}.

Here you will have the following:

>>> c, e, d, f, f2, f2%8, f2/8, n0

(1159, 3, 34, 66, 64, 0, 8.0, 4)

c = 1159

e = 3

d = 34

f = 66

f2 = 64

f2 % 8 = 0

f2/8 = 8

n0 = 4

Now for this record we know the base of the triangles is 10 ((x + n)-1)/2 = ((15 + 6)-1)/2 = 10.

I'm just playing around with the triangles, but I thought of one thing.

When we are filling our triangle with 2d(n-1), then when filling in a single triangle, we should fill inn only 2d(n-1)/8 no?

I played with it and attached the photo. I'm still using this record: (3, 6, 16, 9, 7, 37)

There's two triangles in the photo. The left one is the actual triangle while the right one is our triangle created from n0^2 and multiples of 2*d/8.

Sorry for the colours, but each set of 2d/8 is coloured with it's own colour. In this I used 7 2d/8 parts, which is wrong. It should be 2d5/8 for the case I'm using as n = 6.

It almost overlaps, but it is 1 square too many. Not sure if I'm barking up the wrong tree, but nevertheless I'm still playing with the triangles.

>>4581

Still trying to wrap my head around everything.

In the photo you'll see three squares, all the same. The first one is the one divided into the 8 triangles,

the second one is the ration of squares for (x+n)(x+n) and the last one is for the other equations nn + 2d(n-1) + f - 1.

In the last one I also highlighted each of the 2d we have an named them 2d, 2d2, 2d3, 2d4 and 2d5 to signify the accumulated squares added.

>>4587

Same as the one in the right corner, bottom, but moved f-1 and nn. Looks a lot better.

>>4343

I'm curious about this part:

>That geometry with multiples of 2d will give us the solution, if it exists (which in our example we know it does). If doubling the width of the chunks of (f-2) didn't give us a solution, the number c would be prime, the worst case (doubling the width of (f-2) chunks very quickly determines this as the number of times it doubles is logarithmic in the length of (f-2)).

I mean in a way, this kind of sounds like a solution, no?

Now what does it mean to "double the width"? We know we have a "blue band" that's 5 rows thick. So I'm guessing maybe he means doubling this to 10 width, then 15, then 20 etc until we can determine it is a prime (or find the solution).

I haven't tried anything regarding it yet, but maybe it's worth a shot.

Also, I'm not sure of course, but does this blue band have 5 rows thickness, because we also divided by 5 and not just 8?

I'm guessing VQC will come by later and explain, but I still want to chip away at this, trying to wrap my head around it.

>>4595

Checks out.

>>4598

PMA beat me to it!

Not sure if I'm far off what VQC is getting at, but I've been trying to figure out what he means by adding 2d's.

So I'm again using the record for a=7, b=37.

For this record we will have the following variables:

c=259

d=16

e=3

f=30

f - 2=28

base of triangle = 3

n0 = 1

We can now construct a new square by computing the triangle of the base (triangle(base) = 6). We then multiply it by 8 and add one, giving us 6 * 8 + 1 = 48 + 1 = 49. So our new square is 7*7.

It's here I've been thinking about adding multiples of 2d. I don't see how accurate this is, as what I have now doesn't appear to match along with what VQC has been saying (I think). But attached is a screen shot of my 77 square with triangles highlighted. Around it you will see different multiples of 2d. Each cell is numbered and belong to one part of the 2d, except for the last one which is simply d.

We can see based on the number of d's added that it adds 49 + 2d5 + d. So the 2d5 part is correct, but the square will then miss d (fluke? random? step in the right direction?).

Once we add 49 + 2165 + 16 we have 225, which is 15^2 which is also the (x+n)(x+n) we are looking for w/ regards to this record.

Again though, I'm not sure if this is what VQC has in mind.

>>4614

Doesn't seem to hold up on other numbers. I'm guessing it's a fluke.

>>4615

Also I think we're missing something here, with regards to step 3 part a. If we don't have to use the tree for this specific case, then why bother generating the tree if we can deduce this without computing the tree?

>>4322

So after revising this post I took another look at my a=7,b=37 square and updated it to show the different elements.

The center part is the square composed by (n-1)(n-1). That is 25, which when remove 1 and divide by 8 is (25 - 1)/8 = 24/8 = 3.

So the inner part has a lighter shade, the next part represents the area between the base ((x + n)-1)/2 and (n-1)(n-1) which in this case is 144.

Now 144 mod 2d = 16, which is the square of f-2 % 8 = 4 in this case. Not sure if relevant. It's also (d - 4)*2. Again, not sure if relevant.

The outer layer which has a strong colour is the base of (x + n - 1)/2.

>>4618

I'm kicking my self in the ass for not noticing this until earlier today:

http://mathworld.wolfram.com/TriangularNumber.html

I feel I should probably spend more time reading about this stuff, than just trying to wrap my head around VQC's crumbs and methods.

>>4621

They're great!

>>4669

Great to see you again.

I haven't had much time today, but from what little I could see, at least for n=2 (and n=6) it appears that 2d(n-1) fits half our triangles (4 out of 8) in (x+n)^2.

Not sure if it's a coincidence or a pattern. I'll take another look tomorrow.

>>4718

Hence why I'm using python, a language I never code in ;)

For those who haven't published any code ever, it should be safe. If you have you should consider using a language you don't know (or know ish). Example, if you're a java developer use C# like Chris is. They are similar enough that you won't get your head stuck too much. And again, if you are a C# developer use java etc.

I started trying to make some patterns that matched the group of odd e, even n and odd (x + n).

These are all (3, 2, 1), (3, 2, 2), (3, 2, 3) and (3, 2, 4).

I have no idea if the configuration is correct though. Just playing around with (x+n)^2.

That 2d(n-1) fits half the triangles could simply be a result of n-1 = 1 and 3 being such a low number.

>>4731

Yeah the swastika is unintended. I know Conway and Guy used another way of fitting the triangles in the odd square, which I'm guessing is because they wanted to avoid swastikas.

>>4729

Forgot the Legend of course.

>>4730

Assuming this is some kind of pattern, then yeah, but then you would have too many squares to fit in (x + n)^2 (unless to opt out of using f and n^2 - 1)

>>4735

Thus, if this pattern hold, then d / 4 should be the base of the triangles where 2d(n-1) fits.

>>4736

IF that were the case, then we can simply "grow" the 4 triangles for (x+n)^2, once we have that it's done.

Which sounds too simple, so I don't believe it.

>>4737

Heh, yeah no, d/4 is not the base of the 2d(n-1) triangles. Just got a head of my self here.

Well I still haven't found a counter example of 2*d(n-1) not fitting at least 4 triangles.

The a=7,b=37 record 2d5 will fit more than 4 triangles, though.

I noticed something else. I suspect that (n - 1) appears as a factor in f. When f has multiple factors I can't see an obvious pattern into which of the factors represent (n - 1).

For n - 1 when n = 2, it appears that f is a prime. I've only looked at a few examples so far though. So maybe you can add this in your test cases PMA?

When we are working with a prime (c = prime) (n-1) doesn't appear to be a factor of f, though. Not sure if this is a proper pattern.

Okay, I feel a bit like an idiot.

I was thinking last night about the logic behind f -2 and the 2d(n - 1) part and I realised that we're just adding to perfect squares here.

So the equation we're dealing with:

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

This is the same as (x+n)^2 - n^2 = 2d(n-1) + f - 2 which is also the same as x(2n + x).

Essentially, this equation is just the difference of two perfect squares (n^2 and (x+n)^2) (See more here: https ://en.wikipedia.org/wiki/Difference_of_two_squares

under the "Difference of two perfect squares" . I'm not sure if we were aware of this, but at least I didn't realise this until I started to think more about it.

>>4796

And of course then, what we have is really:

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

Replacing f:

n^2 + 2d(n - 1) + 2*d + 1 - e - 2

n^2 + 2dn + 1 - e - 2 = (x+n)^2 - 1

2dn + 1 - e - 2 = (x+n)^2 - n^2 - 1

Not sure how this would get us closer to finding a solution, though.

>>4808

Hey, sorry I've been busy. Not sure if I have time to look into it today, but I'm definitely going to verify this shit.

>>4822

No, it won't break the internet.

This thing will first turn the next few weeks into a pain in the ass for ops people and browser developers. Browsers will be quickly upgraded to support elliptic curves certificates (if they don't already). Root certificate guys will push out new root certificates using elliptic curves, we'll see a few weeks of stress as this gets deployed. Ops will start demanding elliptic curves on servers (SSH etc) and banks will push heavily for users to update their browsers.

But in short, breaking RSA won't break the internet. It will just turn it into a pre-https system. SSH will be compromised and every public key that's out there (using RSA) will be broken quickly so expect shit to go down.

However, VQC has also said that this thing will break elliptic curves. I doubt we'll be able to do that right away, so it will give everyone a few weeks of peace and quiet until we manage to do that.

When that happens, it will be like an HTTPS reset button. No more HTTPS, no more secure sharing of public keys until they find / change algorithms that can be a drop-in replacement of RSA / Elliptic curves.

But the internet will still chug along as always. It's just that more systems will be open / vulnerable.

Most modern consoles already use elliptic curves for application signature, but once that breaks expect to see an increase in homebrew apps for those :-D.

If this thing is true, though. It's going to take a nice, juicy piece of the financial market with it. Would you use your bank if everyone with a wifi-sniffing tool could see your username and password? What about Amazon? It's going to scare people a lot.

Bitcoin is safe though. If you use it properly it's safe against quantum attacks.

Also, when / if this thing turns out to work / be real expect this board to be flooded with normies. It's going to go down in history.