mersenneforum.org 69660 and 92020
 Register FAQ Search Today's Posts Mark Forums Read

2021-07-11, 19:07   #12
enzocreti

Mar 2018

22×7×19 Posts
...

Quote:
 Originally Posted by rudy235 I never quite understand what you are trying to say. Can you explain to the unwashed masses the function pg? What is it? Paying guest? Picogram?
the last you said

 2022-01-03, 13:06 #14 enzocreti   Mar 2018 53210 Posts 92233=427x6^3+1 92233=6^3=51456 mod 427 ((92233-51456)-1)/3-12592=10^3 12592 divides 541456 PG(51456) and PG(541456) are primes 92233-51456=1=69660 mod 1699 13592=1699*8 43*(((51456+(3^6-1)*4)/4)-10^3)=541456 69660-(487-51456)=69660-(92233-51456)=0 mod (71x1699) (69660-487)=0 mod 313 (69660-486)=0 mod 427 486=162x3 162 divides 69660 -449449-13=64=541456+13 mod 139 so 541456=51 mod 139 (449x13=-1 mod 139) 92020=541456+13-449449 331259=71*6^6=259 mod 331 331259=44 mod 71 331249-44=331215 that Is the concatenation of 331 and 215 331259-44-215 Is a multiple of 331 44+215=259 71x6^6-331259+92020=6^2-6^5 mod 311 (6^2-6^5)=7740 which divides 69660 (7740=1 mod 71) multyplying by 9 both sides 71x6^6x9-331259x9+92020x9=-69660 mod 311 331215 is the concatenation of 331 and 215 331+215=546 546 divides 75894 and 56238 pg(75894) and pg(56238) are primes 75894-92020/2=215x139-1 546=215+331 215x139 can be cancelled in both sides this leaves 46009-46010=-1 56238 can be factorized as (139-6^2)*(331+215)=56238 probably there is something in 79*3^j pg(79) is prime by the way 79*3^2=711 and 69660=71x711+2131x3^2 79*3^3=2131+1 71x79+2131=7740 which divides 69660 79x3^2=1 mod 71 69660-19179=71x79x3^2 239239=-9 mod 12592 541456=239239+9 mod 12592 92020=3867+9 mod 12592 23004*46009-1-(92020-6)=3539x13x23003 i think that there should be some group in action... pg(75894) and pg(56238) are primes with 75894 and 56238 multiple of 139 75894+56238=132132=1 mod (1861x71) 331259=1 mod 1861 75894+56238=331259 mod (1861x107) 331259-132132=107x1861 75894+56238-1=71x1861 107=71+6^2 56238 and 75894 are congruent to +/- 6 mod 132 in particular (75894+6)/132+1=24^2 so 331259=56238+75894+92020+107x1001 pg(394) is prime 1323=-1 mod 331 pg(1323) is prime 1323=72 mod 139 1323=(1001-72) mod 394 i suspect that this is connected to the fact that 331259=-72 mod (1001x331) 331259=-1323 mod (1001-72=929) 331259=-394=-1323 mod 929=1001-72 3x6^3-3=394x(72)^(-1) mod 1001 In particolar 394x431+3-3x6^3=169169 92020=1323-394=929 mod 1001 541456+13=1323-394-13=916 mod 1001 (92020-1323+1)=0 mod 449 (541456+13-1323+1)=0 mod 449 541456-449449=92020-13 i forgot to see that 92020=43*2132+344 344 is the residue of the multiples of 43 (69660, 541456, 92020) mod 559 43*2132 is infact divisible by 559 69660-(46009*23005-1)/(313x49)=2^3x3^4=3x6^3 92020=-2 mod (313*49) 313x49x3x6^3-1=215^3 23005*(2+46009x8)=1 mod 429^2 23005x(2+46009x8)=-1 mod (359xp) where p is a prime p=23586469...i wonder if it has some special property the logic behind these primes requires tools that are far beyond current knowledge... 331259=259=71x6^6 mod 331 6^k=1 mod 259 the order mod 259 of 6 is 4 ord(6)=4 infact 6^4=1 mod 259 92020=2 mod 331 92020=4 mod 6^4 92020=71+4=75 mod 259 (92020-4)=71x6^4 239239=-77=-75-2 mod 259 239239=-78 mod 449 541456-77x5837+13=92020 curious that 449449-(239239+78)=210132=13*6^2x449 which is congruent to (6^4-2) mod 2131 239239+78 is 449x41x... 449x41 is 1840...anything to do with 429^2-1??? 541456=7740 mod 18404 7740 divides 69660 18404=(429^2-1)/10=449x41-5 pg(3336) is prime 3336=24x139=-1 mod 71 (24+71x9)x139=-1 mod 71 (24+71x9)x139+1-138=92020 (24x71x9)x139+1)/71=6^4+2 so 92020=(6^4+2)x71-138 or equivalently 92020=71x6^4+142-138=71x6^4+4 ((24+71*15)*139+1)/71=2132 so for example 19179=3^2*2131 can be rewritten as 3^2x(((24+71*15)*139-70)/71)=19179 71x6^6=6^4-44 mod (331x4=1324) 331259=44+215 mod 1324 331259=44 mod 71 (71*6^6-6^4+44)/(139*18-1)=1324 may be 44 is not random... 69660=(44^2-1)x6^2 331259=44 mod (311x5x71) 331259^2=1936 mod (311x5x71) 1935 divides 69660 6^5=1 mod (311x5) 6^5-6^2 divides 69660 1-36=35 I suspect that there could be some link to the fact that 541456=51456+700^2 700^2 is divisible by 35 curious that pg(75894) is prime 75894 is multiple of 139 -75894=(2^16+16) mod (359x197) 2^16=-2 mod 331 2^16=1 mod (255x257) i think that something big is happening on some field sqrt(71x215x3+1)=214 92020=sqrt(71x215x3+1)x430 92020=(71*215*6+1)+429 92020=214^2+215^2-1 pg(51456) and pg(6231) are primes with 51456 and 6231 multiple of 67 curiously (51456/67-6231/67)=26^2-1 19179=648 mod 23004 331259=(19179-648)/71 mod 359 331259x71=222 mod 359 so 222=19179-648 mod 359 note that 331331-(331259-(19179-648)/71)=333 ((92020*3-6)*3*4-1)/71=6^6+1 ...3312648...i think that 3x6^3 has something to do with these numbers... 331259/71=4665+44/71=6^6/10-6/10+44/71=6^6/10+7/355=6^10/10+7/(359-4) 331259=44 mod 4665 106x44+44/71+1=331259/71 4665=359*13-2 331259 =71 mod 44 331259=44 mod 71 so 331259 is a number of the form 115+(5^5-1)k this is curious 43*(1+sqrt(9x+1))=9x solution x=215 215*9+1=44^2 i think that you can obtain a continued fraction from that 69660=6^2x43x(1+sqrt(1+43x(1+sqrt(1+43x(1+... curious fact: 69660=19179=3x6^3 mod 639 19179/3=(639)3 and 6393=-1 mod 139 I think that something is in action over some field... (429^2-6)=-1 mod 46009=331x139 (429^2-6)=3 mod 639 from this 429^2=80^2-1=79x3^4=711x9 mod (71x139) 6393+((71*4+1)*139+1)=46009 i think that in Z46009 something is in action as well as in Z23004 71x6^6-541456=-1 mod 359 541456+14=-261=-331259 mod 359 69660=14=-6^6 mod 359 541456+69660=611116=-261=-331259 mod 359 611116=131x4665+1 92020-(541456+14-98)/13/359=359x2^8 -331259=98 mod (359x13) -(541456+69660)=-611116=359-98=261 mod (359x13) pg(1323) is prime pg(39699=13233*3) is prime 13233=18^2 mod 331 69660=215x18^2 so 13233x215=69660=150 mod 331 and 13233x3=39699=-150 mod (359x111) (359=331+18) 69660=3x6^3 mod 71 (69660-3x6^3)/71=-1 mod 139 this suggests me that something is in action over Z139 or maybe Z46009 23005*(2+331x139)-1 is divisible by 11503 69660-642(=3x6^3-6) is divisible by 11503 23005x(2+46009)-1-(69660-642) is a multiple of 23003 2*(23005*(2+46009*72)-1)/46009-2=331272x10 331272-13=331259 23005x(2+46009x72-1) is divisible by 449 541456+13-449x1001=331259 something mysterious is boiling in Z46009 46009x72-72=71x6^6 x^2/(6^6-2-sqrt(2x+1))-x*(sqrt(2x+1)-1)/215=0 has solution x=92020 min (x^2/(6^6-2-sqrt(2x+1))-x*(sqrt(2x+1)-1)/215)=-19394.4=-19179-215.4 this is a parabola from 331259x71=19179-17^2 mod 359 we have 222=19179-17^2 mod 359 so (2^9-1)=19179 mod (359x13) curious that 19179-511+1 is divisible by (2^2-1), (2^3-1) and (2^7-1) 331259=-98=-(512-414) mod 359 19179=414 mod 139 19179=138x139-3 69660=19179=6^6 mod 639 the inverse mod 639 of 2131 is 427 69660x427=9 mod 639 69660x427-9=639x46549 46549 is prime =6^6-107 I think this is not random but connected to the fact that 107 divides 92020 19179=7x71+14 mod (359x13) 6^6=-14 mod (359x13) 71x270+9=7x71+14=(2^9-1) mod (359x13) from here 6^6=7x71-19179 mod (359x13) form here after some steps... 331x72=7x71 mod (359x13) curious that 19179x(7^4+1)=1 mod (359x13) 3x6^3+70 is divisible by 359 (331259-5) is divisible by 717x6=6x(3x6^3+69) 6x(3x6^3+69)=-1 mod (331x13) so (331259-5) is divisible by (331x13-1) (331259-5)/(331*13-1)-(92020-5)/239/77=72 331x13=-5 mod 359 359x12=-1 mod 139 331259 has the curious representation: 65^2*77+77^2+5=331259=325325*77+77^2+5 further steps toward a theory of these numbers need super-tools (331259-5)/7-6^6=666 6^6=-666 mod (239x11) anything to do whit 92020=5 mod (239x11) 331259=5 mod (239x11)??? curious fact: (239239-(6^6+666)+2)/111=1729 the Ramanujan number I will call these primes Neme primes (Neighboured Mersenne) Neme(3)=73 Instead of Neme primes I could call them Hopeless primes, no hope to find a logic behind them curious that 69660=-342^2 mod (432^2) I think that starting from 23005x(2+46009x8)=1 mod 429^2 and 23005x8=-1 mod 429^2 one can develop someting useful Using Chinese remainder theory numbers multiple of 23005 (=0 mod 23005) and =1 mod 331x139 should form a ring or something similar...and I think that in that ring one can get something (2+46009x8+2x92020) is divisible by 92019 92019x6-92020x5 is a multiple of 3539 a wagstaff prime... 331259=2132 mod 3539 6^6-3x6^3-1=3539x13=46007 239239=-13 mod (9202) (239239+13)/107=2236 69660+2236x10=92020 331259=331 mod 2236 239239x2236x2=-(331259-331) mod 331 239239x4472=72 mod 331 2236=6^2x239239^(-1) mod 331 22360=(19^2-1)x239239^(-1) mod 331 from here 22360=-148 mod 331 239239=-(148/2) mod 331 22360=257-74 mod 331 239239=257 mod (331x19^2) (331259=257) mod (71x111) 331259x6^2=(22360-257) mod 71 I have the impression that powers of 6 and 71 are involved in the logic behind these neme primes 92020+257-14 (257-14=243 a power of 3) is 359x257 331259-243 is a multiple of 257 1001-((331259-243)/257-359)=72 69660=13 mod 257 69660=14 mod 359 there is a logic but it is so complex that it is almost hopeless to find a pattern 331259+14+84=71x13x359 541456=84 mod 359 69660=-345=-(331+14) mod 359 6^6=-14 mod 359 i cannot put toghether the entire pieces of the puzzle anyway 331259=6^6-541456 mod (359x13) (69660-6^6+331)*2-14=6^6 92020=10 mod 3067 331259=22 mod 139 331259=23 mod 3067 331259-23-92020+10=239239-13 92022x36-6^3=71x6^6 71x6^6=6^3 mod 3067 71x6^3=-1 mod (313x7^2) 71x6^3=1 mod 3067 I would call these prime Neme primes or maybe desperate primes I stronly suspect that the exponents of these neme primes are connected among them with a logic that it is impossible to understand...only a God could find a pattern...or maybe a new Gauss... curious that 541456=353(7)9 mod (3539x13) with that "7" inserted 92020=6 mod (359x13) in other words 541456=3539x10-10-1 mod (3539x13) curious that (541456-3539x10) is divisible by 23003=71x2^2*3^4-1 71x6^6=72 mod (3539x13) -(541456+13)=787 mod 858 -331259=787 mod 858 69660=162 mod 858 92020=214 mod 858 359x239=1 mod 429 541456=0 mod 787 787-13=774 divides 69660 787-429=359-1 so for example 331259=429x774-787 774 divides 69660 429=sqrt(92020x2+1) -541456=344 mod 774 there is a hidden structure it is clear that 331259-774 is a multiple of 4601 and 4601 divides 92020 -541456-13=456+331 541456+13-456=359x11x137 331259=-358 mod (773x429) 358=359-1 773=774-1 331259+773 is divisible by 1297 a prime of the form 6^s+1 331259=(259-215=44) mod 71 331259=259 mod 331 71x6^6=259 mod 331 there is something... 23005*(2+46009*k)-1=N^2 for k=8 for k=3680 ,... 23005*k+1 is a square k=8 k=3680 ... 3680*23005+1=(3x3067)^2=9201 (71x6^6-216)/3/3067=359+1 92020=10 mod (3067x13) (9201^2-1)/4601/23-13=787 Last fiddled with by enzocreti on 2022-03-06 at 12:35
2022-03-06, 12:54   #15
enzocreti

Mar 2018

22·7·19 Posts

Quote:
 Originally Posted by enzocreti 92233=427x6^3+1 92233=6^3=51456 mod 427 ((92233-51456)-1)/3-12592=10^3 [500 lines of excessive quote]
3067 is a prime

there is a very complex hidden structure

331259/3680 is very close to 90

7775*23005+1 Is a Square

-331259=3067-1 mod 7775

(331259+3066)-10001=18^2*(10^3+1)

I think these primes taste very exotic

23005 X+1=Y^2

with X and Y integers

X=92019 is a solution

Elliptic curves???

92016( =71x6^4) x 23005+1=46009^2

3067*5+1=71x6^3

92020-71x6^3 is a multiple of (19179-8=19171)

-331259=(3x3067)^2 mod 359

after some steps (inverse of 3067 mod 359 is 139x2)

-331259x(10^2-1)=9 mod 359

so it is clear that 71 and powers of 6 are involved in these primes

(71*6^6-216-(92020-10))/(71*6^3-1)=210

anything to do with the fact that 541456+13-210*1001=331259??)

71x6^6-216 is a multiple of 3067

92020-10 is a multiple of 3067

3067x6=-1 mod (239x77)

239239=13 mod (3067x6)

-3067x3=1 mod 107

239239=-13 mod 107

331259=-13 mod 107

it seems to be a perfect complex interlocking of modules

I think that only a math-champ chould develop a theory for these numbers...I think that one should know very well Galois theory at least

331259=9203 mod (3067x5+1)
331259=5 mod (3067x6+1)
92020=5 mod (3067x6+1)

239239=0 mod (3067x6+1)

541456=7740 mod (3067x3+1)

7740 divides 69660

71x6^6=6^2 mod (3067x6+1)

19179=777 mod (3067x3)

strange at least curious

19179=3067x6+777

(6^6-3*6^3-3)=46005 is divisible by 3067,5,3

777/3=259

-331259=71x6^3x777 mod 331

359x18^2=1 mod 541

-541456=85 mod 541

-541456=359x(18^2x85) mod 541

-541456=359x490 mod 541

541456-51456=700^2 which is divisible by 490

-700^2=359x490+51456

after some steps:

490x(-1000-359)=51456 mod 541

51456=-480 mod 541

(51456+480)=51936=5x10^4+44^2

(44^2-1)=1935 divides 69660

1935-479=1456

479-394=85

pg(394) is prime

541456=-(44^2-1)+1456+394 mod 541

51456=-490x(10^3+359) mod 541
541456=-490x359 mod 541

51456=129360 mod 541
51456=129359 mod 359

541456=18309 mod 541

541456-51456=490000

curious thta

429^2-1=540x21^2 mod 541

(429^2-1)=3x394 mod 541

pg(394) is prime pg(3x21^2=1323) is prime

92020=-491=3x394x271 mod 541

curious that 69660=-129 mod (541x129)

curious that

(429^2-1)=3x394 mod 541

dividing by 2

92020=3x197 mod 541

92020-3x197=91*10^4+429

92020=14=-331259=(69660+1=69661 prime) mod 257

curious that

69660-6^6+1=23005 so

92020=(69660-6^6+1)x4=(3*6^6-(7^3-1)^2+1)*4

from here we come to the curious:

69660=432^2-342^2 (432 is just apermutation of 342)

or =432^2-(18x19)^2

92020=-67=4 mod (6^4+1=1297 prime) and mod 71

pg(67) is by the way prime

-768x92020=67x768=51456 mod (1297)

so

23^2x92020=51456 mod 1297

curiously

23^2x92020-51456+1=365^3

365^3=1 mod 1297

(92020-4)+6^6=107x6^6

107 is the inverse mod 71 of 215

-331259=1001 mod 449

541456+13-449x1001=92020

331259=92020+239239

-331259x449=541456+13-92020 mod (449^2)

maybe manipulating this you get something

449x740-1001=331259=92020+239x1001=541456+13-210*1001

curious that (541456-6) which is 0 mod 13 is congruent to (1001+13)/13=78 mod 359
the most surprising fact about these primes for me is this: why the inverse concatenation (13 instead of 31, 715 instead of 157, 1531 instead of 3115 does not give patterns???)

71x6^6+72=0 mod 46009

(71x6^6+72)=261x449 mod 359

331259=261 mod 359

((71*6^6+72)*4-331259)/359=35987

331259=3^2*29 mod 359
92020=2^2*29 mod 359

-331259+6^6=541456 mod (359x13)

-331259-14=84 mod(359x13)

-331259=98 mod (359x13)

(541456-84)/359/13=116

116 is the residue mod 359 of 92020

-23004=331 mod (359x13)

-23004x4=-92016=1324 mod (359x13)

pg(1323) is prime

92016+1323 is a palindromic number

331259=13x359 mod 6^6

i think that 13x359 is important for these numbers....and powers of 6...

69660=14 mod 359
6^6=-14 mod 359

-331259=14^2/2=98 mod (359x13)

pure chance???

-331259=2^11x3^12 mod (359x13)

(331259+2^11x3^12)/359/13-1=5x6^6

69660=19179=2131x3^2=6^6 mod 639

69660=7740x9

2131 and 7740 are numbers of the form -3478+5609s (using chinese remainder theorem 2131=-2 mod 79 2131=1 mod 71 7740=-2 mod 7 7740=1 mod 71)

69660=(88^2-4)x3^2

(88^2-2)=0 mod 79

(69660-19179)=(88^2-2) mod (79x541)

69660+541456=611116=17x19x44x43

69660+541456=611116 this strange palindromic number

611115 is divisible by 4665 (=359x13-2) 4665 again shares the same digits with 6^6=46656

i think these numbers are more mysterious than pyramids of Giza

331259=4665x71+44

(359*13)*131-261=611116

261 is the residue of 331259 mod 359

92020-(541456-316)/4665=0 mod 359

69660=-315 mod 4665
541456=316 mod 4665

4665=(3/5)x(6^5-1)

((3/15)*(6^5-1)+1) divides (92020-216)

541456-51456=700^2

(3x13x359-1)x(6^2-1)=700^2=(3x13x359-1)x(394-359)

i think that a key passage is this:

331259=(19179-3x6^3)/71=261 mod 359

-92020=(69660-3x6^3)/284=243 mod 359

222=19179-3x6^3 mod 359

19179=51x359+870

i dont know if this has something to do with the fact that pg(51) and pg(359) are orimes

19179=(2^9-1) mod (359x13)
19179=(2^9+1) mod 51

-69660=2^9 mod 331x2

18^2x213+3x6^3=69660

18^2x213-6^6=22356

(92020-4=71x6^4)-22356=69660

i started the hunt for a new neme prime Ne(k) with k of the form 648+213s. Probably I will never find it

(331259-261)/359=922=-1 mod 71

I think that everything is inset in these numbers, modular congruences,fields,...very very difficult...

no other prime Neme(23005k) found after 92020 (up to 1300000)

I realized that I forgot the most stupid thing: 2131-14^2=1935 which divides 69660

from this

19179x4-84^2-23004=-14 mod (359x13)

after some steps...

19179x4-82^2=-13 mod (359x13)

2131x6^2=82^2-13 mod (359x13)

2131x6^2(=19179x4)+13 is a perfect sqaure!!! (277^2)=1 mod 139

so

277^2=82^2 mod (359x13)

277^2=1 mod (138x139)

277^2-82^2=511x137(=70007)-2

19179=511 mod (359x13)

by the way 19179=-1 mod 137

any possible connection to the fact that (541456-51456)=700^2=-7^2 mod 70007??? (700^2=-35 mod (359x13x35))

it could be a chance...anyway

(277^2-1) is divisible by 23,139 and 24...if you divide by 23 that is (277^2-1)/23=3336 and pg(3336) is prime

3336=1 mod 23

331259=3336x23=261 mod 359

541456-344 is divisible by 559x44

44x559=2236x11

92020-69660=22360

69660=(44^2-1)x6^2

19179=2131x3^2=(3^7-2^7)x3^2+3x6^3

19179=-1=71x3^3 mod (137)

69660=-6^6=14=(71x3^3+1)/(359-222) mod 359

331259x71=222 mod 359

19179-3x6^3 is divisible by 261 (i think it is not chance that 331259-261=0 mod 359)

(19179-648+1) is divisible by 41 and 452 ( pg(451) is prime)

so

(19179-648)=452*41-1

pg(51) is prime

so

452x41-1=222=331259x71 mod 359

452x41-1-222=51x359

(331259*71-452*41+1)/359/71=922=-1 mod 71

(331259-261)/359=922

pg(1323) is prime pg(39699) is prime

39699=9 mod 1323

(39699-9)/2-666=19179

pg(19179=2131*9) is prime pg(2131) is prime

so 19179=2131x3^2=-666 mod 1323

19179=5x63^2-666

19179=-666 mod 1323
19179=-665 mod 451

pg(1323) and pg(451) are primes

pg(451) and pg(1323) are two consecutive pg primes!

this is equivalent to

19179=-666 mod 1323
19179=-214 mod 451

-19179x430=16=92020 mod 451

19179x430+16 is a multiple of 41^2
19179x430+92020 is a multiple of 43^2

(92020-69660)=22360=(19179-3x6^3)/71 mod 451

19179=2x344=3x79 mod 451

69660-19179 is a multiple of 79

71x6^4-69660=-261=-331259 mod 359

71x6^4+4-69660=261 mod 451

71x6^4-14=-261=71x6^4+6^6=-331259 mod 359

71*6^4-69660+331259=0 mod 359
71*6^4-69660+331259=-1 mod (139x106)

I think that in 139Z there is something

(71*6^4-69660+331259+1)/106=3336 and pg(3336) is prime

106=-1 mod 107 (anything to do with the fact that 92020 is a multiple of 107???)

curio:

pg(4) is prime pg(51) is prime pg(451) is prime pg(92020) is prime

92020=4x51x451+16

2x51x261-2x2131=92020-69660=22360

-541456=2^11 mod (3216x13^2)

3216 divides 51456

I arrived to this from

6^6-(331259x71-19179+2^9-1-222)/541=1=51457 mod 3216

so -541456=2^11 mod ((51456/16))

(541456+2^11)=2132 mod (359x13)

(541456+2^11-2132)/359/13=116

92020=116 mod 359

(541456+2^11)=92020=2 mod 331

(541456+(2^11-1)-2131)/(2131x3^2+2^9-1)=29

541456+69660=611116

611116x71 mod (359x13)=137

19179-511=0 mod (359x13)
19179-511-137=19179-648=0 mod 71 and mod 261

611116=-261 mod (359x13)

331259=261 mod 359

(19179-648+611116*71) is divisible by 359x13
(19179-648+611116*71)+1 is divisible by 394 and pg(394) is prime

(611116*71+19179-648)/(331259+359-261) is an integer

541456*71=6^4+1 mod 4667

2131=72 mod 2059=29x71

19179=648 mod (29x71)

541456x71-6^4-1=359x13x8237

8237=1 mod (29x71)

-611116-331259=-6^6-14=69660-14 mod 359

-(541456+69660)-331259=-6^6-14=69660-14 mod 359

71x3^3+1=1918 divides (611116+331259-6^6-14-19179)

(611116+331259)/359=2625=2626-1

239239+92020=331259

239239=-214 mod 359

430x239239=-92020 mod 359

the inverse of 430 mod 359 is 268

(359x499) divides both (92020x268+239239) and (611116+331259-6^6-14)=(541456+69660+92020+239239-6^6-14)

19179+1 is a multiple of 137
19179=511=648-137 mod (359x13)

-19179-1=30003 mod (359x137)

3x6^3-261=387

92020=387 mod (2131x43)

(331259-257)*9/(23004-19179+648)=666

((331259-257)-4-92016)/331=722=2x19^2

23004-19179+648=4473

4472 divides (92020-69660)

331259=92020=5 mod 2629

(2629*5+1) divides (92020+2)

it turns out that 331259 has this curious representation:

(6^6+666)x7+5=331259

(92022=92015+7)/(6^6+666-239239/7+1)=7

2629x7+1=18404

429^2-1=184040

239239 is a multiple of 7 and 2629

331259=92020+239239

429^2=10^2+1 mod (2629x35)

92015 is divisible by 2629x35

(331259*2-2629*7-3)*9/4473=6^4

(2629x7+3)/2=9203=331259 mod 23004

(541456+13-331259=210210=-4472 mod (359x13))
4472 divides (92020-69660)

541456+13-449449=92020

from above

541456-331259=31x449-18404 mod (359x13)

so

(92020-331259)=-970x449+13-18404 mod 4667

-239239-13=-970x449-18404 mod 4667

970x449=12x18404 mod 4667

or

3168=-12x18404 mod 4667

multiplying by 5

3168x5=-12x92020 mod 4667

210210=-4472 mod 4667

1051050=-(92020-69660) mod 4667

16170=-344 mod 4667

344 divides 541456

541456+13-210210=331259

mod 359x13 infact

541456=84
84+13+4472-4667=98

331259=-98=261 mod 359

i think there is something more subtle but it's too hard to understand for me.

manipulating a bit the above equations i obtained

-98=7000-1400x(92020-69660)=331259 mod 4667

but this is:

-98=1400x(-5x7x11x239+69660)=331259 mod 4667

so it seems to pop up 7x11x239 whcih divides 239239=331259-92020

bringing 5 out

-98=7000x(-18403+13932)=331259 mod 4667

-98=-4471x7000=331259 mod 4667

-98=2333x(-18403+13932)=331259 mod 4667

359-98=261

541456-13x16169=331259
541456-13x16169-13x18403=92020

16169,13932, 18403 are not random

maybe this is the reason why 541456=84 mod (359x13)

84=98-14

and 541456+69660=61116=98 mod (359x13)

follows that

(18403-13932) which is 4471 =-14^2 mod (359x13)

this means that
6^12-1=-4472 mod (359x13)

16169-13932=2237=4472/2+1

14^2+14=210

-210210=4472 mod (359x13)

-(14^2+14)*1001=4472 mod (359x13)

-14^2*1001=4472+13 mod (359x13)

from here I think you can derive why 541456+13-210210=331259

mod (359x13) +13-210210=4472+13

359*13*43-196*1001-13=4472

this maybe is the reason why 541456=84=98-14 mod (359x13)

and -(69660+541456)=-611116=261 mod (359x13)

261=359-98

541456=-6^7=84 mod (359x13)

14^2=196

69660=6^2x(2131-14^2)=6^2x(44^2-1)

maybe not a achance???

4x19179-84^2=69660

541456^2=4x19179-69660 mod (359x13)

541456^2+1=239x10 mod (359x13)

6^5=-1 mod 77

541456=-8 mod 77

331259=5=92020 mod 77

5+8=13

i guess that there is something to do with the fact that 541456=331259+210210-13
92020=331259-239239

so

331259=5=92020=(541456+13)=-5x6^5 mod 77

5x6^5=-77 mod 239

i think that here is the key

541456=11^2 mod 239

541456=-8 mod 77

541456 is of the form 6^5-7+18403s

92020 and 331259 are of the form 92020+18403s

(92020-6^5+6)=3370x5^2

pg(3371) is prime

probably i got something

-92020+6^5=2^5-1 mod 3371

this implies

92020=88^2+1 mod 3371

(92020-5)=239x7x11x5=88^2-4 mod 3371

I remember that 69660=6^2x(44^2-1)

so some conclusions can be done

92015x9=69660 mod 3371

92015=71x6^4-1=239x7x11x5

18403=1548 mod 3371

69660=(44^2-1)*36 Is congruenti to 18403 mod 3371

Dividing by 45

18403 congruenti to 43*6^2 mod 3371

18403*18=331259-5 Is congruenti to 43*6^2*18 mod (3371*18)

It utns out that

331259=29x31^2 mod (3371*18)

2x139x331=1001 mod 3371

2x139x331=92020-2

92018=14=331259 mod 51 pg(51) is prime

the inverse mod 51 of 14 is 11

-541456=11 mod 51

92018=14=331259=-449449 mod 51

541456+13-449449=92020

92018=14=331259=10x1001 mod 51

(92018-10x1001) is divisible by 1608

1608 divides 51456 and pg(51456) is prime

-(331259+72)=-331331=16=92020 mod 51

-92092=-92020-72=14=331259=-449x1001 mod 51

449=92=41 mod 51

239=-16=-92020=331331

92020+239 Is also divisibile by 67 PG(67) Is prime After PG(51)

331331-239 Is also divisibile by 541 and 36

i notice that (92020-16) is divisible both by 51 and by 451

pg(51) and pg(451) are primes

by the way famous 23004=69660-6^6 is congruent to 3 mod (451x51)

curious fact

(92020-4^2) is divisible by 4, 51, 451

pg(4), pg(51), pg(451) are primes

pg(215) and pg(541456) pg(2131) are primes

-541456=215= +2131 mod 51

(541456-2131) is divisible by (239+16=255=2^8-1) and by 2115=46^2-1

so 541456=(46^2-1)x(16^2-1)+2131

19179=2131x3^2=-51456 mod (51x5)

541456=2131x3^2x28+4444

239x1001=239239=-2^9=-2 mod 51

-92020x1001=-92020x32=2^9=2 mod 51

so 92020=16 mod 51

(239239+2^9-1)=5^3x(71x3^3+1)

71x3^3+1 divides 19180=2131x3^2+1

23004=3x7667+3

7667 is palindromic in base 10 and 6

(92020-16) is divisible by 451*12
(541456-16^2) is divisible by 451*12

4472 divides (92020-69660)

-4472=16=92020 mod (51x11)

215=22360/2=-541456 mod 51

22360=92020-69660

69660=3 mod 107

92020=0 mod 107

69660-3-23005 is divisible by (108^2-1)

23004=69660-6^6=3 mod (451x51)

so (69660-3) (multiple of 107) -6^6=0 mod (451x51)

69660=3x6^3 mod 23004

69660-3x6^3=9 mod (451x51x3=23001)

69660=3 mod 107
69660=0 mod 215

using chinese remainder theorem

69660 is a number of the form 645+23005k

if k=-1

645-23005=22360=(92020-69660)

also 6^6-1 is a number of this form

I notice the incredible fact that (69660-3)=651x107 where 651 is the product of the first 3 Mersenne primes. 651=3x6^3+3

6^6-19179=0 mod (387x71)

387 divides 69660

I think that something in some field is at work...surely 23004Z has something to do

i don't know if it is even possible for a human beeing to conceive a theory for these numbers

I think that a possible clue could be

18^2=69660=18^2x215=3 mod 107

so for example I notice that

22360=-18^2 mod (106x107)

22360=3 mod 107
22360=4 mod 108

I could think that this has something to do with the fact that 6^6=4 mod (108^2-1) and with the fact that (69660-9) is a multiple of 71 and 109

23008=3 mod 107
23008=4 mod 108

23008-3=23005
23008-4=23004

69660-428 (428 divides 92020) is a number congruent to 3 mod 107 and to 4 mod 108
(69660-428-3)=107x647=107x(3x6^3-1)

107 and 23005 are number of the form 11449+11556k

69660=-6^2 mod 264^2

264 is multiple of 44

69660=(44^2-1)x36

i think that using Lagrange or some primitive root concept one can get something

((139*(47+71*5)-1))=71x787

47 is the order mod 71 that is the least integer such that 139xn=1 mod 71

I suspect that this has something to do with the fact that 787 divides 541456

curious fact

92020=71x6^4+4

331259=92020+239*7*11*13

7,11 and 13 are primitive roots mod 71

-92020=331 mod (7x79)
-69660=18 mod (7x79)
-331259=541 mod (7x79)

(541-331)=210

(541456+13-210210=331259)

-239239=210 mod (7x79) this is equivalent to 239239=7^3 mod (7x79)

-541456=22^2 mod (7^3x79)

210x1001-1=-22^2=541456 mod (7x79)

playing around with this modulus (7x79) which is not random I got

7^3+12+210x1001=-2^7=541456+7^3+13

541456+11+210x1001=-2^7 mod (7x79)

from here

because 541456=(7^3+1)x1574

-140=(7^3+1)x1575

dividing both sides by 7 and by 5

-2^2=(7^3+1)x45 mod (79x7)

from here I got

69660x(7^3+1)=444 mod (79x7)

this reduces to:

-1=86x45 mod (79x7) where 86x45=3870 which divides 69660

i think that this has something to do with the fact that 69660-19179 is a multiple of 79

Last fiddled with by enzocreti on 2022-05-26 at 13:54

All times are UTC. The time now is 21:05.

Thu May 26 21:05:49 UTC 2022 up 42 days, 19:07, 1 user, load averages: 1.77, 1.76, 1.73