>>877

Continuing discussion from yesterday.

I've tried a number of combinations to link the various (0,n) records.

Following is the closest I've come to a workable path which seems to arrive at the desired a prime value.

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

The theory being, we can figure out how to walk backwards from the (1xcc) record, or really figure out the relationship between the various x values.

—————————

15=3x5

1 x c *=(6,5,2) = {6:5:3:2:1:15} = 15;

a x b **=(6,1,1) = {6:1:3:0:3:5} = 15;

c x c =(0,0,1) = {0:0:15:0:15:15} = 225;

aa x bb *=(0,2,4) = {0:2:15:6:9:25} = 225;

b x ca =(0,10,6) = {0:10:15:10:5:45} = 225;

a x cb =(0,24,7) = {0:24:15:12:3:75} = 225;

1 x cc *=(0,98,8) = {0:98:15:14:1:225} = 225;

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

((14-6)/(14-12)) - 1 == 3 = 3; (a matches)

—————————

65=5x13

1 x c *=(1,25,4) = {1:25:8:7:1:65} = 65;

a x b **=(1,1,2) = {1:1:8:3:5:13} = 65;

c x c =(0,0,1) = {0:0:65:0:65:65} = 4225;

aa x bb *=(0,32,21) = {0:32:65:40:25:169} = 4225;

b x ca =(0,104,27) = {0:104:65:52:13:325} = 4225;

a x cb =(0,360,31) = {0:360:65:60:5:845} = 4225;

1 x cc *=(0,2048,33) = {0:2048:65:64:1:4225} = 4225;

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

((64-40)/(64-60)) - 1 == 5 = 5; (a matches)

—————————

145=5x29

1 x c *=(1,61,6) = {1:61:12:11:1:145} = 145;

a x b **=(1,5,4) = {1:5:12:7:5:29} = 145;

c x c =(0,0,1) = {0:0:145:0:145:145} = 21025;

aa x bb *=(0,288,61) = {0:288:145:120:25:841} = 21025;

b x ca =(0,232,59) = {0:232:145:116:29:725} = 21025;

a x cb =(0,1960,71) = {0:1960:145:140:5:4205} = 21025;

1 x cc *=(0,10368,73) = {0:10368:145:144:1:21025} = 21025;

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

((144-120)/(144-140)) - 1 == 5 = 5; (a matches)

—————————

785=5x157

1 x c *=(1,365,14) = {1:365:28:27:1:785} = 785;

a x b **=(1,53,12) = {1:53:28:23:5:157} = 785;

c x c =(0,0,1) = {0:0:785:0:785:785} = 616225;

aa x bb *=(0,11552,381) = {0:11552:785:760:25:24649} = 616225;

b x ca =(0,1256,315) = {0:1256:785:628:157:3925} = 616225;

a x cb =(0,60840,391) = {0:60840:785:780:5:123245} = 616225;

1 x cc *=(0,307328,393) = {0:307328:785:784:1:616225} = 616225;

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

((784-760)/(784-780)) - 1 == 5 = 5; (a matches)

—————————

901=17x53

1 x c *=(1,421,15) = {1:421:30:29:1:901} = 901;

a x b **=(1,5,7) = {1:5:30:13:17:53} = 901;

c x c =(0,0,1) = {0:0:901:0:901:901} = 811801;

aa x bb *=(0,648,307) = {0:648:901:612:289:2809} = 811801;

b x ca =(0,6784,425) = {0:6784:901:848:53:15317} = 811801;

a x cb =(0,22984,443) = {0:22984:901:884:17:47753} = 811801;

1 x cc *=(0,405000,451) = {0:405000:901:900:1:811801} = 811801;

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

((900-612)/(900-884)) - 1 == 17 = 17; (a matches)

—————————

6107=31x197

1 x c *=(23,2976,39) = {23:2976:78:77:1:6107} = 6107;

a x b **=(23,36,24) = {23:36:78:47:31:197} = 6107;

c x c =(0,0,1) = {0:0:6107:0:6107:6107} = 37295449;

aa x bb *=(0,13778,2574) = {0:13778:6107:5146:961:38809} = 37295449;

b x ca =(0,88650,2956) = {0:88650:6107:5910:197:189317} = 37295449;

a x cb =(0,595448,3039) = {0:595448:6107:6076:31:1203079} = 37295449;

1 x cc *=(0,18641618,3054) = {0:18641618:6107:6106:1:37295449} = 37295449;

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

((6106-5146)/(6106-6076)) - 1 == 31 = 31; (a matches)

—————————

20413=137x149

1 x c *=(249,10065,71) = {249:10065:142:141:1:20413} = 20413;

a x b **=(249,1,3) = {249:1:142:5:137:149} = 20413;

c x c =(0,0,1) = {0:0:20413:0:20413:20413} = 416690569;

aa x bb *=(0,72,823) = {0:72:20413:1644:18769:22201} = 416690569;

b x ca =(0,1377952,10133) = {0:1377952:20413:20264:149:2796581} = 416690569;

a x cb =(0,1500424,10139) = {0:1500424:20413:20276:137:3041537} = 416690569;

1 x cc *=(0,208324872,10207) = {0:208324872:20413:20412:1:416690569} = 416690569;

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

((20412-1644)/(20412-20276)) - 1 == 137 = 137; (a matches)

>>879

It gets even simpler.

The solution is here.

(1xcc).x - (aaxbb).x + 1 == (aaxbb).a

Difference in x between the c^2 record and the a^2, b^2 record + 1 == the prime solution a^2.

Who wants to figure out x ?!?!?!?

===

785=5x157

1 x c *=(1,365,14) = {1:365:28:27:1:785} = 785;

a x b **=(1,53,12) = {1:53:28:23:5:157} = 785;

c x c =(0,0,1) = {0:0:785:0:785:785} = 616225;

aa x bb *=(0,11552,381) = {0:11552:785:760:25:24649} = 616225;

b x ca =(0,1256,315) = {0:1256:785:628:157:3925} = 616225;

a x cb =(0,60840,391) = {0:60840:785:780:5:123245} = 616225;

1 x cc *=(0,307328,393) = {0:307328:785:784:1:616225} = 616225;

(1xcc).x - (aaxbb).x + 1 == (aaxbb).a

784 - 760 + 1 = 25.

(1xcc).x - (axcb).x + 1 == (axcb).a

784 - 780 + 1 = 5.

>>882

ok. appologies. this is exciting.

Simplified formula, and removed unnecessary calculations:

sqrt( (1xcc).x - (aaxbb).x + 1 ) == (axb).a

We just need the formula for the x value in (aa x bb).

—————————

65=5x13

1 x c *=(1,25,4) = {1:25:8:7:1:65} = 65;

a x b **=(1,1,2) = {1:1:8:3:5:13} = 65;

aa x bb *=(0,32,21) = {0:32:65:40:25:169} = 4225;

1 x cc *=(0,2048,33) = {0:2048:65:64:1:4225} = 4225;

sqrt( (1xcc).x - (aaxbb).x + 1 ) == (axb).a

sqrt( 64 - 40 + 1 ) = 5. (a matches!!!!)

—————————

145=5x29

1 x c *=(1,61,6) = {1:61:12:11:1:145} = 145;

a x b **=(1,5,4) = {1:5:12:7:5:29} = 145;

aa x bb *=(0,288,61) = {0:288:145:120:25:841} = 21025;

1 x cc *=(0,10368,73) = {0:10368:145:144:1:21025} = 21025;

sqrt( (1xcc).x - (aaxbb).x + 1 ) == (axb).a

sqrt( 144 - 120 + 1 ) = 5. (a matches!!!!)

—————————

785=5x157

1 x c *=(1,365,14) = {1:365:28:27:1:785} = 785;

a x b **=(1,53,12) = {1:53:28:23:5:157} = 785;

aa x bb *=(0,11552,381) = {0:11552:785:760:25:24649} = 616225;

1 x cc *=(0,307328,393) = {0:307328:785:784:1:616225} = 616225;

sqrt( (1xcc).x - (aaxbb).x + 1 ) == (axb).a

sqrt( 784 - 760 + 1 ) = 5. (a matches!!!!)

>>884

d=c yes.

solution can be simplified yet again to

sqrt( c - (aaxbb).x )

not following you on the formula for x.

>>886

if we knew a and b, I think the formula would be x = sqrt(ab - a^2).

We don't have the ab starting position.

The solution depends on figuring out the correct x value for (aa x bb) from only (1xcc).

>>889

anything is possible. but I'm using d, x, and a from completely different records.

>>890

sorry. I'm not seeing it.

>>892

don't we want to figure out x going the other way?

anyone having any luck? going in circles.

>>1013

>>1017

>>1022

still struggling to find any relationships in the (0,1) space.

I have created records for (e,n) of c in (0,1) at the following variations of t just looking for patterns:

t = d

t = 2d

t = 2d + 1

t = x + n

There have been a few interesting connections between (0,1,c[d]), (0,1,cc[t]), and (0,1,c[t]), but they are so confusing I just can't believe there isn't a simple answer.

Also ran into problems walking the (0,1) tree using combinations of p.

Can't seem to find any combination that leads me back to an (a^2 b^2) record.

Attached is a pic of the output I'm looking at. I must be missing something very simple.

>>1099

Not sure if these have been answered:

>Square a value c, where c is the product of two different primes.

>What are the values of n in column zero where these appear?

>What is the value of d where these appear?

For any value c, the cc record can be created via:

e = 0

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

d = c

x = c - 1

>>1310

>they all appear to be in (0, 2), (0, 8), (0, 18), (0, 32) etc

2*(1^2)

2*(2^2)

2*(3^2)

2*(4^2)

etc.

This record is interesting. How did you get to this one?

(0, 2, 255, 30, 225.0, 289.0)

>>1312

>2(x + 1)^2

and where is the x coming from?

>>1318

Very good.

New avenue to explore.

Looks like first record in the a = c^2 series can be created via:

e = 0

x = 2*c

a = c^2

Will iterate these patterns shortly.

>>1309

>>1341

I've updated my test cases to include iterating at a=c^2. See FindValuesOfBFromCSquared.

Interesting is that at m=5, where m is a multiple of x = 2 * c, the square root of b is the prime b squared value we are looking for.

This happens also for different c values, but many multiples lower.

Following is the latest example for c=15:

—————————

15=3x5

a x b *=(6,1,1) = {6:1:3:0:3:5} = 15;

1 x c =(6,5,2) = {6:5:3:2:1:15} = 15;

aa x bb =(0,2,4) = {0:2:15:6:9:25} = 225; (x+n)^2=8x8=64; (d+n)^2=17x17=289; sqrt(n/2) = 0;sqrt(2n) = 1;

1 x cc =(0,98,8) = {0:98:15:14:1:225} = 225; (x+n)^2=112x112=12544; (d+n)^2=113x113=12769; sqrt(n/2) = 7;sqrt(2n) = 14;

TestCalculationsFromC

———————

cc.n = ((c-1)^2)/2 = ((15-1)^2)/2 = 98;

cc.d = c = 15;

cc.x = c-1 = 14;

ccna.d = ((c-1)/2)(c+1) = ((15-1)/2)(15+1) = 112;

TestMLevelJumpingByD

(0,1,7) = {0:1:84:12:72:98} = 7056

(0,1,6) = {0:1:60:10:50:72} = 3600

(0,1,5) = {0:1:40:8:32:50} = 1600

(0,1,4) = {0:1:24:6:18:32} = 576

Matching aabbna record found at m=4

FindValuesOfBFromCSquared

m=1: (0,2,16) = {0:2:255:30:225:289} = 65025;

m=2: (0,8,31) = {0:8:285:60:225:361} = 81225;

m=3: (0,18,46) = {0:18:315:90:225:441} = 99225;

m=4: (0,32,61) = {0:32:345:120:225:529} = 119025;

m=5: (0,50,76) = {0:50:375:150:225:625} = 140625; sqrt(b) == ( p.b * p.b ) Sqrt(b)=25; p.bsquared=25

m=6: (0,72,91) = {0:72:405:180:225:729} = 164025;

m=7: (0,98,106) = {0:98:435:210:225:841} = 189225;

m=8: (0,128,121) = {0:128:465:240:225:961} = 216225;

m=9: (0,162,136) = {0:162:495:270:225:1089} = 245025;

m=10: (0,200,151) = {0:200:525:300:225:1225} = 275625;

m=11: (0,242,166) = {0:242:555:330:225:1369} = 308025;

m=12: (0,288,181) = {0:288:585:360:225:1521} = 342225;

m=13: (0,338,196) = {0:338:615:390:225:1681} = 378225;

m=14: (0,392,211) = {0:392:645:420:225:1849} = 416025;

m=15: (0,450,226) = {0:450:675:450:225:2025} = 455625;

m=16: (0,512,241) = {0:512:705:480:225:2209} = 497025;

m=17: (0,578,256) = {0:578:735:510:225:2401} = 540225;

m=18: (0,648,271) = {0:648:765:540:225:2601} = 585225;

m=19: (0,722,286) = {0:722:795:570:225:2809} = 632025;

m=20: (0,800,301) = {0:800:825:600:225:3025} = 680625;

(1 x cc).x - (aa x bb).x = 14 - 6 = 8

sqrt( (1xcc).d - (aaxbb).x ) == (axb).a

sqrt( 15 - 6 ) = 3. (a matches!!!!)

>>1343

Thanks. I'm using the sqrt big number function vqc originally posted. Do you have a better alternative?

>>1342

Just a reminder, there is a pattern to the n component in these records.

n = 2*(m^2)

If n is the key to breaking this, then we need to figure out the value of m from either our (1xc), (1xcc) starting positions.

>>1349

sorry. c# here. using BigInteger.

>>1365

We're not there yet, as far as I know.

We have been exploring the grid through various combinations of c^2.

Iterating various results in the (0,n) and (0,1) spaces using a multiplier (m) has exposed a few potential matches.

>>1367

To move from (0,n) to (0,1), simply create a new record where e=0, n = 1, and a = n*a.

>>1369

Just want to share an example of an idea that looks promising, but doesn't work when applied to different test cases.

—————————

21=3x7

1 x c =(5,7,2) = {5:7:4:3:1:21} = 21;

a x b =(5,1,1) = {5:1:4:1:3:7} = 21;

1 x cc =(0,200,11) = {0:200:21:20:1:441} = 441;

Searching through the (0,n) space where a = c^2, we find an interesting record where the sqrt(b) equals b^2 from our desired result.

m=14: (0,392,295) = {0:392:1029:588:441:2401} = 1058841; sqrt(exa.b) == ( p.b * p.b ); Sqrt(b)=49; p.bsquared=49; p.b=7;

For this record, we know e = 0, and a = c^2, and a few other things about n. sqrt(n/2) = 14, which equals m (a multiplier used to loop and create records).

We need one more formula or variable in order to construct this record directly, so I have been trying to back into a solution.

Various examples of determining the new n from the (1xc) record:

new.n = (c^2 - n^2) = (2121) - (77) = 392

new.n = 2((c-n)^2) = 2((21-7)^2) = 392

Also tried calculating the new.d from various combinations, but nothing holds up yet.

>>1380

can't wait.

I believe the following examples apply to the na and (n-1)a references for c=145.

145=5x29

a x b *=(1,5,4) = {1:5:12:7:5:29} = 145;

a x b (n-1)a *=(1,1,4) = {1:1:32:7:25:41} = 1025;

1 x c =(1,61,6) = {1:61:12:11:1:145} = 145;

1 x c (n-1)a =(1,1,6) = {1:1:72:11:61:85} = 5185;

.>>1383

would love some of that insight.

>>1389

Been all over the place searching. Still no clear path to a solution.

>>1463

>>1464

MA, not sure I'm following you.

I have been able to run all my test cases against rsa100. See pic attached.

Have you gotten closer to a solution?

>>1466

>>1468

Got it. Thanks.

>>1466

>>1473

>>1372

MA, VA

Just ran another test on RSA100.

If the solution relies on finding out the multiple (m) at a record where e=0, x=2cm, a=c^2, then m would have to be:

m=42489607263661711309596655624896701585610939620632192811974298340236852638535916326429571625162791

Going to continue exploring the na and (n-1)a…

>>1497

MA, do you need an isvalid formula or a method to construct your records? I posted a while back, but happy to share.

>>1498

For small numbers, finding the factors is straight forward. RSA length numbers is a different story. We're missing something here.

VQC dropped 2 hints about na and (n-1)a relationship.

I've run a number of tests comparing the 2 records, and the a and d values are where the movement appears. And specifically at d we start getting new numbers to explore.

Still not seeing a way to relate the the c and ab records.

>>1624

Been following along. Great insights. Are you assuming we know a?