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
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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
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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