20191106, 13:46  #1 
Mar 2018
17×31 Posts 
binary form of the exponents 69660, 92020, 541456
pg(69660), pg(92020) and pg(541456) are probable primes with 69660, 92020 and 541456 multiple of 86
69660 in binary is 10001000000011100 92020 in binary is 10110011101110100 541456 in binary is 10000100001100010000 you can see that the number of the 1's is always a multiple of 5 a chance? 
20191218, 08:26  #2 
Mar 2018
17×31 Posts 
69660, 92020, 541456
69660, 92020 and 541456 are 6 mod 13 (and 10^m mod 41)
69660, 92020 and 541456 are multiple of 43 is there a reason why (696606)/26 is congruent to 13 mod 43 (920206)/26 is congruent to 13 mod 43 (5414566)/26 is congruent to 13 mod 43? 215, 69660, 92020, 541456 are multiple of 43 let be log the log base 10 int(x) let be the integer part of x so for example int(5.43)=5 A=10^2*log(215)215/41 int(A)=227=B 215 (which is odd) is congruent to B+1 mod 13 69660 which is even is congruent to B mod 13 92020 is congruent to B mod 13 and 541456 is also congruent to B mod 13 215 (odd) is congruent to 1215 mod 13 69660 (even) is congruent to 1215 mod 13 92020 (even) is congruent to 1215 mod 13 541456 (even) is congruent to 1215 mod 13 215, 69660, 92020, 541456 can be written as 13x+1763y+769 (54145676913*93)/1763=306 (6966076913*824)/1763=33 (9202076913*781)/1763=46 as you can see 306,33 and 46 are all 7 mod 13 769+13*824 769+13*781 769+13*93 are multiples of 43 (69660(10^3+215))/13=5265 which is 19 mod 43 (92020(10^3+215))/13=6985 which is 19 mod 43 (541456(10^3+215))/13=41557 which is 19 mod 43 10^3+215 is 6 mod 13 now 69660/13=5358,4615384... 92020/13=7078,4615384... 541456/13=41650,4615384... the repeating term 4615384 is the same...so that numbers must have some form 13s+k? 215 (odd) is congruent to 307*2^210^3 or equivalently to  (19*2^61) mod 13 69660 (even) is congruent to 307*2^21001 or equivalenly to (19*2^61) mod 13 92020 (even) the same 541456 (even) the same (19*2^6*(54145692020)/(13*43)+1)/13+10=75215 is a multiple of 307=(215*101)/7=(54145601)/17637 pg(51456) is another probable prime with 51456 congruent to 10^n mod 41 75215=(51456*19+1)/13+10=(19*2^6*(54145692020)/(13*43)+1)/13+10 215 is congruent to 1215=5*3^5 mod 13 69660 is congruent to 1215 mod 13 and so also 92020 and 541456 1215=(51456/22*13*10^243)/19 ...so summing up... 215 (odd) is congruent to 3*19*2^2 mod((41*43307)/(7*2^4)=13) where 41*43307 is congruent to 10^3+3*19*2^2 mod (3*19*2^2=228) 69660 (even) is congruent to (3*19*2^21) mod 13 ... the same for 92020 and 541456 another way is 215 (odd) is congruent to 15*81 mod ((41*43307)/(30715*13)) 69660 (even) is congruent to 15*81 mod ((41*43307)/(30715*13)) the same for 92020 and 541456 215 is congruent to (41*4330710^3)/2 mod ((41*43307)/(7*2^4)) 69660 is congruent to (41*4330710^3)/21 mod ((41*43307)/(7*2^4)) 92020 is congruent to (41*4330710^3)/21 mod ((41*43307)/(7*2^4)) 541456 is congruent to (41*4330710^3)/21 mod ((41*43307)/(7*2^4)) 215 is congruent to 3*19*2^2 mod ((41*43307)/(2*(19*31))) 69660 is congruent to 3*19*2^21 mod ((41*43307)/(2*(19*31))) and so 92020 and 541456 215 is congruent to 3*19*2^2 mod ((3^61)/(3*191)) 69660 is congruent to 3*19*2^21 mod((3^61)/(3*191)) and so 92020 and 541456 51456 (pg(51456) is probable prime and 51456 is 10^n mod 41) is congruent to 19*3*2^4 mod 13 Pg(2131) is probable prime 2131 is prime 227=2131307*426^2 So 215 is congruent also to 2131307*426^2 mod 13 And 69660 is congruent to 2131307*426^2+1 mod 13 and so also 92020 and 541456 69660 is congruent to 1763307*51 mod ((1763307)/112)=13) And so 92020 and 541456 215 which is odd is congruent to  1763+307*5+1 mod 13 215+1763307*51 is divisible by 17 and by 13 And also 5414561763+307*5+1 is divisible by 13 and 17 215 and 541456 have the same residue 10 mod 41 ((5414561763+307*5+1)/(13*17)+1) *200+51456=541456 69660 92020 541456 are congruent to 7*2^6 mod 26 7*2^6(1763307*51)=221=13*17 215+7*2^6 is a multiple of 221 5414567*2^6 is a multiple of 221 Last fiddled with by enzocreti on 20200126 at 17:58 
20200124, 15:31  #3 
Mar 2018
17·31 Posts 
215 , 51456, 69660, 92020, 541456
51456, 69660, 92020, 541456 are even and congruent to 10^n mod 41
pg(51456), pg(69660), pg(92020) and pg(541456) are prp 51456, 69660, 92020, 541456 are all congruent to 7*2^q1 mod 13 with q a nonnegative integer 215 is odd and pg(215) is prp 215 is congruent to 7*2^q mod 13 
20200330, 11:52  #4 
Mar 2018
527_{10} Posts 
69660 and 92020
69660 and 92020 are multiple of 215 and congruent to 344 mod 559
92020=lcm(215,344,559)+69660  denotes concatenation in base 10 2^696601  2^695591 is prime 2^920201  2^920191 is prime!!! Last fiddled with by enzocreti on 20200330 at 11:53 
20200330, 12:19  #5 
Romulan Interpreter
Jun 2011
Thailand
10001110111110_{2} Posts 
please show us a proof that they are prime

20200330, 12:50  #6 
Mar 2018
527_{10} Posts 
...
Well...
actually they are only probable primes... maybe in future they will be proven primes Last fiddled with by enzocreti on 20200330 at 12:52 
20200330, 17:39  #7 
Mar 2018
17·31 Posts 
http://factordb.com/index.php?id=1100000001110801143 http://factordb.com/index.php?query=...%2B2%5E920191 Last fiddled with by enzocreti on 20200330 at 17:55 
20200330, 23:06  #8 
Mar 2018
17·31 Posts 
... I note also...
69660 I note also that
(lcm(215,344,559))^2=4999*10^5+6966060 I notice that lcm(215,344,559)=22360 22360/(18*18)=69.01234567... curious I notice also that 541456 (multiple of 43),is  215 mod (18*181) and 69660 (multiple of 43) is 215 mod (18*181) I note that the polynomial X^2X*429^2+7967780460=0 has the solution x=69660 If you see the discriminant of such polynomial you can see interesting things about pg primes with exponent multiple of 43 I note that 429^2 is congruent to 1 mod 215 and to 1 mod 344. I note that 92020*2+1=429^2 The discriminant of the polynomial is 429^44*7967780460 which is a perfect square and lcm(215,344,559) divides 429^44*79677804601 Pg(331259) is prime and pg(92020) is prime. 92020+(92020/2151)*559+546=331259 Again magic numbers 559 and 546 strike! So 331259 is a number of the form 215*(13s1)+(13*b2)*559+546 For some s, b positive integers So there are pg primes pg(75894) and pg(56238) with 75894 and 56238 multiple of 546 and pg(331259) with 331259 of the form 215m+559g+546 for some positive m and g. Pg(69660) is prime. 69660=(3067*8546)*311# where # is the primorial and 3067 is a prime of the form 787+456s notice that lcm(215,344,559)=22360 22360/(18*18)=69.01234567... curious I notice also that 541456 (multiple of 43),is  215 mod (18*181) and 69660 (multiple of 43) is 215 mod (18*181) So we have pg(215) is prime pg(69660) is prime pg(92020) is prime With 215 69660 and 92020 multiple of 43 69660=215+(18*181)*215 92020=69660+18*18*69+4 215 is congruent to 108 mod (18*181=323) 541456 is congruent to 108 mod (18*181) 92020 is congruent to (17*171) mod (18*181) 69660 is congruent to 215 mod (18*181) 108=6^318^2 17*171+36216=17*171+6^26^3=108 215=6^31 To make it easier 215, 69660, 541456 are congruent to plus or minus 215 mod 323 92020 is congruent to (17*171) mod (18*181) curious that 289/215 is about 1.(344)... and 541456 92020 69660 215 are congruent to plus or minus 344 mod 559 215, 69660, 541456 are congruent to plus or minus 6^31 mod 323 92020 is congruent to (12/9)*6^3 mod 323 92020*9/12 is congruent to 6^3 mod 323 92020 is congruent to (17^21) mod 323 and to  (6^21) mod 323 92020 is a number of the form 8686+13889s 13889=(6^3+1)*64 215 69660 92020 541456 are + or  344 mod 559 lcm(215,344,559)86*(10^2+1)+6^31=(6^3+1)*2^6+1 92020=69660+lcm(215,344,559) so you can substitute 92020=69660+86*(10^2+1)6^3+1+(6^3+1)*2^6+1 86*(10^2+1) mod 323 is 17^21 215, 69660, 541456 are multiple of 43 and congruent to 10 and 1 mod 41 They are congruent to plus or minus 215 mod 323 92020 is congruent to 2^4 (not a power of 10) mod 41 92020 is congruent to (2^4+1)^21 mod 323 288 is 17^21 288 in base 16 is 120 120=11^21 also 323=18^21 in base 16 is 143=12^21 344*((14444561763*2^9) /3441)=541456 1444456=lcm(13,323,344) 541456=lcm(13,323,344)344*(41*2^6+1) 215 69660 92020 541456 are congruent to plus or minus (3^a*2^b) mod 323 215 is congruent to  108 mod 323 541456 is congruent to 108 mod 323 69660 is congruent to  108 mod 323 92020 is congruent to 288 mod 323 108 and 288 are numbers of the form 3^a*2^b So exponents multiple of 43 are congruent to plus or minus 344 mod 559 and to plus or minus a 3smooth number mod 323 108 and 288 are both divisible by 36 Last fiddled with by enzocreti on 20200818 at 21:00 Reason: notice that lcm(215,344,559)=22360 22360/(18*18)=69.01234567... curious I notice also that 541456 (multiple of 43),is  21 
20200819, 05:48  #9  
Mar 2018
1017_{8} Posts 
Q77I
Quote:
215 69660 92020 541456 are congruent to plus or minus (6^k1) mod 323 for k=3,2 69660=(2^5*3^7)(2^3*3^4) so it is the difference of two 3 smooth numbers (2^a*3^b)(2^(a3)*3^(b3)) 69660 is multiple of 3 and congruent to 0 mod (6^21) 215, 92020, 541456 are not multiple of 3 and multiple of 43 and are congruent to plus or minus 2^k mod 36 for some k 1763*323(6^3+1)*(2^7+1)=541456 or (42^21)*(18^21)(6^3+1)*(2^7+1)=541456 I also note that 69660=(2^7+1)*(6^3+1)+(2^7+1)*(18^21) And by the way 215=(42^21)*4^2(2^7+1)*(6^3+1) 27993=(217*129)=(2^7+1)*(6^3+1) I notice that  541456 mod 27993=9202*2 92020=10*9202 And 9202*2*10+1=429^2 27993=3/5*(6^61) 27993 in base 6 is 333333 27993 has also the representation: (42^21)*4^2(6^31)=27993=2*43*(18^21)+(6^31) x/5+(42^21)*(18^21)3/5*(6^61)=20*3/5*(6^61) The solution of this equation x=92020 this identity: (42^21)*(18^21)=(10^3+18^2)*430+43*3 Maybe it is not a chance that pg(10^3+18^21=1323) is prime pg(1323), pg(215), pg(69660), pg(92020), pg(541456) are primes 1323, 215, 69660, 92020, 541456 are congruent to plus or minus (2^a*3^b1) mod 323 where 2^a*3^b is a 3 smooth number with 2^a*3^b<323 or 1323, 215, 69660, 92020, 541456 are congruent to plus or minus (p1) mod 323 where p is a perfect power 92020 has the factorization 2*(6^31)*(6^32) lcm(215,344,559)=(6^31)*(2*(6^32)18^2)=(2*(6^32)18^2)*(2*(6^32)18^2+111) 69660=92020lcm(215,344,559) 215=2*(6^32)+2*(6^3+1)2*18^2+1 92020=(2*(6^32)+2*(6^3+1)2*18^2+1)*(2*(6^32)+2*(6^3+1)2*18^2)*2 so 215=2*(6^32)+2*(6^3+1)2*18^2+1 69660==(2*(6^32)+2*(6^3+1)2*18^2+1)*(2*(6^32)+2*(6^3+1)2*18^2)*2((6^31)*(2*(6^32)18^2)) 92020=(2*(6^32)+2*(6^3+1)2*18^2+1)*(2*(6^32)+2*(6^3+1)2*18^2)*2 541456=(42^21)*(18^21)(7^36^3+2)*(6^3+1) 69660=1111115*11111+8*(42^21) 344=7^3+1=2*(1111115*11111)/(18^21)=A so lcm(215,344,559)=lcm(215, A, A+215) 541456=5*111112(42^21)*8 (12^21) divides (1111115*111116^3+1) 559=(1111115*111116^3+1)/(12^21)+(1111115*11111)/(18^21) Quite clear that 215, 69660, 92020,541456 multiple of 43 are congruent to plus or minus (18^26^k) mod (18^21) where k is 2 or 3 and 3 indeed is the maximum exponent such that 18^26^k i 1s positive (54145618^2+6^3)/323=41^25=1676 (9202018^2+6^2)/323=17^25=284 (69660+18^26^3)/323=6^3 (215+18^26^3)/323=1 There is clearly a pattern!!! I notice also that (1+5)=6 is a semiprime (6^3+5)=13*17 is a semiprime (1676+5)=41*41 is a semiprime (284+5)=17*17 is a semiprime (6,221,1681,289) are either squares of primes or product of two consecutive primes so when pg(k) is prime and k is a multiple of 43, then k can be expressed in this way: (p*q5)*(18^21) plus or minus (18^26^k) with 6^k<18^2 p and q are primes when pg(k) is prime and k is a multiple of 546, then k is congruent to 78 mod (18^26^3) as in the cases k=56238 and k=75894 Curio of the curios: pg(331259) is probable prime and 331259 is prime magic: 71*6^6331*(10^43*331)=331259 71*6^6 and 331259 have in common the first five digits 33125=182^2+1 If you consider 71*6^n for n>3 For n even 71*6^n is congruent to (11^21) mod (13^21) and for n odd Is congruent to (7^21) mod (13^21) For n=4 71*6^4=92016 for n=6 71*6^6=3312576 As you can see Pg(92020) is prime and pg(331259) is prime 92020 has the last two digits 20 different from 92016 The same for 3312576 and 331259 Moreover 331259 mod (71*6^3) =9203 Both 331259 and 92020 are 5 mod 239 Is there something connected with the fact ord (71*6^k)=4 I mean the smallest value k for which 71*6^k is congruent to 1 mod 239 is k=4??? 71 and 6 are both quadratic residues mod 239 92020 and 331259 are congruent to 71*3^3 mod (239*13) I wonder if this concept could be generalized pg(51456), pg(92020), pg(331259) are probable primes 51456 is congruent to 71 mod (239*(6^31)) 92020 is congruent to 71*3^3 mod (239*13) 331259 is congruent to 71*3^3 mod (239*13) 92020 and 331259 are congruent to 6 mod 13 51456 and 331259 are congruent to 2^3 mod 109 71 and 71*3^3 are both congruent to 6 mod 13 so 51456, 92020, 331259 are congruent to 13*(5+71*k)+6 either mod (239*215) or mod (239*13) are these primes infinite? the odd thing is that 51456, 92020, 331259 are either congruent to 10^m mod 41 or prime (331259 infact is prime) So I think it is no chance that 51456, 92020, 331259 are either congruent to 2^j mod 71 or to 13*2^i mod 71 No chance at all! There is a file Rouge! Pg(541456) is probable prime as pg(51456) And 541456 mod (239*215)=9202*3, that is 3*( 92020/10 )and pg(92020) is prime 541456 is congruent to (3/20)*(429^21) mod (239*215) 541456 is congruent to 14*71*3^3+3*2^8 (mod (239*215)) Maybe there is some connection to the fact that 215, 69660, 92020 and 541456 are plus or minus 344 mod 559 lcm(344,559)=4472=71*(2^61)1 I suspect that something in field F(239) is in action! 331259^(1) mod (215*239)=49999=(10^52)/2 ((71*(2^61))1)/2=lcm(344,559)=92026966 Multiplying both sides by 10 you have 92020=69660+lcm(215,344,559) I would suggest to study these exponents in field F(51385=239*215) (239*10*215+3*(429^21)/10541456) /3=9202 So 10* (239*10*215+3*(429^21)/10541456) /3=92020 In this equation we have 215, 92020, 541456 multiple of 43 and not of 3 The other multiple of 43 is 69660 which is multiple of 3 And 92020=69660+lcm(344,215,559) 239*10*215+3*(429^21)/10 is a multiple of 559 239*10*215+3*(429^21)/10=(42^21)*(18^22)+43*2^5 This identity (239*10*215+(3/10)*(429^21)541456+92020)/559=92020/(215*2) One can play around with this expression containing 215 and 559 And substitute for example 92020 with (429^21)/2 Pg(331259) is prime and also 331259 is prime The inverse modulo (215*239) of 331259 is the prime 5*10^41=49999 (331259*499991)/(215*239)=7*(18^21)*(12^21)10^3+1 So 331259=((((18^21)*(10^3+1)10^3+1)*239*215)+1)/(5*10^41) Using Wolphram Alpha i considered this equation: ((323*(10^x+1)10^x+1)*239*215+1)/(5*10^(x+1)1)=y wolphram say that the integer solution is x=3, y=331259 wolphram gives an alternate form: 84898302/(5*(2^(x+1)*5^(x+2)1))+1654597/5=y the number 1654597=69660+(30^21)*(42^21) so 69660 and 331259 (both 6 mod 13 and pg(69660) pg(331259) primes) are linked by this equation: 84898302/(5*(2^(3+1)*5^(3+2)1))+(69660+(30^21)*(42^21))/5=331259 541456 in field F(239*215) and in field F(239*323) 541456 mod (239*215)=9202*3 541456 mod (239*323)=359*3 Pg(359) is prime, pg(9202*10) is prime 9202359 is a multiple of 239 Pg(92020+239239=331259) is prime I note also 331259=71*6^6331*(10^43*331)=6^2*920213=92020+239239 239239 is congruent to 13 mod 107 and mod 43 22360=(10*(239239+13))/107 92020=69660+22360 92020=331259239239 92020 is a multiple of 107 lcm(215,344,559)*10=22360 multiple of 86, that is 69660, 92020 and 541456 are congruent to a square mod 428 so the muliple of 86=k for which pg(k) is prime are numbers for which there is a solution to this modular equation: y is congruent to 36*x^2 mod 428 infact 92020 is congruent to 36*0^2 mod 428 69660 is congruent to 36*3^2 mod 428 541456 is congruent to 36*1^2 mod 428 There are pg(k) primes with k multiple of 215 and k multiple of 546 probably that numbers 215 and 546 are not random at all look at this equality: 71*6^6182^23^2331*(10^43*331)10=546^2 546215=331=546(6^31) by the way 182^2+3^2+546^2 is prime 546^2 is congruent to 6^3 mod 331 541456((3*239) ^2239+(429^21)/20)=(429^21)/10 541456=(18/5)*(429^21)/2+13*((429^211)/102235) where 2235=lcm(215,344,559)1 541456*10=3*331259+239*(136^2+1) 92020 and 331259 are both 5 mod 239 71*6^k is congruent to 1 mod 239 for k=4 But 71*6^4 is 920... And 71*6^6 is 33125... 92020 mod 71 is 4 and mod 331 is 3 71*6^4 is congruent to 1 mod (385*239) 541456 is congruent to  239 mod 385 I notice that 92020 and 331259 are congruent to 5 mod 239 but also to  72 mod (1001) I notice that 92020 is congruent to 146 mod 71 541456 and 331259 are congruent to 146 mod 703 I notice that 215, 69660, 541456 are plus or minus 215 mod 323 92020 which is congruent to 16 mod 41, is congruent to 288 mod 323 and mod 71 288 and 92020 are both 4 mod 71 so 92020 is congruent to 4 mod 71 and is congruent to 4+284 (mod 323) where 284 is the residue mod 323 of 71*6^4 215, 69660, 541456 are congruent to plus or minus 215 mod 323 where 215=4+211 211 is the residue of 71*6^4 mod 215 in particular 92020 is congruent to 288 mod 323 and mod 284, infact 92020=71*6^4+4 I think that also 331259 has something to do with 71*6^6 so multiple of 43, that is 215, 69660, 92020, 541456 are either of the form 323k+108, or 323k108, or 323k288 I notice that (323108)=6^31 and (323288)=6^21 211 and 284 are also the residues of 71*6^4 and 71*6^6 mod 323 the 18th pg prime is pg(1323), the 36th is pg(360787) modulo (18^21=323), 1323 is 31 and 360787 is 319 the difference between 319 and 31 is again 288 1323 is congruent to 31 mod (6^44) 360787 is congruent to 6^2*2^3+31 mod (6^44) 319 is congruent to 31 mod 6^2 infact here: 215, 69660, 541456 are congruent to plus or minus 108 mod 323 92020 is congruent to (6^41008=288) mod 323 so multiple of 43 are either congruent to plus or minus (18+90) mod (18^21) or to 6^4(18+90+900) mod (18^21) 92020 is congruent to (11*6^2108) mod (18^21) and to  11*6^2 mod (304^2) so great fact 215, 69660, 92020, 541456 are either congruent to plus or minus (108*21) mod (108*31) or to plus/minus (108/31) mod (108*31) pg(331259) and pg(92020) are probable primes. 331259=92020+239239 as said 331259 and 92020 are congruent to 5 mod 239 331259 and 92020 are congruent to 6 mod 13 using chinese remainder theorem it yields something like 239x(1)+13y(1)=1 using Euclidean algor you have the solution y(1)=92 92 are the first two digits of 92020 now 331259 can be rewritten as 239*10^3+92*10^3+331711 331259 and 92020 have the property that they are congruent to the last two decimal digits modulo 9200 331259 is congruent to 59 mod 9200 92020 is congruent to 20 mod 9200 (239*77+1)*5=92020 239*77+1 is a multiple of 215 541456 is congruento to 3/2*(239*77+1) mod (239*215) 331259 is congruent to 5 mod (239*77) and also 92020 (5414569202*3)/215/239=10 (13*3311)*7*11+57*11*13*239=92020 541456 is congruent to (9203+(71*6^41)/5) (mod (239*5*43)) 9203 is prime (5414569203(71*6^41)/5)/215/239=10 33125913*(9202*21)=92020 9202013*1720=69660 lcm(215,344,559)=13*1720 67*(16683/67+22360/431)=51456 where 16683=9202*211720 and 22360=lcm(215,344,559) 331259 has the representation 331259=36*(7/180+(920204)/10) Let be Floor(x) the floor function floor(5.5)=5 for example Floor(239*(331259/(71*6^4))/4)=215 So i suspect that there are infinitely many pg(k) primes with k multiple either of 215 or multiple of something transformed by the Floor function in 215 and in these Cases k is alway 6 mod 13 215=(239*6^22^2)/40=(239*6^22^2)/(6^2+2^2) 331259=(3*71*6^5+7)/5 71*6^6(3^3*71*6^57)/5=331259 (3^3*71*6^57)/5=(10^43*331)*331 so we have 92020=71*6^6(3^3*71*6^57)/513*(71*6^41)/5 (69660/3860)/13=1720 I see where 215 comes from (33125923666)/(12^21)=9*2391=215*10 23666 I used CRT x is congruent to 6 mod 13 and to 5 mod (239*7) As you can see Both 331259 and 92020 are congruent to 23666 mod 143 But this simply means that both 331259 and 92020 are congruent to 71 mod 143 By the way also 215 is congruent to 71 mod 143 so 215, 92020, 331259 are congruent to plus or minus 71 mod 143 215 is 2 mod (71) 92020 is 4 mod (71*6^3) 331259 is 9203 (prime) mod (71*6^3) there is clearly a pattern note that also 541456 is congruent to (9203*3) mod (7*71) i have no idea how to develop these ideas but i strongly suspect that there is a structure It should be clear that 541456 215 69660 331259 92020 are congruent to plus or minus (19+13s) mod 143 for some non negative s. 331259 is also congruent to sqrt(239*9215) mod 71 215*9=44^21 So 331259 is congruent to sqrt(215*9+1) mod 71 another way to see the same thing is that 215 92020 and 331259 are congruent to plus or minus 72 (mod 143) and 92020 and 331259 are even congruent to  72 (mod 143*7) infact 92020+72=92092 and 331259+72=331331 It's easy to see that 9203 conguent to 44 mod 71 and 331259 is as well congruent to 44 mod 71 and 9203 has the same residue 259 mod 344 and mod 559 ((541456/921)((41*43*10+7)/30))^(1)7=9203 note that 921=3*307 9210=30*307=((541456/921)((41*43*10+7)/30))^(1) 2150=307*7+1 92020*(5879)/99971.107107107=541456 or 92020*(1763*10+7)/299771.107107...=541456 541456(92020*(1763*10+7)/58790)=10*215*239 331259 is congruent to (71*6^3+307*10)/2 (mod (71*7*6^4)) 9203=(331259*27*71*6^4)/2 331259=(71*6^3*43)/2+307*5 331259=(2^10+1)*(18^21)+184 I would conjecture that if pg(k) is prime and k is multiple of 43, then k is a multiple of (72+143s) for some positive integer s the multiple of 215=72+143 are 215 itself, 69660, 92020 the other multiple of 43 is 541456 which is a multiple of 787=72+143*5 multiple of 43 are 215, 69660, 92020, 541456 they are either congruent to (plus/minus) 215 mod 323 or to 288 mod 323 215 and 288 are integers for which a integer solution exists for the equation x^2+71*y^2=z (x,y,z positive integers) multiple of 43 are 215 69660 92020 541456 Now consider the equations: x^2+71*y^2=215 x^2+71*y^2=69660 x^2+71*y^2=92020 x^2+71*y^2=541456 there are Always non zero integer solutions x and y 541456 is about 271*999*2 i wonder Why??? so multiple of 86, 92020 69660 and 541456 are congruent to (271*9991) mod 172 this implies (92020 69660 541456 are also 6 mod 13) that 92020 69660 and 541456 are congruent to 344 mod 2236 and I remember that 92020=69660+22360 this could suggest why multiple of 43 are congruent to plus or minus 344 mod 559 infact (271*9991) is congruent to 172 mod 559 (271*9991) is congruent to 172 mod 215 pg(51456) and pg(541456) are probable primes observing that 51456 is congruent to 508 mod 542 and to 507 mod 999 using CRT: solutions are 51456+541458n infact lcm(542,999)=541458=541456+2 331259= 11*2^10+39*2^13+507 51456*9 is congruent to 700^2 mod (164^2) 541456=51456+700^2 (700^2164^2)/51456=3^2 (10*700^2164^2)/3^2=541456 So i wonder inf there are infinitely many primes pg(k) with k multiple of 86 with the property that k is congruent to 164^2 mod (268) as pg(541456) and pg(92020) More generally if k is multiple of 86 and k is multiple of 3, 69660 is the example k/3 is congruent to  164^2 mod 268 if k is not multiple of 3 and k is multiple of 86, then Examples are 92020 and 541456 are congruent to 164^2 mod 268 541456=271*9*999*218 271*9=2439 Maybe this explain the fact that 541456/41 and 51456/41 have a repenting term 2439 and so they are 10^m mod (700^2+239)*2^8/(271*9)=51456 (2439*1111)*2(2439*201239)=51456 From above reasoning it results that 541456 and 51456 are both congruent to 239 mod 245=7*25 In subastance here the formulas 541456=245*2211239 51456=245*211239 Where 211 is prime and 2211 is a multiple of 67 as 51456 (54145692020) is a multiple of 559 and 67 (54145692020)/(67*3)=2236 69660+22360=92020 But ed can substitute 67*3 with 2211/11 So (54145692020)*11/2211=2236 i note also that (541456210*1001+13)=331259 92020 is so congruent to (245*2211239) mod (3*67*43*13) and (245*2211239=541456) is congruent to (331259+13) mod 559 i think that it shuould be possible to prove that when pg(k) is prime and k is congruent to 6 mod 13, (examples known 215, 69660, 92020, 331259, 541456) then k is congruent to plus or minus (34413s) mod 559 with s=1 in the case of 331259 which infact is congruent to 331 mod 559 and s=0 in all the other cases where k is a multiple of 43 i think that something interesting could be found examining this formula: f(x)=((71*6^x+4)/10)*6^213 for x integer x=4 f(4)=331259 331259 congruent to 13 mod 9202 i think that the study of this function could shed light on these numbers 331259 is congruent for example to 7 mod 55211=f(3) 331259=92020+239239 note that also 239239 is congruent to 13 mod 9202 239239 is congruent to 546 mod 559 215*(239239546)/559+215=92020 I suspect that 546 is not a random residue infact there are primes pg(k) with k multiple of 546 for example pg(75894) is prime and 75894 is multiple of 546 modulo 559 75894 is 429 (92020=(429^21)/2) (239239546)/559+2=429 so( ((239239546)/559+2)^21)/2=92020 (239239+(12^21)*4)=429 Given the attention this question has received, it is disappointing that it is closed. I have a more concise answer that I cannot post other than as a comment: 245⋅22…11−239=245(2000*(10^m−1)/9+211)−239. Using the congruences: 245≡−1, 2000≡−9, 451≡0mod41 we get (−1)(−9)10m−19+(−1)(211)−239≡10m−1−211−239=10m−451≡10m 245 is congruent to 1 mod 41 modulo 41 we have 245*211=211 2392111 =451 congruent to 10^m mod 41 pg(451) is prime ((71*6^4+4)344)/559=164 but we saw before that 51456=(700^2164^2)/9 541456=(10*700^2164^2)/9 so we arrived to this: (700^2(((71*6^4+4)344)/559)^2)/9=51456 (10*700^2(((71*6^4+4)344)/559)^2)/9=541456 So multiple of 43 215 69660 92020 and 541456 are of the form s(71*6^4340)/52 +r with r being the residue 1763s+r such that is congruent to 10^m mod 41 34052=288 maybe this explain why 92020 is congruent to 288 mod 323??? I think yes because 92020=52*1763+344 (71*6^4+4) is congruent to 344 both modulo 1763 and modulo 559 (92020344)/559+(92020344)/1763=6^3 92020=(71*6^4+4) 331259=(71*6^6+4+10)/10 71*6^6 is congruent to 12^2 mod (71*6^4+4) 2236=(331*100181*143)/143 Anyway we have a fact: 92020 and 331259 are congruent to 71*3^3 mod (239*13)..so I wonder if there are infinitely many of such exponents By the way I note that 541456 is congruent to 71*3^3 mod (1001) 541456 is congruent to 331*1001+71*3^3 mod (1001*13) Or 541456 is congruent to 71*3^3 mod (11*1001) so 541456=71*3^3+11011n where 11011 is 3^3 in base 2 71*3^3 is congruent to 344 mod 143 I think that this could be useful It holds 541456(331259)+13210*1001=0 So this could be useful if you have in mind that 331259 is congruent to 71*3^3 mod (239*13) 92020=54145613*(449*771) 331259=92020+239*1001 it is clear that there is a relationship 541456 is congruent to (331259+71*3^3+111) mod (239*13) 331259=92020 mod(239*13) so 541456 is congruent to 2*71*3^3+111 mod (239*13) 92020 and 331259 are congruent to 71*3^3 mod (239*13) and to 71*3^2+13 mod (1001) 92020 and 331259 are congruent to 929 mod (1001) 541456 is congruent to (92913) mod 1001 69660*559 is congruent to (331259559331) mod (1001) 69660*559 is so also congruent to (92020559331) mod (1001) so 69660*559 is either congruent to (x890) mod 1001 or to (x903) mod (1001) where x=541456, 331259, 92020 numbers congruent to 6 mod 13 331259 and 92020 are 72 (mod 1001) and so 69660*559 is congruent to (x890) mod (1001) with x=331259, 92020 if x=541456 not congruent to  72 mod 1001, then 69660*559 is congruent to (x903) mod 1001 but this is equivalent to say 69660*559 is congruent to either (x +111) or (x+111+13) mod (1001) x=331259, 541456, 92020 541456, 331259, 92020 are congruent to either (13*2111) or to (13*3111) mod 1001 69660 is congruent to 591 mod 1001 591*559 is congruent to 39 mod 1001 69660/3 is congruent to (14^2+1) mod 1001 92020 and 331259 are congruent to (14^2111) mod 1001 541456 is congruent to (14^211113) mod 1001 so 69660 is congruent to 591 mod (1001) 331259 is congruent to 559*591111 mod (1001) 541456 is congruent to 559*59111113 mod (1001) 92020 the same as 331259 if i am not wrong so 69660 is (14^2+1)*3 mod 1001 331259 is congruent to 559*3*(14^2+1)111 mod 1001... (69660/81) is congruent to 141 mod 1001 so 541456 is congruent to (4*(14^2+1)86013) mod (1001) 92020 and 331259 to (4*(14^2+1)860) mod 1001 69660 is congruent to 3*(14^2+1) mod 1001 14^2+1=197 is a prime 69660/81=860 541456 is congruent to 3*86041*5*13 mod (1001) 331259 and 92020 are congruent to 3*86041*5*13+13 mod (1001) 69660 is congruent to 2*41*5 mod 1001 541456 is congruent to 34413*2^513 mod 1001 331259 and 92020 are congruent to 34413*2^5 mod 1001 541456 is congruent to (344+13*2^513*2^613) mod 1001 92020 and 331259 to (344+13*2^513*2^6) mod 1001 69660 to (344+13*2^5+13*2^6) mod 1001 (344+3*13*2^5) is congruent to 410 mod 1001 (34413*2^5) is congruent to 72 mod 1001 (34413*2^513) is congruent to 85 mod 1001 69660 is 410 mod 1001 331259 and 92020 are 72 mod 1001 541456 is 85 mod 1001 pg(2131) pg(2131*9=19179) and pg(92020) are probable primes 2131 is congruent to 1^2 mod 164 19179 is congruent to  3^2 mod 164 92020 is congruent to 4^2 mod 164 92020 is congruent to 43*2^3 mod (2132) and congruent to 43*3^2 mod 2131 19179 is congruent to 81 mod (2131+9) 2131 is congruent to 9 mod (2131+9) 92020 is a multiple of (2131+9) This implies that (920202131) is congruent to 9 mod 4280 (920202131*9) is congruent to 81 mod 4280 92020 is congruent to 2140 mod 4280 92020 s congruent to 2140 mod 8988 2131=p is prime pg(2131) is prime and pg(2131*9=19179) is prime pg(92020) is prime pg(69660) is prime 92020=69660+2236*10 92020 can be written both as (2236*43+344) and (2132*41+344) so 92020 leaves the same residue 344 mod (41*2236) and mod (2132*43) so 92020 is also congruent to 344 mod 1763 Also 541456 and 69660 multiple of 86 are 344 mod 2236 So the exponents multiple of 86 are 69660 92020 and 541456 They are congruent to 344 mod 2236 But only 92020 is congruent to 344 modulo 2236, mod 1763 and modulo 2132=p+1=2131+1 where this prime p 2131 gives the other two probable prime pg(2131) and pg(19179) 19179 is congruent to 214 mod 1763 92020 is a multiple of 214 2131*3^2*430 is congruent to 344 mod 1763 92020=214*430 so modulo 1763 2131*3^2=214 92020=214*430 congruent to 344 mod 1763 pg(331259) is prime 331259 is congruent to 6 mod 13 331259 is congruent to 344559*3 mod (2132) pg(56238) and pg(331259) are probable primes the difference 33125956238 is congruent to  7 mod (2132*43) 331259 is 7 mod 13, 56238 is 0 mod 13 331259+756238=43*2132*3 this is equivalent to 92020+239*1001+756238=43*2132*3 92020+239239=331259 239239 is congruent to (562383447) mod (2132*43) 92020 is congruent to 344 mod (2132*43) 331259 is congruent to (562387) mod (2132*43) 92020 is congruent to 344 mod 2132 331259 is congruent to 344*2+111 mod 2132 562387 is congruent to 344*2+111 mod 2132 pg(451=11*41) is prime pg(2131) is prime pg(92020) is prime 92020 is congruent to 2131*9*430 both mod 1763 and mod 451 2131*9 is 214 mod 451 and mod 1763 92020 is congruent to 215 mod (6149=11*43*41) 541456 is congruent to 344 mod 6149 (215+2131*91)/43=451 11*43*41 mod 2131 is 214 92020=214*430 Mod 2131 214*430 is 387 So 92020 is congruent to 387 mod 2131 (1763*11) mod 2131 is 71*3^3 331259 and 92020 have something tondo with 71*3^3 In fact 92020 and 331259 are congruent to 71*3^3 mod (239*13) 71*3^3=2131*1041*43*11 This means that 331259 and 92020 are congruent to (2131*10451*43) mod (239*13) and to (21310451*43+13) mod (1001) 541456 is congruent to (21310451*43) mod (1001) Consider this set of congruences: x==13 mod 214 x==6 mod 13 x+13=344 mod 43 Using CRT calulator I found the solution x=92007+119626k 92007+13=92020 and pg(92020) is prime 92007+119626*2=331259 and pg(331259) is prime another curio: pg(1323) is prime pg((1323*10+3)*3=39699) is prime 1323 is congruent to 215*3 mod 984 39699 is congruent to 215*3 mod 984 Pg(69660) is prime 69660 is multiple of 215*3=645 1323/3=441 69660 is congruent to 215*3+441 mod 984 Or to 215*3+21^2 mod 984 modulo 984, the number 69660 is a perfect square (42*9)^2 modulo 984 69660 is 1323*108=(42*9)^2 as you can see 69660 is 1323*108=(42*9)^2 mod (6^3) now i consider this modular equation 1323x is congruent to 6^3 mod 984 the soluton wolphram gives me is (40+328n) for n=1 you get 288 92020 is congruent to 288 mod 323 and 215, 69660 and 541456 are + or  108 mod 323 modulo 328: 21^2 is 215 69660=21^2*42^2 modulo 328 92020=107*42^2 modulo 328 215 and 69660 in a certain sense are perfect squares (21^2 and 378^2) 378^2 is congruent to 21^2 mod (42*323) 92020=428*215 428 is congruent to 13 mod 21^2 92020 is congruent to 215*10^2 mod (215*328) Ah ah this is weird Also 92020 is a perfect square 210^2 mod 328 Infact 428 is 10^ 2 mod 328 So mod 328 92020 is 210^2 1323 and 39669 are 11 mod 328 69660 modulo 328 is 11*108 Modulo 328 92020 is 11*4*(108^21) 69660 is congruent to 204 mod 984 204=64521^2 69660 is a multiple of 645 1323=42^221^2 1323*108 mod 984 is 204 as 69660 mod 984 69660 is congruent to 21^2645 mod 984 92020 is congruent to 13^2645 mod 984 1323 is 11 mod 328 39699 is 11 mod 328 69660 is (11+21^2) mod 328 1323 is 339 mod 984 39699 is 339 mod 984 69660 is (339+21^2) mod 984 69660 mod 984 is 780 42^2 also is 780 mod 984 this because 1323 is congruent to 339 mod 984 69660 is congruent to (339+21^2) mod 984 so 69660 is congruent to (3*21^2+21^2=42^2) mod 984 239*7*11 is 1 mod 172 69660 92020 and 541456 are congruent to (239*7*11171) mod 2236 I notice that 92020 and 331259 are both congruent to 5 mod (239*7*11) 331259 is congruent to (239*7*1117113) mod 2236 modulo 2236 infact 239*7*11171 is 344 239*7*11 is 515 mod 2236 so 239*7*11+1 so is 516 mod 2236 69660 is divisible by 516 92020, 541456 are 172 mod 516 331259 is 13 mod 516 69660 is congruent to 1323 mod (239*11) 69660 is congruent to 6 mod 13 using wolphram alpha solution is : 69660=1306+34177*2 92020=1306+22360+34177*2 331259=1306+22360+34177*9 what is quite clear is that there are k such that pg(k) is prime with k congruent to 6 mod 13 and k of the form 215+1001s287 (case 92020 and 331259) and 541456 which is of the form 215+1001s28713 so k is of the form 72+1001s or 85+1001s 215 is congruent to (92911*13) mod (7*11*13) 92020 is congruent to (929) mod (7*11*13) 331259 is congruent to (929) mod (7*11*13) 541456 is congruent to (92913) mod (7*11*13) 929 and 331259 are primes congruent to 72 mod (7*11*13) so mod 143 215 331259 92020 ...are plus or minus 71 mod 143 541456 is (7113=58) mod 143.. 215 is congruent to 786 mod (11*13*7) 786 is a number of the form 71+143*s 331259 and 92020 are congruent to 929 mod (11*13*7) 929 is of the form 71+143s so these exponents leading to a prime with k congruent to 6 mod 13 have this property 215 is congruent to (71+143s) mod 1001 541456 is congruent to (58+143s) mod 1001 331259 is congruent to (71+143s) mod 1001 92020 also I dont know 69660??? maybe not but 69660 is the only one multiple of 3 i note that 215 541456 331259 92020 69660 are congruent to plus or minus (71+r) mod 1001 where r is a 13smooth number which is not a 11 smooth number r infact can be one of these numbers: 845, 858 or 520 520=71+18^2+14^2 845=71+18^2+14^2+18^2+1 858=71+18^2+14^2+18^2+14 Last fiddled with by enzocreti on 20210115 at 14:35 Reason: 3 
