 mersenneforum.org 69660 and 92020
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enzocreti

Mar 2018

22·7·19 Posts ...

Quote:
 Originally Posted by rudy235 I never quite understand what you are trying to say. Can you explain to the unwashed masses the function pg? What is it? Paying guest? Picogram?
the last you said   2021-07-11, 20:35 #13 enzocreti   Mar 2018 22·7·19 Posts 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 Chinese Remainder Theorem Calculator 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   2022-01-03, 13:06 #14 enzocreti   Mar 2018 22·7·19 Posts 92233=427x6^3+1 92233=6^3=51456 mod 427 ((92233-51456)-1)/3-12592=10^3 12592 divides 541456 PG(51456) and PG(541456) are primes 92233-51456=1=69660 mod 1699 13592=1699*8 43*(((51456+(3^6-1)*4)/4)-10^3)=541456 69660-(487-51456)=69660-(92233-51456)=0 mod (71x1699) (69660-487)=0 mod 313 (69660-486)=0 mod 427 486=162x3 162 divides 69660 -449449-13=64=541456+13 mod 139 so 541456=51 mod 139 (449x13=-1 mod 139) 92020=541456+13-449449 331259=71*6^6=259 mod 331 331259=44 mod 71 331249-44=331215 that Is the concatenation of 331 and 215 331259-44-215 Is a multiple of 331 44+215=259 71x6^6-331259+92020=6^2-6^5 mod 311 (6^2-6^5)=7740 which divides 69660 (7740=1 mod 71) multyplying by 9 both sides 71x6^6x9-331259x9+92020x9=-69660 mod 311 331215 is the concatenation of 331 and 215 331+215=546 546 divides 75894 and 56238 pg(75894) and pg(56238) are primes 75894-92020/2=215x139-1 546=215+331 215x139 can be cancelled in both sides this leaves 46009-46010=-1 56238 can be factorized as (139-6^2)*(331+215)=56238 probably there is something in 79*3^j pg(79) is prime by the way 79*3^2=711 and 69660=71x711+2131x3^2 79*3^3=2131+1 71x79+2131=7740 which divides 69660 79x3^2=1 mod 71 69660-19179=71x79x3^2 239239=-9 mod 12592 541456=239239+9 mod 12592 92020=3867+9 mod 12592 23004*46009-1-(92020-6)=3539x13x23003 i think that there should be some group in action... pg(75894) and pg(56238) are primes with 75894 and 56238 multiple of 139 75894+56238=132132=1 mod (1861x71) 331259=1 mod 1861 75894+56238=331259 mod (1861x107) 331259-132132=107x1861 75894+56238-1=71x1861 107=71+6^2 56238 and 75894 are congruent to +/- 6 mod 132 in particular (75894+6)/132+1=24^2 so 331259=56238+75894+92020+107x1001 pg(394) is prime 1323=-1 mod 331 pg(1323) is prime 1323=72 mod 139 1323=(1001-72) mod 394 i suspect that this is connected to the fact that 331259=-72 mod (1001x331) 331259=-1323 mod (1001-72=929) 331259=-394=-1323 mod 929=1001-72 3x6^3-3=394x(72)^(-1) mod 1001 In particolar 394x431+3-3x6^3=169169 92020=1323-394=929 mod 1001 541456+13=1323-394-13=916 mod 1001 (92020-1323+1)=0 mod 449 (541456+13-1323+1)=0 mod 449 541456-449449=92020-13 i forgot to see that 92020=43*2132+344 344 is the residue of the multiples of 43 (69660, 541456, 92020) mod 559 43*2132 is infact divisible by 559 69660-(46009*23005-1)/(313x49)=2^3x3^4=3x6^3 92020=-2 mod (313*49) 313x49x3x6^3-1=215^3 23005*(2+46009x8)=1 mod 429^2 23005x(2+46009x8)=-1 mod (359xp) where p is a prime p=23586469...i wonder if it has some special property the logic behind these primes requires tools that are far beyond current knowledge... 331259=259=71x6^6 mod 331 6^k=1 mod 259 the order mod 259 of 6 is 4 ord(6)=4 infact 6^4=1 mod 259 92020=2 mod 331 92020=4 mod 6^4 92020=71+4=75 mod 259 (92020-4)=71x6^4 239239=-77=-75-2 mod 259 239239=-78 mod 449 541456-77x5837+13=92020 curious that 449449-(239239+78)=210132=13*6^2x449 which is congruent to (6^4-2) mod 2131 239239+78 is 449x41x... 449x41 is 1840...anything to do with 429^2-1??? 541456=7740 mod 18404 7740 divides 69660 18404=(429^2-1)/10=449x41-5 pg(3336) is prime 3336=24x139=-1 mod 71 (24+71x9)x139=-1 mod 71 (24+71x9)x139+1-138=92020 (24x71x9)x139+1)/71=6^4+2 so 92020=(6^4+2)x71-138 or equivalently 92020=71x6^4+142-138=71x6^4+4 ((24+71*15)*139+1)/71=2132 so for example 19179=3^2*2131 can be rewritten as 3^2x(((24+71*15)*139-70)/71)=19179 71x6^6=6^4-44 mod (331x4=1324) 331259=44+215 mod 1324 331259=44 mod 71 (71*6^6-6^4+44)/(139*18-1)=1324 may be 44 is not random... 69660=(44^2-1)x6^2 331259=44 mod (311x5x71) 331259^2=1936 mod (311x5x71) 1935 divides 69660 6^5=1 mod (311x5) 6^5-6^2 divides 69660 1-36=35 I suspect that there could be some link to the fact that 541456=51456+700^2 700^2 is divisible by 35 curious that pg(75894) is prime 75894 is multiple of 139 -75894=(2^16+16) mod (359x197) 2^16=-2 mod 331 2^16=1 mod (255x257) i think that something big is happening on some field sqrt(71x215x3+1)=214 92020=sqrt(71x215x3+1)x430 92020=(71*215*6+1)+429 92020=214^2+215^2-1 pg(51456) and pg(6231) are primes with 51456 and 6231 multiple of 67 curiously (51456/67-6231/67)=26^2-1 19179=648 mod 23004 331259=(19179-648)/71 mod 359 331259x71=222 mod 359 so 222=19179-648 mod 359 note that 331331-(331259-(19179-648)/71)=333 ((92020*3-6)*3*4-1)/71=6^6+1 ...3312648...i think that 3x6^3 has something to do with these numbers... 331259/71=4665+44/71=6^6/10-6/10+44/71=6^6/10+7/355=6^10/10+7/(359-4) 331259=44 mod 4665 106x44+44/71+1=331259/71 4665=359*13-2 331259 =71 mod 44 331259=44 mod 71 so 331259 is a number of the form 115+(5^5-1)k this is curious 43*(1+sqrt(9x+1))=9x solution x=215 215*9+1=44^2 i think that you can obtain a continued fraction from that 69660=6^2x43x(1+sqrt(1+43x(1+sqrt(1+43x(1+... curious fact: 69660=19179=3x6^3 mod 639 19179/3=(639)3 and 6393=-1 mod 139 I think that something is in action over some field... (429^2-6)=-1 mod 46009=331x139 (429^2-6)=3 mod 639 from this 429^2=80^2-1=79x3^4=711x9 mod (71x139) 6393+((71*4+1)*139+1)=46009 i think that in Z46009 something is in action as well as in Z23004 71x6^6-541456=-1 mod 359 541456+14=-261=-331259 mod 359 69660=14=-6^6 mod 359 541456+69660=611116=-261=-331259 mod 359 611116=131x4665+1 92020-(541456+14-98)/13/359=359x2^8 -331259=98 mod (359x13) -(541456+69660)=-611116=359-98=261 mod (359x13) pg(1323) is prime pg(39699=13233*3) is prime 13233=18^2 mod 331 69660=215x18^2 so 13233x215=69660=150 mod 331 and 13233x3=39699=-150 mod (359x111) (359=331+18) 69660=3x6^3 mod 71 (69660-3x6^3)/71=-1 mod 139 this suggests me that something is in action over Z139 or maybe Z46009 23005*(2+331x139)-1 is divisible by 11503 69660-642(=3x6^3-6) is divisible by 11503 23005x(2+46009)-1-(69660-642) is a multiple of 23003 2*(23005*(2+46009*72)-1)/46009-2=331272x10 331272-13=331259 23005x(2+46009x72-1) is divisible by 449 541456+13-449x1001=331259 something mysterious is boiling in Z46009 46009x72-72=71x6^6 x^2/(6^6-2-sqrt(2x+1))-x*(sqrt(2x+1)-1)/215=0 has solution x=92020 min (x^2/(6^6-2-sqrt(2x+1))-x*(sqrt(2x+1)-1)/215)=-19394.4=-19179-215.4 this is a parabola from 331259x71=19179-17^2 mod 359 we have 222=19179-17^2 mod 359 so (2^9-1)=19179 mod (359x13) curious that 19179-511+1 is divisible by (2^2-1), (2^3-1) and (2^7-1) 331259=-98=-(512-414) mod 359 19179=414 mod 139 19179=138x139-3 69660=19179=6^6 mod 639 the inverse mod 639 of 2131 is 427 69660x427=9 mod 639 69660x427-9=639x46549 46549 is prime =6^6-107 I think this is not random but connected to the fact that 107 divides 92020 19179=7x71+14 mod (359x13) 6^6=-14 mod (359x13) 71x270+9=7x71+14=(2^9-1) mod (359x13) from here 6^6=7x71-19179 mod (359x13) form here after some steps... 331x72=7x71 mod (359x13) curious that 19179x(7^4+1)=1 mod (359x13) 3x6^3+70 is divisible by 359 (331259-5) is divisible by 717x6=6x(3x6^3+69) 6x(3x6^3+69)=-1 mod (331x13) so (331259-5) is divisible by (331x13-1) (331259-5)/(331*13-1)-(92020-5)/239/77=72 331x13=-5 mod 359 359x12=-1 mod 139 331259 has the curious representation: 65^2*77+77^2+5=331259=325325*77+77^2+5 further steps toward a theory of these numbers need super-tools (331259-5)/7-6^6=666 6^6=-666 mod (239x11) anything to do whit 92020=5 mod (239x11) 331259=5 mod (239x11)??? curious fact: (239239-(6^6+666)+2)/111=1729 the Ramanujan number I will call these primes Neme primes (Neighboured Mersenne) Neme(3)=73 Instead of Neme primes I could call them Hopeless primes, no hope to find a logic behind them curious that 69660=-342^2 mod (432^2) I think that starting from 23005x(2+46009x8)=1 mod 429^2 and 23005x8=-1 mod 429^2 one can develop someting useful Using Chinese remainder theory numbers multiple of 23005 (=0 mod 23005) and =1 mod 331x139 should form a ring or something similar...and I think that in that ring one can get something (2+46009x8+2x92020) is divisible by 92019 92019x6-92020x5 is a multiple of 3539 a wagstaff prime... 331259=2132 mod 3539 6^6-3x6^3-1=3539x13=46007 239239=-13 mod (9202) (239239+13)/107=2236 69660+2236x10=92020 331259=331 mod 2236 239239x2236x2=-(331259-331) mod 331 239239x4472=72 mod 331 2236=6^2x239239^(-1) mod 331 22360=(19^2-1)x239239^(-1) mod 331 from here 22360=-148 mod 331 239239=-(148/2) mod 331 22360=257-74 mod 331 239239=257 mod (331x19^2) (331259=257) mod (71x111) 331259x6^2=(22360-257) mod 71 I have the impression that powers of 6 and 71 are involved in the logic behind these neme primes 92020+257-14 (257-14=243 a power of 3) is 359x257 331259-243 is a multiple of 257 1001-((331259-243)/257-359)=72 69660=13 mod 257 69660=14 mod 359 there is a logic but it is so complex that it is almost hopeless to find a pattern 331259+14+84=71x13x359 541456=84 mod 359 69660=-345=-(331+14) mod 359 6^6=-14 mod 359 i cannot put toghether the entire pieces of the puzzle anyway 331259=6^6-541456 mod (359x13) (69660-6^6+331)*2-14=6^6 92020=10 mod 3067 331259=22 mod 139 331259=23 mod 3067 331259-23-92020+10=239239-13 92022x36-6^3=71x6^6 71x6^6=6^3 mod 3067 71x6^3=-1 mod (313x7^2) 71x6^3=1 mod 3067 I would call these prime Neme primes or maybe desperate primes I stronly suspect that the exponents of these neme primes are connected among them with a logic that it is impossible to understand...only a God could find a pattern...or maybe a new Gauss... curious that 541456=353(7)9 mod (3539x13) with that "7" inserted 92020=6 mod (359x13) in other words 541456=3539x10-10-1 mod (3539x13) curious that (541456-3539x10) is divisible by 23003=71x2^2*3^4-1 71x6^6=72 mod (3539x13) -(541456+13)=787 mod 858 -331259=787 mod 858 69660=162 mod 858 92020=214 mod 858 359x239=1 mod 429 541456=0 mod 787 787-13=774 divides 69660 787-429=359-1 so for example 331259=429x774-787 774 divides 69660 429=sqrt(92020x2+1) -541456=344 mod 774 there is a hidden structure it is clear that 331259-774 is a multiple of 4601 and 4601 divides 92020 -541456-13=456+331 541456+13-456=359x11x137 331259=-358 mod (773x429) 358=359-1 773=774-1 331259+773 is divisible by 1297 a prime of the form 6^s+1 331259=(259-215=44) mod 71 331259=259 mod 331 71x6^6=259 mod 331 there is something... 23005*(2+46009*k)-1=N^2 for k=8 for k=3680 ,... 23005*k+1 is a square k=8 k=3680 ... 3680*23005+1=(3x3067)^2=9201 (71x6^6-216)/3/3067=359+1 92020=10 mod (3067x13) (9201^2-1)/4601/23-13=787 Last fiddled with by enzocreti on 2022-03-06 at 12:35   2022-03-06, 12:54   #15
enzocreti

Mar 2018

22·7·19 Posts Quote:
 Originally Posted by enzocreti 92233=427x6^3+1 92233=6^3=51456 mod 427 ((92233-51456)-1)/3-12592=10^3 [500 lines of excessive quote]
3067 is a prime

there is a very complex hidden structure

331259/3680 is very close to 90

7775*23005+1 Is a Square

-331259=3067-1 mod 7775

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

I think these primes taste very exotic

23005 X+1=Y^2

with X and Y integers

X=92019 is a solution

Elliptic curves???

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

3067*5+1=71x6^3

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

-331259=(3x3067)^2 mod 359

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

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

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

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

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

71x6^6-216 is a multiple of 3067

92020-10 is a multiple of 3067

3067x6=-1 mod (239x77)

239239=13 mod (3067x6)

-3067x3=1 mod 107

239239=-13 mod 107

331259=-13 mod 107

it seems to be a perfect complex interlocking of modules

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

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

239239=0 mod (3067x6+1)

541456=7740 mod (3067x3+1)

7740 divides 69660

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

19179=777 mod (3067x3)

strange at least curious

19179=3067x6+777

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

777/3=259

-331259=71x6^3x777 mod 331

359x18^2=1 mod 541

-541456=85 mod 541

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

-541456=359x490 mod 541

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

-700^2=359x490+51456

after some steps:

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

51456=-480 mod 541

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

(44^2-1)=1935 divides 69660

1935-479=1456

479-394=85

pg(394) is prime

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

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

51456=129360 mod 541
51456=129359 mod 359

541456=18309 mod 541

541456-51456=490000

curious thta

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

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

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

92020=-491=3x394x271 mod 541

curious that 69660=-129 mod (541x129)

curious that

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

dividing by 2

92020=3x197 mod 541

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

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

curious that

69660-6^6+1=23005 so

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

from here we come to the curious:

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

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

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

pg(67) is by the way prime

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

so

23^2x92020=51456 mod 1297

curiously

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

365^3=1 mod 1297

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

107 is the inverse mod 71 of 215

-331259=1001 mod 449

541456+13-449x1001=92020

331259=92020+239239

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

maybe manipulating this you get something

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

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

71x6^6+72=0 mod 46009

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

331259=261 mod 359

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

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

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

-331259-14=84 mod(359x13)

-331259=98 mod (359x13)

(541456-84)/359/13=116

116 is the residue mod 359 of 92020

-23004=331 mod (359x13)

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

pg(1323) is prime

92016+1323 is a palindromic number

331259=13x359 mod 6^6

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

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

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

pure chance???

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

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

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

69660=7740x9

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

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

(88^2-2)=0 mod 79

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

69660+541456=611116=17x19x44x43

69660+541456=611116 this strange palindromic number

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

i think these numbers are more mysterious than pyramids of Giza

331259=4665x71+44

(359*13)*131-261=611116

261 is the residue of 331259 mod 359

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

69660=-315 mod 4665
541456=316 mod 4665

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

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

541456-51456=700^2

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

i think that a key passage is this:

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

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

222=19179-3x6^3 mod 359

19179=51x359+870

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

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

-69660=2^9 mod 331x2

18^2x213+3x6^3=69660

18^2x213-6^6=22356

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

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

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

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

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

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

from this

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

after some steps...

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

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

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

so

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

277^2=1 mod (138x139)

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

19179=511 mod (359x13)

by the way 19179=-1 mod 137

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

it could be a chance...anyway

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

3336=1 mod 23

331259=3336x23=261 mod 359

541456-344 is divisible by 559x44

44x559=2236x11

92020-69660=22360

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

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

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

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

331259x71=222 mod 359

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

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

so

(19179-648)=452*41-1

pg(51) is prime

so

452x41-1=222=331259x71 mod 359

452x41-1-222=51x359

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

(331259-261)/359=922

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

39699=9 mod 1323

(39699-9)/2-666=19179

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

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

19179=5x63^2-666

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

pg(1323) and pg(451) are primes

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

this is equivalent to

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

-19179x430=16=92020 mod 451

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

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

19179=2x344=3x79 mod 451

69660-19179 is a multiple of 79

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

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

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

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

I think that in 139Z there is something

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

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

curio:

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

92020=4x51x451+16

2x51x261-2x2131=92020-69660=22360

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

3216 divides 51456

I arrived to this from

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

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

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

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

92020=116 mod 359

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

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

541456+69660=611116

611116x71 mod (359x13)=137

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

611116=-261 mod (359x13)

331259=261 mod 359

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

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

541456*71=6^4+1 mod 4667

2131=72 mod 2059=29x71

19179=648 mod (29x71)

541456x71-6^4-1=359x13x8237

8237=1 mod (29x71)

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

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

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

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

239239+92020=331259

239239=-214 mod 359

430x239239=-92020 mod 359

the inverse of 430 mod 359 is 268

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

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

-19179-1=30003 mod (359x137)

3x6^3-261=387

92020=387 mod (2131x43)

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

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

23004-19179+648=4473

4472 divides (92020-69660)

331259=92020=5 mod 2629

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

it turns out that 331259 has this curious representation:

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

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

2629x7+1=18404

429^2-1=184040

239239 is a multiple of 7 and 2629

331259=92020+239239

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

92015 is divisible by 2629x35

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

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

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

541456+13-449449=92020

from above

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

so

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

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

970x449=12x18404 mod 4667

or

3168=-12x18404 mod 4667

multiplying by 5

3168x5=-12x92020 mod 4667

210210=-4472 mod 4667

1051050=-(92020-69660) mod 4667

16170=-344 mod 4667

344 divides 541456

541456+13-210210=331259

mod 359x13 infact

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

331259=-98=261 mod 359

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

manipulating a bit the above equations i obtained

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

but this is:

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

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

bringing 5 out

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

-98=-4471x7000=331259 mod 4667

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

359-98=261

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

16169,13932, 18403 are not random

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

84=98-14

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

follows that

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

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

16169-13932=2237=4472/2+1

14^2+14=210

-210210=4472 mod (359x13)

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

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

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

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

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

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

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

261=359-98

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

14^2=196

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

maybe not a achance???

4x19179-84^2=69660

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

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

6^5=-1 mod 77

541456=-8 mod 77

331259=5=92020 mod 77

5+8=13

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

so

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

5x6^5=-77 mod 239

i think that here is the key

541456=11^2 mod 239

541456=-8 mod 77

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

92020 and 331259 are of the form 92020+18403s

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

pg(3371) is prime

probably i got something

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

this implies

92020=88^2+1 mod 3371

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

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

so some conclusions can be done

92015x9=69660 mod 3371

92015=71x6^4-1=239x7x11x5

18403=1548 mod 3371

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

Dividing by 45

18403 congruenti to 43*6^2 mod 3371

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

It utns out that

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

2x139x331=1001 mod 3371

2x139x331=92020-2

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

the inverse mod 51 of 14 is 11

-541456=11 mod 51

92018=14=331259=-449449 mod 51

541456+13-449449=92020

92018=14=331259=10x1001 mod 51

(92018-10x1001) is divisible by 1608

1608 divides 51456 and pg(51456) is prime

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

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

449=92=41 mod 51

239=-16=-92020=331331

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

331331-239 Is also divisibile by 541 and 36

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

pg(51) and pg(451) are primes

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

curious fact

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

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

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

-541456=215= +2131 mod 51

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

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

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

541456=2131x3^2x28+4444

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

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

so 92020=16 mod 51

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

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

23004=3x7667+3

7667 is palindromic in base 10 and 6

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

4472 divides (92020-69660)

-4472=16=92020 mod (51x11)

215=22360/2=-541456 mod 51

22360=92020-69660

69660=3 mod 107

92020=0 mod 107

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

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

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

69660=3x6^3 mod 23004

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

69660=3 mod 107
69660=0 mod 215

using chinese remainder theorem

69660 is a number of the form 645+23005k

if k=-1

645-23005=22360=(92020-69660)

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

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

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

387 divides 69660

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

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

I think that a possible clue could be

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

so for example I notice that

22360=-18^2 mod (106x107)

22360=3 mod 107
22360=4 mod 108

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

23008=3 mod 107
23008=4 mod 108

23008-3=23005
23008-4=23004

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

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

69660=-6^2 mod 264^2

264 is multiple of 44

69660=(44^2-1)x36

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

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

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

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

curious fact

92020=71x6^4+4

331259=92020+239*7*11*13

7,11 and 13 are primitive roots mod 71

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

(541-331)=210

(541456+13-210210=331259)

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

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

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

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

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

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

from here

because 541456=(7^3+1)x1574

-140=(7^3+1)x1575

dividing both sides by 7 and by 5

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

from here I got

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

this reduces to:

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

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

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

Last fiddled with by enzocreti on 2022-07-01 at 11:51   Thread Tools Show Printable Version Email this Page

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