Thread: 69660 and 92020
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Old 2022-03-06, 12:54   #15
enzocreti
 
Mar 2018

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Quote:
Originally Posted by enzocreti View Post
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


curious that 71 239 359 have 7 as smallest primitive root


another curio about these crazy numbers:

pg(451=11x41) is prime

pg(2131) is prime

451+41^2-1=2131

this could be connected with the fact that it seems to exist infinitely many pg(k) primes with k multiple of 41. pg(181015) for exampe is prime and 181015 is a multiple of 41

it seems that there are infinitely many pg(k) primes with k multiple of 41 and infinitely many with k multiple of 43.

When k is multiple of 43, then k is of the form 41x43xk+r


manipulating a bit the previous things I got

(92020-16) is divisible by (71x108-1)=7667 a palindrome
7667=11x41x17

pg(451) and pg(17x3) are primes

71*108*12-1=92015=239x5x11...

(92020-6) is divisible by (71*108*6-1)

(92020-10) is divisible by (71x108x2-1)

pg(19179=2131x9) is prime

19180 is divisible by (71x27x2+2)

181015 and 92020 are numbers of the form 51s+16

92020 is 0 mod 43

using crt

92020 has the form 2107+2193s

allowing negative s

-4472 is a number of the form 2107+2193s

4472 divides (92020-69660)

4472=-16=-92020 mod (51x11x4)

i wonder if there are infinitely many primes pg(k) with k of the form 16+51s

pg(67), pg(92020), pg(181015) are primes with k of the form 16+51s

are there infinitely many pg(k) primes with k of the form 14+51s?

pg(79) and pg(331259) are primes and k is of the form 14+51s




331259=-13 mod 18404
239239=-13 mod 18404
541456=7740 mod 18404


7740 divides 69660


541456x9=69660 mod (18404x261)


69660=14448 mod 18404


(69660-14448)*6-13=331259


an extension of the conjecture could be:


there are infinitely many primes pg(k) with k of the form +/- 14+51s.
394 for example is of the form -14+51s




pg(181015=16+51s) is prime


181015=-1 mod 22^3 curious




curious that also 67=51+16 is congruent to 1 mod 11 pg(67) is prime


11 is one og the factors of 451


92020=16+51s


92020-16 is divisible by 11


92020 is even , 67 and 181015 are odd


(92020-16)=-11 mod 239x5x7 so 92020=5 mod 239x5x7


331259=5 mod 11




92020=5 mod 11


92020/5=18404


18404=1 mod (239x11...)




-181015=(5x11)^2+1 mod (429^2)


92020=(429^2-1)/2


i think that this could explains something

mod (429^2-1)/2 for example 92020=0




-181015=55^2 mod (429^2-1)


this is equivalent to


-181015=(5x11)^2 mod 92020




181015 and 92020 have the same form 16+51s




181015, 67, 92020 are of the form 16+51s




-181015=(5x11)^2 mod 92020
-92020=0 mod 92020
-67=71x(6^4+1) mod 92020




curious that


181015, 67 and 92020 (with the form 16+51s) are congruent to +/- j^2 mod 71


67 and 92020 are +/- 4 mod 71


181015=6^2 mod 71




-71x(6^4+1)=5=92020=331259 mod 11


-181016=(5x11)^2 mod (429^2)


i think that here is the rub....


-181016, 5x11 and 429 have 11^2 as divisor


so you can divide by 11




it turns out thst


-1496=5^2 mod 39^2


181015=16 mod 51 and mod 39^2


39^2*7+1=22^2




67 92020 and 181015 are of the form 16+51s


residues mod 11 and mod 17 and mod 13 are not random I think


as you can see


-67=-1 mod 17 and -67=1 mod 11 67=2 mod 13






92020=4^2 mod 41

-92020=5^2 mod 41

-92020=5^2 mod 449

541456+13-449449=92020

541456=-5^2-13 mod 449

inverse of 25 mod 449 is 18

92020x18=-1 mod 449

92020x18*(1/5)=331272

331272-13=331259

92020*(1/5)=18404

18x18404-13=331259

becasue 18404=1 mod (239x7x11)

18x18404-13-5 is a multiple of (239x7x11)


429^2x18=16 mod (449x17)

92020=(429^2-1)/2

92020=16 mod 17


331259=18404*(5+13)-13

Mod 449


18404*5=-25 mod 449
18404*13-13=239239

331259-239239=92020


-18404*13=65=69660 mod 449


-331259-65=-541456=--331259-69660 mod 449


92020x18=-1 mod 449

this is the starting point
331259*5-92020*18=65

69660=65 mod 449

92020x18-13x5=0 mod 331259

5x(18404x18-13)=0 mod 331259


(331259+13+65)x18=-1 mod 449

331259+13=0 mod 18404x18

-(331259+13+65)=5^2=-92020 mod 449

92020, 331259 and 541456 are congruent to -25-13s mod 449 for some nonngeative s

331259+13=-90 mod 18409=41x449

90-65=25


69660=541456-331259 mod 449



18404x18=359 mod 449
-18404x18=90 mod 449

18404x18-13=331259

-18404=5 mod 449

331259=-(449-359)-13 mod 449
pg(359) is prime

69660=(449-359)-5^2 mod 449


331259=(359-13) mod 449

331259+13=0 mod 18404

-18404=5 mod (449)

so 92020=5x18404=-5x5 mod (449)

-18404x18=90=-359 mod 449

331259=18404x18-13

so 331259=-90-13=-103=(359-13) mod 449

5x359=-1 mod 449

92020=-(1/359^2) mod 449

so 92020=-25 mod 449

92020=-1/18=-(1/359^2)=-25 mod 449


359^2x18404=-90 mod 449
18x18404=-90 mod 449

90^2x18404=-90 mod 449
(1/25)x18404=-90 mod 449
90 is the inverse of 5 mof 449

331259-90-13=0 mod 449

-5x18x18=(331259+13)/5^2=359/5^2 mod 449


-5x18=331259+13=359 mod 449

5x18x18=-(331259+13)/5^2=90/5^2 mod 449

331259=-90-13=-103 mod 449

92020=-(1/359^2)=(-1/90^2)=-25 mod 449





18404x90=-1 mod 449

the invers mod 449 of 5 is 90

18404x90x90=-90 mod 449
this means

331272=-90=359 mod 449

331259=-103 mod 449

331272x444=1 mod 449
331272x(-5)=1 mod 449

331272x(-5)=-92020x18

from here follows necause inverse of 18 mod 449=25

92020=-25 mod 449

but the question I think is more subtle than I think


429^2=-7^2 mod 449

541456+13=-7^2-1-(429^2-1)/2

429^2=-7^2 mod 449

(429^2-1)/2=92020

mod 449

(429^2-1)/2=-5^2 mod (449)

541456+13=-5^2 mod 449

541456=-5^2-13 mod 449

from thsi follows that

541456+13-92020=0 mod 449

92020=-5^2 mod 449

18404x359=1 mod 449

92020=18404x5


359x5=-1 mod 449


-541456-13-239239=-331259=103 mod 449

-541456-13+78=-331259=103 mod 449

so

-541456+65=103 mod 449

103-65=38=13+25
541456=-25-13 mod 449
69660=65 mod 449

so

541456-331259=65=69660 mod 449


210210=-239239=78 mod 449

541456+13-210210=331259

331259-239239=92020

92020=359+65 mod 449

so 22360=92020-69660=359=331259+13 mod 449

the numbers are clearly structured, but unfortunally there is no elementary method to solve the puzzle of the giant mega-structure that generetes these primes.
Beeing structured, no surprise we do not find any prime of this type congruent to 6 mod 7.


exponets leading to such type of primes appear to assume only certain particular forms. This maybe obstrues the possibility of a 6 mod 7 prime of this form


look at this crazy curio:

427x428x429x430=1 mod 449
33712999320=427x428x429x430 is the concatenation of 3371 (pg(3371) is prime) and 2999320 which is divisible by 449

i think that with new tecnologies just for recreational purposes it would be worth to find other exponents leading to a prime of this type

429^2x394=1 mod 449

pg(394) is prime

becasue 429^2=92020x2+1 (pg(92020) is prime)

92020x2x394=-3x131 mod 449

92020=-3x131x300^2 mod (449x359)

i think that these exponents leading to a prime are connected to each other in a very deep and mysterious way


exist pg(K) primes with k multiple of 215 (3 found)


exist pg(k) primes with k multiple of 43 (4 found three of which are multiple of 215)


exist pg(k) primes with k multiple of 139 (2 found)


exist pg(k) primes with k of the form 16+51s (3 found)...


it seems clear that the exponents leading to a prime are not random at all.




Incredible:


pg(181015) is prime pg((429^2-1)/2=92020) is prime


181015=429^2-1-55^2!!!


429 and 55 have 11 as common divisor


11x(16731-275)-1=181015


181015=11^2x(39^2-5^2)-1




181015=92020=67=55^2=4^2 mod 51

Neme(k) this is the name of these numbers




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


1323=11=39699 mod 41


this is another case in which exponents leading to a prime seem to have a certain form, in this case 41s+11


39699=11 also mod 11x41=451


pg(6231) and pg(2131) are primes


6231 and 2131 have the form 41s+40



I notice that 1323 and 39699 have also the same residue 10 mod 13


so 1323 and 39699 have the form 257+533s


incredibly 1323 and 39699 are of the form 257+41x(t^2+1) for some t.


1323=257+41x(5^2+1)
39699=257+41x(31^2+1)


and remarkably 5^2 and 31^2=-1 mod 13

These numbers contain a lotq of surprises because they are structured...but the problem Is that only an alien of type 5 civilisation could solve this kind of problems I think that when Riemann hypotesis Will be solved this kind of problems Will be still open and for many other centuries

(449*41-6) divides both 92015 and 331254


A Little pompously I could call these numbers numbers for the end of the world or at least for the next geologic era

It's simpler to say that 1323 and 39699 are of the form 298+41xp^2


neme(176006) (or pg(176006) is prime)

(176006+2) is divisible by (92020-69660-359)

176006=-2=-451 mod 449

pg(451) is prime

also pg(2) or neme(2) is prime






pg(92020), pg(67) pg(51) and pg(451) are primes


92020-67=-51 mod 451


pg(181015) is prime

11^2*(39^2-5^2)-1=181015

11^2x39^2=429^2=2x92020+1

because (a^2-b^2)=(a+b)x(a-b)

11^2x(39+5)x(39-5)-1=181015


39+5=44 divides (92020-16)
39-5=34 divides (92020-16)


181015 has the curious representation 44x4114-1 with 4114 a palindromic number









92020x18=-1 mod 449

92020x18=17^2-1 mod (451x51x6^3)


92020 has the curious representation 828180/9

82-81-80...


mod 449

the inverse of 359 is 444

444=-5 mod 449

22360=-90 mod 449

90x5=1 mod 449

90x5+1=451

I think that probably there is a connection to the fact that:

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

331259=-14^2/2 mod 359

but I am not sure

a curious thing I noticed is this:

6231, 19179, 39699, 51456, 56238, 69660, 75894. Seven multiples of 3 in a row.


pg(19179=2131x3^2) is prime

69660=9 mod 71
19179=9 mod 71

(69660-19179) is a multiple of 79*3^2

pg(79) is prime

79*3^k =1 mod 71 for k=2+70xs

(69660-19179)/79+9=3x6^3=2^nx3^(n+1)

(7740-2131)/71=79

19179 and 69660 arebof the form -31302+50481s



6^6=-2 mod 41

69660=1 mod 41

69660-1-6^6-2=23001 divisibile by 51 and 451




69660=1=51456 mod 41 and mod 4551=111x41


92020=10^3 mod 41 and mod 4551=111x41




(69660-51456) divides (92020-10^3)


10^3 mod 41=16

6^6=-14=-69660 mod 359


77x6^6=-98x11=-69660x77 mod 359

77x98x6^6=-98=-69660x77x98=331259 mod 359

77x6^6+98*11=3593590

7x6^6=-98=-7x69660=331259 mod 359


7x6^6=-2x7^2=7^4+14=331259=7^4+69660=7^4-6^6 mod 359

I think that 6^6 69660 are the way to beat...
the role of 71 and 359 are mysterious.

I think that it needs an extremely complex tool for understanding these numbers.

PG(3336) and PG(75894) are primes wirh 3336 and 75894 multiple of 139

I suspect that somerhing Is happening in Z139

3336=1=75894 mod 29

75893/29=2617 which Is a wagstaff prime exponent.

I conjecture that there are infinitely many PG(3336+29*139s) primes.

It seems that exponents leasing to a PG prime can assume only certain forms


92020=2 mod 139
92020=3 mod 29
I think that there Is something in 139Z and 29Z a mysteriius force in these fields generaring the exponents


I think that there Is a connection wirh the fact that the modular multiplicative inverse of 71 mod 139 Is 47

71*47-1=3336


71x6^4+2=0 mod 139
71x6^4+2=1 mod 29


75894=-19X3 mod 87

3336=-19x3 mod 87

i think that gigantic groups are at work in these numbers, but it's very very hard to find our rosebud

Last fiddled with by enzocreti on 2022-09-26 at 13:31
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