mersenneforum.org  

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

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

10228 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 online now   Reply With Quote
Old 2021-07-11, 20:35   #13
enzocreti
 
Mar 2018

2×5×53 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

Last fiddled with by enzocreti on 2021-10-26 at 11:08
enzocreti is online now   Reply With Quote
Reply

Thread Tools


All times are UTC. The time now is 11:46.


Tue Oct 26 11:46:35 UTC 2021 up 95 days, 6:15, 0 users, load averages: 2.96, 2.83, 2.45

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, 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.