>>883

Doesn't d=c=ab=a^2-x for the record {0,n,c,x,a^2,b^2}

So x=a^2-ab

Sorry

ab=a^2+x

So x=ab-a^2

>>885

For record {0,N,D=c=a*b,X,a^2,b^2}

ab=a^2+x so X=ab-a^2

a and b being from record

{e,n,d,x,a,b}

Where capital represent C^2 record and small letters represent the a*b=c record

>>891

For a and b primes

and a*b=c

there is a record for C =c^2

where A=a^2 and B=b^2

Using cap for C^2 record

{0,N,D=c=ab,X,A=a^2,B=b^2}

So D=c=a*b=A+X

a*b=a^2+X

X=a*b-a^2

For a and b primes

a*b=c

There is a special record

a^2*b^2=c^2

where you can solve for X of c^2 record in terms of a and b

D=c=ab

D=A+X

D=c=ab=a^2 +X

X=ab-a^2

a and b being from the a*b=c record

>>893

the goal is to figure out a and b

We got rid of a veriable

You can do N in terms of a and b also.

X=ab-a^2=a(b-a)

X^2=2AN=a^22N=a^2(b-a)^2

N=((b-a)^2)/2

just for the a^2*b^2=c^2 record

we got rid of X and N

so now

{0,N=((b-a)^2)/2,D=c,X=a*b-a^2,a^2,b^2}

So c^2 is fully described by a and b eliminating e,n,d,x variables

seems like progress

It gets real hairy but you can factor the c^2 record entirely in a and b

The a and b we are trying to solve.So I use caps for c^2 record. A of the c^2 record is a^2. B of c^2 record is b^2. a and b is the goal doesn't make sense to change that.

>>903

There are 5 factorizations of c^2 if a and b are prime. One is a^2*b^2

>>905

Wow you guys seem to think I'm off my rocker! I just tested my formula for A and N for {0,2,35,10,25,49} based on 57=35 squared and for {0,8,21,12,9,49} based on 37=21 squared and for {0,98,51,42,9,289} based on 3*17=51 squared and it worked every time

X = a*b-a^2

N = ((b-a)^2)/2

where ab=c for record a^2b^2=c^2

so for {0,8,21,12,9,49} based on 3*7=21 squared.

X=37-33=12

N=(7-3)*(7-3)/2=8

>>913

By substitution for a^2*b^2=c^2

e is 0

d=c

N=((b-a)^2)/2

X= b*a - a^2

So you can describe C^2 using only the original a and b. It may be possible to substitute and solve for a or b but it kind of hairy. I'm still working with the equations. I'm just a GO player with little math training. I'd love some new eyes. This has been a blast watching and learning.

>>915

There are more factors of c^2 than c

but the a^2*b^2=c^2 factoring is obviously special

>>917

It may be dead end. But eliminating 4 variables seems like progress and maybe there is more that can be done. Potentially even solving for a in terms of b

>>921

>>920

>>919

It was VCQ who suggested the a^2*b^2=c^2

>>924

Creating an e,1 from e,n is huge.

>>924

You said

"You can switch from an (e,n) record to (e,1) by setting n = 1 and a = na."

Does it only work for e=0 still huge

You guys must be close look at bitcoin price!

Below $13,000

>>951

Not knowledgeable but QVC thinks cracking RSA problem for bitcoin.

>>974

thanks

>>924

Is finding element t = (c^2+1)/2 a short cut? We can find n for any c for the c^2 record

{0, n, c, x, 1, c^2}

for e 0 x^2=2na

x=c^2-1

n=((c^2-1)^2)/2

Now create record in 0,1

using n=1 a =na

{0,1, sqrt(A*B), c^2-1, A=((c^2-1)^2)/2, B=c^2}

for any c=a*b prime

Now find the factors which have a pattern of repeating in this column

>>935

Thanks for show how to calculate a record

{0, n, t} for any t.

Even easier for {0, 1, t}

x=2t

d= triangular sum of h=t * 4

d = 4*(t(t+1)/2

Crazy stuff for record {0, 1, t}

If a number m has only prime factors

If t >=m then m is a factor of t{a} if t has those factors. t doesn't have to have duplicate factors. E.G. 225 for any t multiple of 15 and greater than 225 t{a} will be divisible by 225.

>>979

a of {0,1,t}

a number m will show up as a factor of a for {0,1,t}

if t >= m and has the same factors as m duplicate factors not necessary for t

>>986

I was confused

I'm afraid converting to c^2 record to a (0,1) record is a dead end because the factors show up only when they are a factor of t. When there is a remainder it changes and if i understand QVC they show up twice as frequently.

>>993

In an (e,1) record

K fags

for a*b=c

a and b prime

c^2= b^4/(b^4-b^2=1)

Is that useful?

c completely described by b

Minecraft how about a picture.

for a*b=c a and b prime

a^2*b^2=c^2

using caps for the squared expression

{0,N,c,X,a^2,b^2}

X=a^2*(b^2-1)

N=a^2*((b^2-1)^2)/2

see earlier posts

a^2+2X+2N=b^2

a^2+2a^2(b^2-1)+a^2(b^2-1)^2=b^2

a^2(1+b^2-1+b^4-2b^2+1)=b^2

a^2(b^4-b^2+1)=b^2

a^2=b^2/(b^4-b^2+1)

c^2=b^4/(b^4-b^2+1)

>>1097

Please ignore made mistake

>>1098

tested results thanks QVC

for a*b=c a and b prime

a^2*b^2=c^2

using capitals to designate squared record

{0, N, D=c=a*b, X, a^2, b^2}

X=ab-a^2 from X=D-A

N=((ab-a^2)^2)/(2*a^2) from 2AN=X^2

N=a^2/2-ab+b^2/2 square and factor above

c= (D+N)^2-(X+N)^2

D+N = ab+a^2/2-ab+b^2/2=b^2/2+a^2/2

D+N large side = b^2/2 +a^2/2

X+N = ab-a^2+b^2/2-ab+a^2/2=b^2/2-a^2/2

X+N small side equals b^2/2-a^2/2

>>1115

Didn't realize that what you were doing. Got some good ideas anyway. Thanks

>>1169 Amazing body of work. You seem to have all the pieces!!

I'm having trouble verifying your new.d formula. Could you give an example?

Chris is gone from twitter. So is Julian Assange on same day. Remember there are no coincidences!

>>1187 You start with b=1 a=cc record so you have n and x

>>1189

I'm sure q has some good pedo intel on @Jack.That would be awesome to see him perp walk!

>>1179

So how much for you first RSA cracking program!

>>1171

Getting to the aa bb record is ballgame! I'm working on understanding each of your steps.

Goodnight fags.

This is epic!

Merry Christmas!

>>1207

>>1198

It won't be as fast as Julian oops Chris says so there is work to do. However the numbers in those cells grow fast. PMA may have already done enough to endanger RSA for those with ill intent.

>>1196

b = 1 was typo. Its the a =1 b = cc record which is solvable for x and n because e is 0. PMA has put together all the steps you need to move around grid to solve problem. Now it needs speeding up.

>>1222

Second that.

Pray for Chris!

>>1198

Not sure if you know how far you need to move within (0,1) to find aa*bb but within (0,1) it is pretty easy to move to arbitrary new t from existing record since we know t.

x(t-1) =x(t) -2 for any x and t in (0,1)

d(t-1) = d(t) - 4*(t)

So x(t-m)= x(t) - 2m

d(t-m) = 4*sum of (t+t-1…t-(m-1))

Seems you could iterate more accurate guesses of how far you need to move.

>>1231

By x(t-1) i mean x for (0,1,t-1) etc.

Sum can be cleaned up by pulling t out.

>>1231

>>1198

d of (0,1,t-m)=

d of (0,1,t) - (4*t-4(sum of 1+ 2 … m-1))

d of (0,1,t-m) = d of (0,1,t)-4*(t-((m-1)^2)/2)

>>1234

sorry

d of (0,1, t-m)=d of (0,1,t)-4*(t-m(m-1)/2)

Tested!!!

Tested!!!

>>1235

>>1198

I'll always be a fag. Posting formulas before testing!

d of(0,1,t-m) = d(0,1,t)- 4tm+2m(m+1)

for t = 73 d (0,1,73) = 10512

m=12

t-m=61 d(0,1,61)= 7320

10512-47312+212(12+1)=7320

x(0,1,t-m)=x(0,1,t)-2m

x(0,1,73)=144 -2*12=120=x(0,1,61)

>>1243

thanks

>>1259

Since we were starting from 1 cc record the formula was designed for reducing m. It should work for increasing m just use -m in formula.

In (0,1) n is always odd.

Thanks for testing formula.

>>1264

I meant formula is designed for reducing t.

PMA was right yesterday we have more work to do! Even with ability to make large jumps t +- m we don't know where to jump to.If we make a jump the results as far as I know don't tell us if we are any closer to the answer. Its either right or it isn't.

Remember no such thing as coincidence. Chris's twatter is back. So is …

Julian Assange!

I believe white hat spooks had to protect certain twatter accounts from new twatter spying

>>1271

100%

>>1168

Doing some algebra on a+n and knowing it was a perfect square and kept getting /2 inside the sqrt. Thanks for insight.

>>1304

>>1239

m is for Moving through t. The formula is for (0,1) records.

>>1169

PMA check out this record!

{0,8,21,12,9,49}

It doesn't fit the the new.d formula

there are lots more!

When you start at 1 cc you get different patterns

>>1309 excellent that's what VQC was talking about.

For all (0,n) where n =2*(i)^2 where i is integer

all a and b are perfect squares.

All aa*bb records are (0,n) fitting 2,8,28,32,50 pattern

In (0,1) all a and b fit 2,8,18,32,50 also

We are going to crack this shortly

>>1355

Did you or someone else post a derivation for the (0,n) -(0,1) transformation using na.

I missed it.

>>1373

>>1119

For a and b prime

a*b=c

There is a record (0,n,t) {0,n,c,x,aa,bb}

n = 2*tt

2t=sqrt(bb)-sqrt(aa)

any t

d of (1,1, t) = 2*tt

a of (0,1, t) = 2*tt

all (0,n) n = 2*tt a and b perfect squares.

for {0, n, c, c-1, 1, cc) n=2*TT is very large because N is huge.

We know T and VQC will show us how to use

(1, 1 ,T) record and (0, 1, T) record to find

small t (0, 2*tt, ?) {0, 1, c, x, aa, b} record

>>1380

>>1381

>>1386

I'm a fag again. Mis explained what I meant.

For (0,n ) when n equals 2*mm for and integer m then all a and b of those records are perfect squares.

The {0, 2MM , c, c-1, 1 , cc} record is one such record with a very large M. were are looking for the {0, 2mm, c, x, aa, bb} record with a much smaller m.

I believe Chris is going to use the

a of (0,1, M) {0,1, d, x, 2*MM, b} and

d of (1,1, M) {1, 1, 2*MM, x, a, b}

to derive the smaller m with the aa bb record

>>1427

Your right m is doing double duty!

I am using m to count down the perfect square records in (0,n).

Amazingly it turns out that m is the difference between sqrt(a) and sqrt(b)

when n = 2*mm

It is still being used to represent the t in (1,1) records where d equals 2mm and its the t in (0,1) records a equals 2mm so m is MOVEMENT usually up and down t in a cell.

>>1447

how long?

Is that the so called darkness?

>>1451

fag alert

I meant 2m is difference between sqrt(a) and sqrt(b)

>>1422

LOL! Anyone hanging out here obviously has way to many of their own problems to be able to solve math(s) problems

>>1427

The difference between M and m.

M record is {0, 2*MM, c, c-1, cc} with huge M

m record is {0, 2*mm, c, aa, bb} much smaller

Chris's twitter keeps reappearing and disappearing. It was here earlier this morning now gone again.

>>1459

@chrisrootodavid

>>1447

Chris quotes Trump "Keep tweeting"

Chris's twitter profile disappears

>>1473

If we know m problem solved.

But we know big M

and for t = m, M in (0,1) a equals 2mm, 2MM

and for t = m, M in (1,1) d equals 2mm, 2MM

and we know (0, n=2MM) {0, 2MM, c, c-1, 1, cc}

we want (0, n=2mm) {0, 2mm, c, x, aa, bb}

we also know Sqrt(b) - Sqrt(a) = 2m

>>1478

bankster stocks down

Gold silver and their miners up

>>1535

You forgot ab*ab

There are 5 ways

Tested!

For all records {0, n, c, c-aa, aa, bb}

A Goal record!

There is m such that n=2mm

and …

b-a = 2m

{0, 24200, 669, 660, 9, 49729)

m=110

b=223 a=3 a - b = 2m = 220

{0, 8, 77, 28, 49, 121}

m=2

b=11 a=7 a - b = 2m =4