mersenneforum.org  

Go Back   mersenneforum.org > Extra Stuff > Blogorrhea > enzocreti

Reply
 
Thread Tools
Old 2021-07-11, 19:07   #12
enzocreti
 
Mar 2018

2·269 Posts
Default ...

Quote:
Originally Posted by rudy235 View Post
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
enzocreti is offline   Reply With Quote
Old 2021-07-11, 20:35   #13
enzocreti
 
Mar 2018

2·269 Posts
Default

215*107*12^2 is congruent to -71*6^6 which is congruent to 72 mod (331*139)


331*139*8 is congruent to -8 mod (215*107)


s^2 is congruent to 1 mod 23005

the first not trivial solution is

s=429

the second is s=9201

the third is s=13374

(13374^2-1)/23005 is congruent to -1 mod 6^5


215*107*(2+331*139*8)-1 is a multiple of 331*139 and 429^2


mod 429^2 we have

23005*184033-1 which is a multiple of 429^2 and 71


((184033*23005-1)/71-429^2)/429^2=18^2-1



92020 is congruent to 4*(2+331*139*8)^(-1) mod 429^2 and mod (331*139) so


92020 is congruent to 4*(368074)^(-1) mod 429^2 and mod (331*139)

the inverse of 368074 so is 23005

92020 is congruent to 4*(429^2-2^2)^(-1) mod (331*139*429^2)


maybe it is useful


431=(427)^(-1) mod 46009???




331259 for example is congruent to -(9203*4+1) mod (331*139)


and 331259 is congruent to 9203 mod 23004




(92020)^(-1)=23005 mod (331*139)




92020*23005=(46010)^2




92021 divides 215*107*(2+331*139*4)-1


pg(69660), pg(19179) are primes

maybe something useful can be derived from this:

69660 is congruent to 19179 which is congruent to 429^2 which is congruent to 9 mod (71)

69660 and 19179 are of the form 648+213s

probably there are infinitely many pg(648+213s) which are primes


in particular

6^6 is congruent to 19179 which is congruent to 429^2 which is congruent to 861 mod (71*43)

6^6 is congruent to 860 mod (214^2)


92020 is congruent to -2^0 mod (17*5413)

92020 is congruent to -2^1 mod (3*313*7)

92020 is congruent to -2^2 mod 11503

23005*(2+331*139*2^0)-1 is a multiple of 11503
23005*(2+331*139*2^1)-1 is a multiple of 3*313*7
23005*(2+331*139*2^2)-1 is a multiple of 17*5413


The inverse of 9203 mod (331*139) is x

9203*x is congruent to 1 mod 429^2


331259 is congruent to 9203 mod 23004


23005 Is binomiale(214+1,2) so 23005 Is the 214th triangolare Number


92020=(858^2*139*331+4)/(2+331*139*8)


3371 and 331259 are primes pg(3371) and pg(331259) are primes

3371 and 331259 leave the same remainder 59 mod 3312

3371 and 331259 are primes of the form 59+ ((71*6^6-24^2)/10^3)*10^m, for some nonnegative integer m

(46009x-1)/(y^3-1)+y^3=(x^2-1)/2 over positive integers


x=429 y=6




(23005*(2+46009*(4))-1)/92021=46009

215^2 is congruent to 46009 mod 216


11503=71*2*3^4+1


92020 is congruent to -4 mod 11503


69660 is congruen to 3*6^3 mod 11502

92020 is congruent to 2^2 mod 11502


92020*23005 is congruent to 4 mod (71*11503)




23005*(2+331*139*(8+184041*(8+184041*(8+184041... is congruent to 1 mod (429^n)

23005*(2+139*331) is congruent to (429^2-1)/10 mod 230051=31*41*181



92020 Is 4 mod 11502 and -4 mod 11503


92020 Is congruent to (71*6^6/11502=288=17^2-1) mod 323=18^2-1


6^6/162=288

162 divides 69660


(2*(46009*6)^2+144*(2+46009*2))/313/49/3-6^3=71*6^6


46009+2 Is a multiple of 313*49*3


2*46009+2=92020




69660=3/2*(6^6-6^3)


92020-69660=22360 which is divisible by 860

22360=860*(3^3-1)

Z46009 is isomorphic to Z331XZ139


429^2 Is congruent to 2^2 which Is congruent. To (139*331*8)^2 mod (431*427)

71*6^3 is congruent to -23005*(2+46009*2) mod (7^2*313)


71*107*6 is congruent to -430 mod 11503


69660 mod 11503=642=107*6

642=6*(3*6^2-1)=6*107


11476*2*860+428 is congruent to 429 mod (3*313*49)

Consider

71*6^6 is congruent to -72k mod j

for k=1 j=46009
for k=2 j=4601
for k=3 j=(313*7^2*3)
for k=4 j=11503
for k=7 j=9203

71*6^6 is congruent to -72*7 mod 9203

46008 (=331*139-1) is congruent to -7 mod 9203

331259 is congruent to -7^2 mod 9203

or 9203=331259-46008*7



(139*331)^2 is congruent to 1 mod (92020) and mod (23004)


69660 is congruent to 92020-(15229*(2+46009*2)-216)/313/49=648=3*6^3 mod (23004)

23005*(2+46009*2) is congruent to 1 mod (313*7^2)

71*6^6 is congruent to -6^3 mod (313*7^2)

23005*6^3 mod (313*7^2)=15229


(6^3/2)*(2+46009*2) is congruent to -6^3 mod (313*7^2)

69660*313*7^2 mod 23004=3*6^3

6^6-(69660*49*313-648)/23004=3*71




71*6^6 is congruent to 67 mod 139 and 259 mod 331


chinese remainder theorem to the rescue:


45937+46009k...allowing negative k, you have -92090 which is -2 mod 1001 and to 92020

331259 is 259 mod 331
331259 is 4588 mod 4601
MathCelebrity
START HERE
OUR STORY
VIDEOS
PODCAST
Upgrade to Math Mastery
Enter math problem or search term (algebra, 3+3, 90 mod 8)
Invia








Using the Chinese Remainder Theorem, solve the following system of modulo equations: x=259 mod 331 x=4588 mod 4601
Enter modulo statements
x=259 mod 331
x=4588 mod 4601


Using the Chinese Remainder Theorem, solve the following system of modulo equations
x ≡ 259 mod 331
x ≡ 4588 mod 4601

We first check to see if each ni is pairwise coprime
Take the GCF of 331 compared to the other numbers
Using our GCF Calculator, we see that GCF(331,4601) = 1

Since all 1 GCF calculation equal 1, the ni's are pairwise coprime, so we can use the regular formula for the CRT

Calculate the moduli product N
We do this by taking the product of each ni in each moduli equation above where x ≡ ai mod ni
N = n1 x n2
N = 331 x 4601
N = 1522931

Determine Equation Coefficients denoted as ci
ci = N
ni

Calculate c1
c1 = 1522931
331

c1 = 4601

Calculate c2
c2 = 1522931
4601

c2 = 331

Our equation becomes:
x = a1(c1y1) + a2(c2y2)
x = a1(4601y1) + a2(331y2)
Note: The ai piece is factored out for now and will be used down below

Use Euclid's Extended Algorithm to determine each yi
Using our equation 1 modulus of 331 and our coefficient c1 of 4601, calculate y1 in the equation below:
331x1 + 4601y1 = 1
Using the Euclid Extended Algorithm Calculator, we get our y1 = 10

Using our equation 2 modulus of 4601 and our coefficient c2 of 331, calculate y2 in the equation below:
4601x2 + 331y2 = 1
Using the Euclid Extended Algorithm Calculator, we get our y2 = -139

Plug in y values and solve our eqation
x = a1(4601y1) + a2(331y2)
x = 259 x 4601 x 10 + 4588 x 331 x -139
x = 11916590 - 211089292
x = -199172702

Now plug in -199172702 into our 2 modulus equations and confirm our answer
Equation 1:
-199172702 ≡ 259 mod 331
We see from our multiplication lesson that 331 x -601731 = -199172961
Adding our remainder of 259 to -199172961 gives us -199172702

Equation 2:
-199172702 ≡ 4588 mod 4601
We see from our multiplication lesson that 4601 x -43290 = -199177290
Adding our remainder of 4588 to -199177290 gives us -199172702

Share the knowledge!

Chinese Remainder Theorem Video


Tags:
equationmodulustheorem


Add This Calculator To Your Website
<!— Math Engine Widget Copyright MathCelebrity, LLC at www.mathcelebrity.com. Use is granted only if this statement and all links to www.mathcelebrity.com are maintained. --><a href="https://www.mathcelebrity.com/chinese.php" onclick="window.open('https://www.mathcelebrity.com/chinese.php?do=pop','Chinese Remainder Theorem Calculator','width=400,height=600,toolbar=no,menubar=no,scrollbars=yes,resizable=yes');return false;">Chinese Remainder Theorem Calculator</a>



Run Another Calculation





Email: donsevcik@gmail.com
Tel: 800-234-2933
MembershipMath AnxietyCPC PodcastHomework CoachMath GlossarySubjectsBaseball MathPrivacy PolicyCookie PolicyFriendsContact UsMath Teacher Jobs


331259/(2*12^2=288=17^2-1) is about 11502/10...


71*6^6 is congruent to - 12^2 mod 4601 and mod 774


331259=11502*(17^2-1)-9007*331


-9007*331 cogruent to 9203 congrue
nt to 331259 mod 11502

(9007*331+9203)/11502=259+1


331259*11502 is congruent to 4473*666 mod (23004*331)

4473=4472+1


maybe something useful can be derived by this:

139^(-1)=331 mod 23004


for example

331259*139 is - 9007 mod (1001*23004)



(6^6)^(-1)=22 mod 331

331259 is congruent to 22 mod 139



331259 is -72 mod 1001
92020 is -72 mod 1001

71*6^6 is -72 mod 331


331259, 92020 and 71*6^6 are numbers y that satisfy this congruence equation

y is congruent to (-72+331*(10^x-1)) mod 1001 for some nonnegative integer x


(71*6^6+72-331*999)/9-72=331259=(71*6^6+72-331*999-3*6^3)/9


(359+71)*(6^6+11502)/359=69660

pg(359) is prime

69660 is congruent to 14 mod 359
6^6 is congruent to -14 mod 359

-23004 is congruent to 331 mod 359


92020 and 331259 are 5 mod 239

92020 is congruent to -331259 which is congruent to -3^5 mod 257

92020+3^5=359*257


1001-((331259-243*2-92020)/257)=72


331259 and 92020 are -72 mod 1001


28 is congruent to (429^2-1) mod 257

-239239 is congruent to 28 mod 257


92020+239239=331259


14 is congruent to 129*(429^2-1) mod 257

so

107 is congruent to 69660*92020*2 mod 257

from this follows

331259 is congruent to -14 mod 257

69660 is 13 mod 257

92020 is 14 mod 257



92020 Is congruente to 1 mod 829 and mod 37

331259=92020+239239 Is congeuent to - 3*2^11 mod 829 and mod 37


541456 Is congeuent to -2 mod 37 (and also 331259 Is -2 mod 37)


541456+3*2^11 Is a perfect square


PG(359) Is prime

331259*5 Is congruente (23004+331=multiple of 359)=4667 mod 6^6


71*6^6-(331259*5-(23004+331))=6^8


541456=43*(10^4+2^5*3^4)
280 Is congeuent to -2592=2^5*3^4 mod 359

69660 Is 28/2 mod 359 23004 Is 28 or -331 mod 359

The inverse of 10 mod 359 Is 36

69660*10^3 Is -1 mod 359 so 69660 Is -6^6 mod 359

-6^6 =14 =69660=-2592*18 mod 359

From here

3870=-2592 mod (359x18)

Dividing by18


215=-12^2 mod 359

12^2=(71^2-1)/(6^2-1) mod 359

So (6^3-1)=-(71^2-1)/(6^2-1) mod 359



PG(541456) PG(331259) and PG(92020) are primes

541456 92020 and 331259 are Numbers of the form

-a+1001*s where a is a Number congruente to 7 mod 13
a=72 and a=85

Because a=13d+7 and 1001=7*11*13

541456 92020 and 331259 are of the form -13d-7(1-143f) for some dnand f

So

(541456+7)/13 Is 71 mod 77

(92020+7)/13 and (331259+7)/13 are 72 mod 77


((X^2-1)*(46009+1/4)-1)/46009-x^2=0

This Is a parabola for x=+ or - 429 this goes to zero...

429^2-1=2*92020

I dont know of from that equation One can derive something more general

parabola
focus | (0, -33870353513/736148)≈(0, -46010.2)
vertex | (0, -184041/4) = (0, -46010.3)
semi-axis length | 1/184037≈5.43369×10^-6
focal parameter | 2/184037≈0.0000108674
eccentricity | 1
directrix | y = -33870353521/736148


This Is the parabola

((X^2-1)*(46009+1/4)-1)/46009=y


429^2 Is congruente to 6^6 which Is congruente to 69660 which Is congruente to 9 mod 71


(429^2-9)/71=2592=2^5*3^4

Curious that 541456 Is divisible by (10^4+2592)

43*2592 Is 111456...the last digits 1456 are the same as in 541456

92020/2592/5-1/3240=71/10

324 divides 69660....i think there Is something involving 18^2

2*92020/2592-1/324=71

(429^2-1)/2592-1/324=71


1/69660=(1/215)*((184040/2592-71))



2592=72^2/2


215/10*(20000+72^2)=541456


71*6^6/331259+1/(239*99) Is about 10...

1/(1/(71*6^6/331259-10)/99+239)=-277.199999...


The inverse of 5 mod 46009 Is 9202


429^2-5 Is a multiple of 46009


(429^2-5) Is then congruent to 6 mod (239*7*11)

92020 Is 10 mod 3067

239239 Is 13 mod 3067

So 331259=92020+239239 Is 23 mod 3067

71*6^6 Is 6^3 mod 3067...








71*6^6=429^2=3=-239239=3*6^4 mod 37


92020 for example =1 mod (37*3*829)


429^2=3 mod (37*829)




331259=92020+239*1001


so


331259 is -2 mod 37

331259 is congruent to 9203 mod 23004

9203 mod 71=44

(331259-44)/71=4665

4665 are the first four digits of 6^6=46656

4665 in base 6 is 33333 a repdigit


331259 is congruent to 9203 mod 23004

9203 is 5 mod 7
331259 is 5 mod 7

9203 is 44 mod 71
331259 is 44 mod 71

so 331259 and 9203 are numbers of the form 19143+497k

curious that 19143+6^2=19179 and pg(19179) is prime

curious that allowing negative numbers k -1234 is a number of the form 19143+497k


9203 and 331259 are also congruent to 131 mod 648=3*6^3

so using CRT they are numbers of the form 9203+322056k

71*6^6 is congruent to -(9203-131)/648 mod 331259


(331259-131)/648=2^9-1

71*6^6/648-4601=2^9-1

4601 divides 92020

Numbers of the form 512, 5112, 511...12,...

The difference 5112-512, 51112-5112,...Is a multiple of 46

(331259-9203)/5112=2^6-1

331259/5112 is about 64,8...=648/10

71*6^6=5112*3*6^3

331259/648=511,20216...

1/216=46/10^4+1/(10*15^3)


from above

370=92020X648 mod 511

370=92020X137 mod 511

138010=92020 mod 511


6^6=(138010-92020) mod 666

(20*71*6^6-4601*648*20)/511=10*6^4

370=92020X648 mod 511

370=40*648=10*2^5*3^4 mod 511

138010=92020 mod 511
69005=46010 mod 511
pg(69660) is prime

69660-69005 is a multiple of 131

pg(331259) is prime

331259=(6^2-1) mod 13801

370=92020X((69660-9-155)/511+1)=92020X137=92020X(70007)x511^(-1) mod 511^2

155 is 6^6 reduced mod 511

so

370X511=92020X70007 mod 511^2....92020 reduced mod 511 is 40

(40*70007-370*511)/511^2=10



9203-131=7*6^4

5112=71*72


(331259-131)/14-6^4+4=22360=92020-69660


92020+(6^4-4)=0 mod 6^6


9203=5=331259 mod 14



774*(1301-131)/13=69660

71*6^6=-14 mod 331259

71*6^6=-15 mod 43


(71*6^6+15)=4472 mod (43*107)

4472 divides (92020-69660)

(71*6^6+15-4472)/43/107=719

719=-1 mod 72

The inverse of 15 mod 4601 Is 1227
15*1227=18404

18404/2=9202

71*6^6 Is also =-15 mod 129

4472+129=4601

maybe It Is for that readon that

71*6^6=-144=-129-15 mod (4601*5)


-15*1227=(331259+13)/2 mod 429^2


19179=2131*3^2=-6=-429^2=-71*6^6=331259*6=9 mod 15


from 23005*(2+331*139*8)=1 mod (429^2*331*139) we obtain:

23005X2X431X7X61=1 mod (429^2*331*139)

7X61=427
so
431X7X61 is the factorization of (429^2-2^2)=(429+2)x(429-2)

71*6^6=-90 mod (431X7X61)

curious that

-69660=-90X774=71X6^6*774=(7^3-2)x7^3 mod 431

69660+(7^3-2)x7^3=431X433=432^2-1

pg(92020), pg(331259) are probable primes

92020=2^5=215=71x6^6=-331259 mod 61

69660=-2 mod 61 i think it is not chance


there are two probable primes pg(56238) and pg(75894) where 56238 and 75894 are multiple of 546
75894=139(again this 139!)*546 and the other 56238=103*546

103=139-6^2


71*6^6*429=-lcm(429,546)=-6006 mod 331259

69660=-2 mod 61

46009X8=-2 mod 61 (23005X(2+46009X8))=1 mod 429^2

so

69660=46009X8 mod 61



71*6^6=-6^3 mod (46011)

69660=2^8X3^9 mod (46011)

92020=-2 mod 46011


71*6^6=259=331259 mod 331

71*6^6=-72 mod (46009=331*139)

331259=9203 mod (23004)

331259=(9203-7) mod (46009x7)

331259=(9196+7) mod 23004x7

331259=(9196-7) mod (4601*7)

4601 divides 92020 and 4601x5=23005


(331259-9196)=7*139*331

331x7=2317, whose last digits are 317

71*6^6-331259=2981317, whose last three digits are 317


71*6^6-331259-(331*7)=331*3^2*10^3


331259-9196 is a multiple of 331*139


pg(69660) and pg(2131*3^2=19179) are primes, pg(92020) is prime

69660=19179 mod 639

(69660-19179)=4472 mod 46009
4472 divides (92020-69660)

curious that pg(2131), pg(19179=2131*9) and pg(69660=19179 mod 213) have this property: pg(2131) is the 19-th pg prime, pg(19179) is the (19+4)=23th pg prime and pg(69660) is the (19+4+4)=27th pg prime


we know that

23005*(2+46009*8)=1 mod 429^2
i noticed that

23005*(2+46009*8)=-1 mod 359

pg(359) is prime

it is not clear yet but maybe there is a connection to the fact that 6^6=-14 mod (359x13)

infact 17x13^2=1 mod 359

23005*(2+46009*8)=359*17^(-1)-1 mod 359^2

follows

29*(2+97)=359x17^(-1)-1 mod 359
99=(359*13^2-1)x260 mod 359

331259=-98 mod (359x71x13)

331259=(260-1) mod 331 by the way

331259=-46009x8-1 mod 359

331259=-(260x358^(-1)-1) mod 359
because 358x(10^2-1)=260 mod 359

71x6^6=83 mod 359

358x(10^2-1)=1 mod 83


so 331259=-98 mod (71*13*359)

71*6^6=-98-7x2^7=-994 mod (71*13*359)

71*6^6=-14*71 mod 359


331259=-98=+7*6^6 mod 13x359

359-99=260

331259=261 mod 359

99*(331-1)=1 mod (359x13)

331*99=1 mod (2^15)

331259=(1-2^15) mod (359x13)

7*6^6+2^15-1=359*1001 so 71x6^6=1-2^15 mod (359x13)

2^15=-1 mod 331
((2^(15+330*(1+138*k))+1)) is a multiple of 139x331


(71*6^6+2^15-1-359359)=12^6


12^6=(71*6^6-331259) mod (359x13)

12^6=-7*2^7 mod (359x13)

(71*6^6-331259)=-7x2^7 mod (359x13x71)


6^6=-14 mod 359 pg(359) is prime
71*6^6=-14 mod 331259 pg(331259) is prime
the difference 71*6^6-6^6=2^7x3^6x(6^2-1) where 2^7x3^6 is a 3 smooth number 3^(n+1)x2^n. 92020=2^7x3^6-(6^4-4)


(6^4-4)=215=6^3-1 mod 359

2^7x3^6=331 mod 359

331+215=546 and it is curious (but I think it is not a chance) that there are two pg(k) probable primes ( and perhaps infinitely many) with k multiple of 546

the multiplicative inverse mod 359 of 3^6 is 98

331259=-98 mod 359

so 331259=-(3^6)^(-1) mod 359

I think that there is a giant structure under these exponents but it is so complicated that no simple tool can shed even the slightest light on it

after few calculaions I found:

331259=-98=-(123*111)^(-1) mod 359


form here I could find that

3^6x324=333 mod 359

and so

333x98=324=18^2 mod 359

and

69660=324*215=333*98*215=14=-6^6 mod (359)


331259=-98 mod 359

69660=6^6 mod 23004 and -6^6 mod 359 infact

prime 359 is generating something, but I have no tools except some modular procedure to catch something

I could notice that

6^6=-14 mod (359x13)

331259=98 mod (359x13)


(69660-14)/359-(6^6+14)/359=2^6

infact 23004=28 mod 359 or equivalently 23004=-331 mod 359

in particular 23004=-331 mod (359x13)

and -23004=6^2 mod (2^6)

(331259+98)/359-(69660-14)/359=3^6

(69660-14)/359-(6^6+14)/359=2^6

some other possible ideas:

3*6^3=-70=17^2 mod 359

648x72=6^6=-14=17^2*72=-69660 mod 359

17*13^2=1 mod 359


92020=-3^5 mod 359 but also mod 257

331259=-(-3^6)^(-1) mod 359
and 331259=3^5 mod 257

257-14=3^5

i think that there should be an explanation why this number 14 appears so often

6^6=-14 mod 359
69660=14 mod 359
71x6^6=-14 mod 331259




i think that mod (23x5x3) something interesting can be found


so for example


331259=59 mod (23x5x3)


-3371=79 mod (23x5x3) pg(79) is prime


-359=331 mod (23x5x3) pg(359) is prime ...


but these are just ideas they have to be developed




mod 69 for example


331259=-79=-10=3371 mod 69


pg(331259) pg(79) pg(3371) are primes...


pg(359) is prime


359=-55 mod (23x3)


331 reduced mod 69 is 55




I could conjecture that there are infinitely many pg(k) primes with k=+/- 10 mod 69 and when it happens k is prime


-79=59=331259=3371 mod 138

i think this could be connected in some way to the fact that

ord (71*6^k) mod 23 =6

71*6^6 infact=1 mod 23

curious that 71x6^6=83 mod 359

and 359=83 mod 138


331259-3371 is divisible by 138 and by 6^3


71*6^6=(331259-59)/13800=24 mod 138


6^6+14 is a multiple of 359

69660-14 is a multiple of 359


6^6+15 is a multiple of 331x3

6966-15 is a multiple of 331x3

6966=69660/10


curious that 69660+6^6=-14^2 mod 331

6^6=-14=-69660 mod 359



The inverse of 14 mod 331259 Is a multiple of 359...why?



92020=5=331259 mod (239x7x11x13)

-92020=29=331259 mod (61x3)


648=-14 mod 331

the inverse mod 331 of 648 is 71

-(6^6+14)=1 mod 331 (6^6+14) is a multiple of 359

-(69660-14)=(14^2-1) mod 331 (69660-14) is a multiple of 359


pg(69660) and pg(19179=2131*3^2) are primes

69660=19179=-18 mod 79



69660=9=19179 mod 71

curious that 69660=-18 mod 79 and mod (21^2)

pg(3*21^2=1323) is prime and also pg(79) is prime

on the other hand 19179=6^3=-15^2 mod 21^2


(69660+18)/79+21^2=1323


From 23004=-331 mod (359x13x5)

I derived 23004x141=-1 mod (359x13x5)

10011x18^2=-1 mod (359x13x5)

18^2 divides 69660


10011^(-1)=23011 mod (359x13x5)


Curious that pg(10011/3-1=3336) Is prime


-3336=19999 mod (359x5x13)


14*10^3=-1 mod (359*13)

6^6=-14 mod (359*13)


The inverse of 1000 mod 359 Is 345=359-14


From

(10^4+1667)x3335=10^4 mod (359x13x5)

I arrived After some steps

-10^4x5x667+8191x5=-10^4 mod 359

And so


3810+8191=-2*10^3 mod 359


71*6^6=-14 mod 331259

71x6^6x359x725=-1 mod 331259


From this follows

70984x71x6^6=1 mod 331259

70984=261 mod 359

331259=70984=261 mod 359

In particolare (70984-261) Is divisible by 359 and (14^2+1)


so 70984=-14^(-1) mod 331259
70984=261 mod 359

so you can apply CRT here

and find the form of 70984

70984x5=92020=331259 mod (239x11)



Curious that pg(1323) PG(69660) are primes 1323=-1 mod 331

69660=150=70984 mod 331 and 70984-69660=1324 PG(1323) Is prime




92020-69660=22360

(92020=4) mod 71 (92016=71*6^4). 92016-69660=22356


-331259=22356 mod (197*359)


22356=18^2*69

197*359 Is 70984-261

curoius that -331259=22356 mod (359*197) and -331259=22357 mod 139

maybe something to do with 46009=331*139???

maybe...(331259+22357)=0 mod 139 (331259+22357)=108 mod 331

i have no tools anyway no advanced skills to make progress


(331259+22357)=7^3 mod 541


probably this is connected to the fact that 541456=-85 mod (541x1001)




(331259+22356)/359=444 mod 541

(116315+1)/359=18^2

maybe 331259+22356=107 mod 331 is connected in sojme way to the fact that 23005 is divisible by 107

maybe there is something in F46009 and F23005


331259+22356=-107 mod (3337x106)


pg(3336) is prime


3337=47x71


216x104=2x11x111=22356+108=22360+104=-331258
mod 3337

Curious that pg(75894) Is prime and -75894=(2^16+16) mod (359x197)




I notice that 22356=-23 mod (23x139)...


i think that something very complicated is under these exponents...
331259=59+23k


3371=59+23k pg(3371) and pg(331259) are primes...the numbers 23 and 139 are in some way involved but I have no idea how it happens


69660=648 mod 972

22356=-23 mod 973=139x7


i think that something very very hard to understand is happening in Z139 and in Z107


69660=648 mod 23004=71x972

2^2x3^5=-1 mod (139x7)

22356=0 mod 972 (22356=972x23)

973 divides 75894 and pg(75894) is prime


69660=16=239239=71x6^4 mod 23

(239239-71*6^4)/23-1=80^2 by the way

92020=69660+239239


22360=92020-69660=4 mod 23

4 is the square root of 16 by the way


(239239-16)/23+3=102^2 by the way

22360^2=16=69660=71x6^4=239239 mod 23


PG(359) Is prime PG(3336) Is prime PG(92020) are primes

92020=-(3336-359) mod 331






92020=21297 mod (359*197)


21297=21296+1
21297=11*44^2+1


44^2-1=1935 divides 69660






mod (359x197=359x(14^2+1)) I can see:


239239=27070 mod (359x197)


27070=541456/20-14^2/70


331259=92020+239239


(6^3-1)*(6^2-1)+1 divides (331259+22356+107) which is also divisible by 3337


pg(3336) is prime


this because 215x35=-1 mod (2x53x71)


(22356+107+3337)+331259 is a multiple of 71x107

215x106=-1 mod (71x107)

maybe is not chance that 22356+107+331259+3337=-1 mod 541


(107+22356-3337+331259)/10011=6^2-1

(107+22356-3337+331259)=1 mod 359



22357=22360-3 is a multiple of 139

(331259+22357)/106=3336 and pg(3336) is prime

22357=-8=79 mod 71

from here I have (because 9 is the multiplicative inverse of 79 mod 71)

201213=-72=-711 mod 71

in particular

201213=-711 mod (71x79)

but 71x79 =5609 divides (69660-19179) where 69660=9=19179 mod 71




201213+711=71x79x6^2


I don't understand why powers of 6 are involved in these numbers!


pg(3336) and pg(75894) are primes and 3336 and 75894 are multiple of 139

it holds:

-(75894-3336)=4=92020 mod 71

curiously

((75894-3336)+4)/(72^2-1)=14


3336=139x24

75894=139x546

I suspect that when pg(k) is prime and k is a multiple of 139 then k is of the form 139x(29xs-5) with s some integer.

An observation:

3336 and 75894 are multiple of 139

PG(3336) and PG(75894) are primes

3336=1=75894 mod 29

Maybe all PG(k) primes with k multiple of 139 k=1 mod 29?

Are there infinitely many PG(k) with k of the form 3336+139x29s?


92020-69660=1=75894=3336 mod 29

I think that there Is some ccomplicated connection among these exponents


In particolare 92020=2 mod 139
92020=3 mod 29

So 92020 Is a Number of the form 3338 (=71x47+1)+139x29s

Whereas 3336 and 75894 are of the form (71x47-1)+139x29s

139x29x2-3338/2=2131x3

pg(2131) is prime and also pg(2131x9)

indeed


2131=(46x139-1)/3






3336=-1 mod 71
75894=-5 mod 71



75894=3336x5 mod (71x139)

-75894=17^2 mod (71x29)

3336=-1 mod 71
3336=1 mod 29
75894=1 mod 29

so 75894=3336=92020-69660=1=-17^2=-(3x6^3-359) mod 29

75894=22360=-17^2=-(3x6^3-359) mod (29x71)

I think that that is the rub

because 69660=3x6^3 mod 23004

22360=92020-69660

359 mod 71=4

92020=4 mod 71

because 359=-1700 mod (71x29) and because the inverse of 1700 mod (17x29) is (14^2-1)=195

75894=22360=-3x6^3+195^(-1) mod (71x29)

75894x195=6^4+1 mod (71x29)

92020=2 mod 139
92020=3 mod 29

331259=21 mod 29
331259=22 mod 139

3336=0 mod 139
3336=1 mod 29


75894=0 mod 139
75894=1 mod 29


as you can see there are classes of exponents that are 1 unit far away mod 29 and mod 139 (21,22-0,1-2,3)


92020 (not multiple of 139)=2 mod 139
92020=2x2 mod 71

331259 (not multiple of 139)=22 mod 139
331259=22x2 mod 71

the not multiple of 139 (92020 and 331259)=d mod 139 and =2xd mod 71 for some d

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

19179=9 mod 71
19179=8 mod (19x1009)

but 19x1009 is a divisor of 23005*(2+331x139*5)-1

it seems that there is a connection between the exponents leading to a prime and the divisors of 23005*(2+139x331xs)-1 for some s

but for more general results it takes a deep knowledge of field theory that I don't have



92020=71x6^3 mod (19x1009)
92020=71x6^3=-11503x2=-3835 mod (19x1009)




after some steps (dividing 3835 by 5 and 92020 by 5) I came to


7x11x239=-3x2^8 mod (19x1009)


13x7x11x239=239239


331259=92020+239239


...


331259=53x101-1 mod (19x1009)


92020+2=6^6-69660 mod (19x1009)




92020+69660=6^6-6 mod 11503


331259=44 mod (311x71)

curious that (23005*(2+46009*5)-1)=276054 is a number in Oeis sequence A007275, walks on hexagonal lattices

276054x12=3312648=3x6^3 mod 3312

3312648=72 mod (23004)
3312648=-72 mod (23005)

In my opinion something interesting should be found studyng
Z23005xZ46009


(71x6^3+1) for example divides (23005*(2+46009*2)-1) and 92020+2




23005x46009=3 mod (3539x11503x13) (by the way 3539 is a Wagstaff prime)


92020=-4 mod 11503
92020=6 mod (13*3539)


curious that 331259=2131+1 mod 3539 a chance???


92020=71x6^3 mod 19171

i think that theory of ideals should help 46009Z




curious that 541456=(13*359+1) mod 19171


23005x92015=-1 mod 92019


92015 is multiple of 239


curious that 92020=15336 mod 19171 and 92020=15335 mod (313x49)

-75894 (pg(75894) is prime))=790 mod 19171 and -75894=791 mod (313x49)

75894=-790 mod 19171 75894=-791 mod (313x49)

there are pg(k) primes (as pg(92020) and pg(75894) I have not yet checked if there are others) such that k=a mod 19171 and k=a-1 mod (313x49) where a is a certain integer belonging to the set Z

i checked also pg(69660) is one of these


69660=-7024 mod 19171
69660=-7025 mod (313x49)




14377x92020=1 mod 19171


14377x71x6^3=23005x(2+46009x5) mod 19171


follows:


14377x15336=3834x19166 mod (1917x19171)

(23005*(2+46009*20)-1)/138027-71*6^3*10=7

138027 divides also

(23005*(2+46009*8)-1) and 23005*(2+46009*5)-1


71x6^3x10=-5 mod (829x37)

829x37 divides (92020-1)

71x6^3x10=-8 mod (19171x8)


92020-19171x4=71x6^3


92020-19171x4=-1 mod (313x49)
92020-19171x4=1 mod (3067)

92020=10 mod 3067

153367-(92020-19171*4+1)=138030


consider 429^2-19171x

the maximum value of x such that 429^2-19171x >0 is x=9

429^2-19171x9=11502
429^2-19171x8=37x829

37x829 divides (92020-1)


429^2-19171x6=69015
69015+645=69660
645 divides 69660


92020x2=429^2-1

19171x6=1=429^2 mod 23005


69660=645 mod 23005

69660=6^2*71*3^3+3*6^3


46011=19171*12-429^2

i observe that 19171=3^9-2^9


541456+13-449449=92020=46010x2=331259-239239

(46010x2-19171x2-2222)=51456
Pg(51456) and pg(541456) are primes

46010-1111+1=449x10^2

-19171x2=51456+2 mod 449

51456+2+19171x2+1 is a multiple of 1009

1111=71x3 mod 449

46010=71x3-1 mod 449

4601=111 mod 449

so 4601x20=92020=111x20=2222-2 mod 449

92020+2=2222 mod 449

(92020+2) is divisible by (71x6^3+1)


92020=-5^2=2220 mod 449


331259=-(4601x3-2) mod 23004


331259+2 is a multiple of 37 whereas 92020=331259-239239=1 mod 37

71x6^3=1 mod 3067 (3067x13=9201)

71x6^3=-1 mod (313x7^2)

92020=10 mod 3067
92020=-2 mod (313x7^2)

239239=13 mod (3067x13)


429*46009=-1 mod (214x92233)

92233 is a prime. 92233-92020=71x3

Last fiddled with by enzocreti on 2022-01-03 at 10:30 Reason: Added observation
enzocreti is offline   Reply With Quote
Old 2022-01-03, 13:06   #14
enzocreti
 
Mar 2018

2·269 Posts
Default

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
enzocreti is offline   Reply With Quote
Old 2022-03-06, 12:54   #15
enzocreti
 
Mar 2018

10328 Posts
Default

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

139 is the inverse of 546 mod 139

i think that there should be a connection with the other number 56238 which is divisible by 546 but I don't understand how


3335=(4X10^3+2)X(5/6)

just curious

I conjecture that there are infinitely many pg(k) primes with k of the form (2x10^n+16)/6

the two found are pg(36) and pg(3336)


the next 333...6 congruent to 0 mod 139 and 1 mod 29 is

(2x10^648+16)/6 ....648=3x6^3

I wonder if pg(333...6) is prime ...


i think that something periodic is at work...maybe applying Dirichlet characters???


75894 and 56238 are multiples of 546

the sum is 132132=546x(3^5-1)=546x242
546x242=1 mod 71


92020=2+92018

662x139=2 mod 71

23x139=2 mod 71

48x139=2x3336=-2 mod 71

3336=-1 mod 71




75895/5=15179=-4000 mod 2131

1776x(19179-667x6)=-667=260x1776 mod 2131

3335=667x5

541456-51456=700^2

I suspect that there Is something to do with

Numbers 546-56=490

7*(1+7*111...) 56 546 5446 5444...6....

546=7*(1+7*11)
490=546-56

92020 and 541456 are of the form 8643s-3053.


541456+13-449449=92020

449449=13 mod 8643

8643/43=201 which divides 51456

3053 mod 449=359


541456-10*51456=164^2



430*(71*(108+7667*3)-1)/451/17=92020

Amazing that 71*6^4*5-331259=128821 a palindromico Number

71*6^4=1 mod 239
331259=5 mod 239


Amazing that

PG(39699) and PG(79798) are primes

39699=401*99
79798=401*99+401*100-1



92020 is congruent to 0 mod 215

92020=3338 mod 4031


Using chinese remainder theorem

92020+866...5s are numbers =0 mod 215 and 3338 mod
4031

Allowimg negative s

92020-215*139*29=774645

774645 Is the concatenation of 774 and 645 which both divide 69660

29Zx139Z something here Is in action!


331259=717=239x3 mod (29*139)


331259=5=92020 mod 239


43*((429^2-9)/71+10^4)=541456

Is there a meaning??


(92020x2-8)/71 Is a 3 smooth Number 2592. 2592+10000=12592 which divides 541456

43*(429^2-9)/71 Is divisible also by 6966=69660/10









429 mod 139=12

PG(1323) Is prime

(429^2-12^2)=139x1323


I think that a super tool Is needed for these primes...something much more difficult than permutarions in 3x+1 problems or dirichelet characters

331259=92020+239239

I think that 429 Is involved but i canot clearly see how

239239/13=18403

429^2=11 mod 239

Dividing by 11

16731=1 mod 239


239x70=-1 mod 429

92020=331259=5 mod 1673


239239=-13 mod (2236x107)

2236 divides (92020-69660)

429^2=1 mod 107

I think that here Is the Key

429 is the 8-th catalan number

I suspect that partitions are involved in these exponents, in a way, however, that is not easy


there is this identity:

((44^2-1)x6^2-6^6+1)x4=(429^2-1)/2

i think that manipulating this identity and knowing that gcd(44,429)=11 one can get something but I have no idea how


Maybe an intuition but the idea has to be devoped:

541456+13-210210*2-29029=92020

210x2+29=449


29Z in action?


i think that ring 4031Z is in some mysterioius and hard way involved

92020=-77x3^2 mod 4031

331259=239239+92020

3338+4031*22=4*((44^2-1)*36-6^6+1)


Manipulating this identity

You get a 2 degree equation in x

4031x+3338=4x144x^2-4x46691

One solution Is x=22

The discriminant DELTA=(21313)^2 where 21313 looks like 2131 so I suspect that there Is some connection with these exponents

in the ring 215Z the zero divisors have the form q*172

69660 92020 541456 are all divisble by 172

the starting point is

215x107=1 mod 71

this is equivalent to

215x36=1 mod 71

215x36=7740 which divides 69660

pg(36) is prime pg(215) is prime pg(69660) pg(215x107x2=92020) are primes




92020=2x(6^3-1)x(6^3-2)

215=2 mod 71

69660=3 mod (6^3+1)

6^6=1 mod (6^3+1)

69660=3x6^6 mod (6^3+1)


92020=16 mod (451x51=23001)

23001 is a Poulet number Fermat pseudoprime in base 2



215x107=1 mod 71
139x331=1 mod 71

92020=2 mod 139x331

139x331-215x107=69660-6^6=23004

215x107=1 mod 71 139x331=1 mod 71


pg(1323) and PG(39699) are primes

1323 and 39699 have the form 98+y^2

For y=35 and y=199


1323=98+35^2
39699=98+199^2

35^2 and 199^2=6^2 mod 41

35=199=-6 mod 41


(239239+243)/29=1 mod 359

so

239239=-214 mod 359

92020=430x214

430 mod 359=71 or =-288

so

288x239239=-243 mod 359
288x239239=-242 mod 451

92020=-243 mod 359
these are only simple modular considerations, I think that something more complex is needed to fully undertand what is going on...

243-29=214

(451-359)x1001-72=92092-72=92020...

Last fiddled with by enzocreti on 2022-12-06 at 17:31
enzocreti is offline   Reply With Quote
Reply

Thread Tools


All times are UTC. The time now is 00:31.


Wed Dec 7 00:31:01 UTC 2022 up 110 days, 21:59, 0 users, load averages: 0.62, 0.76, 0.86

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2022, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.

≠ ± ∓ ÷ × · − √ ‰ ⊗ ⊕ ⊖ ⊘ ⊙ ≤ ≥ ≦ ≧ ≨ ≩ ≺ ≻ ≼ ≽ ⊏ ⊐ ⊑ ⊒ ² ³ °
∠ ∟ ° ≅ ~ ‖ ⟂ ⫛
≡ ≜ ≈ ∝ ∞ ≪ ≫ ⌊⌋ ⌈⌉ ∘ ∏ ∐ ∑ ∧ ∨ ∩ ∪ ⨀ ⊕ ⊗ 𝖕 𝖖 𝖗 ⊲ ⊳
∅ ∖ ∁ ↦ ↣ ∩ ∪ ⊆ ⊂ ⊄ ⊊ ⊇ ⊃ ⊅ ⊋ ⊖ ∈ ∉ ∋ ∌ ℕ ℤ ℚ ℝ ℂ ℵ ℶ ℷ ℸ 𝓟
¬ ∨ ∧ ⊕ → ← ⇒ ⇐ ⇔ ∀ ∃ ∄ ∴ ∵ ⊤ ⊥ ⊢ ⊨ ⫤ ⊣ … ⋯ ⋮ ⋰ ⋱
∫ ∬ ∭ ∮ ∯ ∰ ∇ ∆ δ ∂ ℱ ℒ ℓ
𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎𝜍 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔