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2021-07-11, 19:07   #12
enzocreti

Mar 2018

32×59 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 32·59 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 32×59 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 Last fiddled with by enzocreti on 2022-01-17 at 09:32

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