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
1017_{8} 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
17·31 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
2×5×937 Posts 
please show us a proof that they are prime

20200330, 12:50  #6 
Mar 2018
17×31 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
17·31 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 multiple of 43 (215, 69660, 92020, 541456) are congruent either to phi(323)=288 mod 323 or to +/ phi(324)=108 mod 323 phi is the Euler function 215eulerphi(324)=107 92020 is a multiple of 107, 69660 is 3 mod 107...the thing becomes too difficult multiple of 215=6^31 are 69660 and 92020 69660=324*215 92020=428*215 324 and 428 are numbers n such that neulerphi(n)=6^3 is it a chance that the other multiple of 43 (and not multiple of 215), that is 541456 is congruent to 6^2 mod 428? Another way to see the problem : 215 69660 92020 541456 are congruent to plus or minus K mod 323 Where K is either 215 or 288 215 and 288 are numbers of the form 41s+r where r is in the set (1,10,16,18,37) So 215 69660 92020 541456 are of the form 41a+b (b in the set 1,10,16,18,37)and congruent to + (41s+r) mod 323 215 and 288 are numbers of the form (3+n)*71+2^(n+1) for n non negative integer note that 12345679*323 divides (10^2881) 12345679*323*81 divides (10^(phi(323)1) phi(323)=288 92020/324 gives a repeating decimal period of 012345679 92020 is congruent to 288 mod 323 215, 69660, 541456 are of the form k*(41s+r)*323 + or minus 215, where r is in the set (1,10,16,18,37) and k some nonnegative integer 541456=323*1677215 215=0*(41s+r)+215 69660=323*215215 the only exception is 92020 which is congruent to 16 mod 41 92020 is of the form 323*284+phi(323) now pg(1323) and pg(39699) are probable primes 1323=441*3 39699=4411*3^2 numbers of the form 441...4411...are (397*10^n1)/9 1323 and 39699 are congruent to 645 mod (41*3*2^3) (397*10^n1) is divisible by 9209 for a certain n... 397*10^5((397*101)/9126)*10015^3 is a multiple of 92020 39699 and 1323 are of the form 328n^2+11 maybe there are infinitely many of such exponents 541456 is a multiple of 787 69660 is a multple od (78713=774) 541456(222*2021+78713)=92020 331259118*2021(78713*2)=92020 from the equation above 541456(222*2021+78713)=92020 if we work mod 787 0+715+13=728 so 92020 is congruent to (3^61) mod 787 92020 is also congruent to 43*2^n mod (78713) 9202069660=22360 is also congruent to 43*2^n=688 mod (78713) 69660 is congruent to (3^6118^2) mod 787 22360 is congruent to 18^2 mod 787 92020=69660+22360 92020 is congruent to 3^61 mod 787 69660 is divisible by 18^2 and it is of the form (2^j*3^k)18^2 with (2^j*3^k) congruent to 3^61 mod 787 69660 is also congruent to 18^2 mod (107*2) 92020 instead is a multiple of (107*2) 92020 is a multiple of 428 69660 is congruent to 18^2 mod 428 541456 is 0 mod 787 curious that pg(787428=359) is prime 215 69660 92020 541456 are +/ 344 mod 559 lcm(215,559,344)=22360 69660+22360=92020 22360 is of the form 787n+324 maybe working in the field F(787): the inverse mod 787 of 107 is 559 92020 is congruent to (107*73) mod (107*787) 541456=107*2942+787*288+6 541456=107*(17*19*10phi(17*19))+787*phi(17*19)+6 215 (because odd so signus ), 69660, 92020, 541456=G are solutions of the diophantine equation: 107x+787y+6=G I think that we think that the inverse mod 239 of 428 is 43... 43*4281=7*11*239 and that lcm(428,559)=239*7*11*13+13 then maybe we could explain why 541456+(72)+13 is congruent to 0 mod 1001 331259+72 is congruent to 0 mod 1001=7*11*13 92020+72 is congruent to 0 mod 1001 ... A strategy could be working in the field F(239) 107*172 is congruent to 1 mod (239*7*11) so the inverse of 172 mod 239 is 107 172=344/2 215 69660 92020 and 541456 are all congruent to +344 mod 559 92020 is a multiple of 107 331259=92020+239239 69660 92020 and 541456 are multiple of 172...but 172 in the field 239 is (107^(1)) multiple of 86 are 69660 92020 541456... 69660*(11/172) is congruent to 0 mod (3^2*5*19) 92020*(11/172) is congruent to 0 mod (3^2*5*19=855) 541456*(11/172) is congruent to 513=19*3^3 mod (3^2*5*19) 855 is congruent to 1 mod 107 69660*(11/172) is congruent to 9^2 mod 107 92020*(11/172) is congruent to 0^2 mod 107 541456*(11/172) is congruent to 3^2 mod 107 2^97^3 is congruent to 6^31 mod 107 2^9=19*3^3+1 6^3+7^3 is the famous 559 344=7^3+1 lcm((6^3+7^3),(7^3+1),(6^31))=22360 69660+22360=92020! curious that 559 is congruent to (12^9) mod 107 and to (1+2^9) mod 67 maybe this is not chance because there are primes pg(67s) like pg(51456) but i dont know 92020 for example is congruent to (6^3+2^9) mod 787 541456 is a multiple of 787 69660 is congruent to (6^3+2^918^2) mod 787 6^3+2^9=3^61 so 3^6 congruent to 27^3 mod 107 541456=344*1574=2^9*1574 mod 107 2^9 mod 107 is 84 1574 mod 107 is 76 so 541456=84*76 mod 107 but incredibly 541456=84*76 mod (107*10*2^9) and the difference 541456107*10*2^9 is divisible by 456 in other words: 541456=(6^3+7^3+2^91)*2^984*76 if you reduce the right side mod 107 6^3=2 7^3=22 2^91=83 2+22+83=107 69660 is congruent to (6^3+1)*(6^31) mod (5*43*107) infact 324 is congruent to (6^3+1) congruent to 3 mod 107 69660=324*215 so 69660 is congruent to (6^61) congruent to 3 mod 107 69660 is congruent to (6^61) mod (215*107) 215*107 is congruent to 1 mod (71*6^2) is this connected to the fact that for example 92020 is about 71*6^4??? 69660(6^61) divides 92020 92020=69660+22360 69660 is a multiple of 645 22360=215*107215*3 69660 is congruent to 645 mod (215*107) 69660 is congruent to 22360 mod (215*107) using wolphram (chinese remainder theorem) this leads to numbers of the form: 645+23005n 6^61 and 69660 are so nummbers of the form 645+23005n 92020 is divisible by 23005 pg(331259) is probable prime 331259 is congruent to 9203 (prime of the for 107n+1) mod (107*2151) ((215*1071)*12^2+12^2)/10(239239+13)=92020 and 541456 is congruent to the curious 12341 mod (215*107), 12341=7*41*43, this remainds me the famous 1763 69660 is congruent to 6^6 mod (71*6) congruent to 222 mod (71*6) 69660/3222/3786=22360 a rapid calculation from above leads 69660+1 congruent to 2^6*(6^3+1)+2^15 mod (215*107)... this yields 69660+1 is congruent to (13888+9763) mod (215*107) 69661138889763=46010 46010 is a divisor of 92020 69660=3^2*(71^2+71*36+71*2+1) so this leads to a polynomial x^2+36x+2x+1 71+6^2=107 so I arrived to 69660=3^2*(71^2+71*36+71*2+1) feom this I found this polynomial: (x+19)^29^2*10^2=0 whcih has solutions: x=71 x=109 71*109+1=6^2*(6^31) from this we obtain (9^2*10^219^2+1)=6^2*(6^31)=7740, which divides 69660 69660 has also the factorization 6^2*(44+1)*(441) pg(331259) is probable prime 331259 is congruent to (71*18^21)=215*1072 mod (456) and mod (26^2) maybe there is a link 513=2^9+1=19*3^3 and 456=2^3*3*19, the difference is 15*19 congruent to 1 mod 71 215*10715*19+6^2*(6^31)90^2=22360 where 215*107 15*19 and 6^2*(6^31) are 1 mod 71 pg(331259) is probable prime 331259=(13*(71*11*18^2+1)+71*18^2+1)/10 71*18^2+1=23005 which divides 92020 6966071*18^2=6^6 and pg(69660) is prime (13*(71*11*18^2+1)+71*18^2+1) is congruent to 13*10 mod (215*107) maybe it has something to do with the fact that 23005*143 is congruent to 1 mod 71 infact 23005 is 1 mod 71 and so 23005*143 is 1 mod 71 I don't know how to connect things but 107*73 is congruent to 1 mod 71 and so all the numbers of the form (73+71s)*107 are congruent to 1 mod 71... numbers of the form 73+71s are 215 and 428 maybe it is not a chance that there 92020 is divisible by 428 (and also 215) 92020 is also divisible by 23005 which is of the form 73+71s+1 215 and 428 in the field Z 71 have 36=6^2 as inverse maybe it is not chance that 541456 is 36 mod 107 and 92020 is (32436=288) mod 323 pg(359) is prime 35971=288 the famous 288=phi(323) pg(359) comes after pg(215) which is probable prime 359 and 215 are numbers of the form 71+144s 69660=71*324+6^6 if we rearrange 324*(71+144)=69660 I note that 359=19^22 pg(359) is prime 215, 69660, 92020, 541456 are congruent to plus or minus (19^273s) mod 323 infact 215=19^273*2 and 288=19^273 pg(51456) is prime 2^4*(73*49361)=51456 pg(67) is prime 215 is a number of the form 67+2+73s also 288 is a number of the form 67+2+73s maybe it is not a chance that also 1456 the final digits of 51456 and 541456 is of the form 69+73s (1456691)=1386 1386*239+5=331259 and pg(331259) is prime (13861001)*239+5=92020 and pg(92020) is prime 215 288 and 1456 in the field Z 73 have 18 as inverse...curious that 1456*181=73*359 and pg(359) is prime by the way 541456 is congruent to 1456 mod 288 108 and 288 are numbers of the form 54+18s 215 69660 92020 541456 are congruent to +/ (108 or 288) mod (18^21) also 774 is a number of the form 54+18s and 774 divides 69660 69660 is divisible by 18^2 which is a number of the form 54+18s, by 774 which is again a number of the form 54+18s, by 90 which is a number of the form 54+18s and by 108 which is of the form 54+18s 288 and 108 are two smooth numbers congruent to 18 mod 90 71*18+224^2+700^2+2=541456 71*18+224^2+2=51456 224^2+700^2 is congruent to 215*2=43*5*2 mod (71*18) 224^2=(67*35)*2^8 51456=67*2^8*3 curious that 224^2+1 and 700^2+1 are both prime 69660=(9*71+6^4)*6^2 645 divides 69660 92020=69660+22360 22360=23005645 23005=215*107 divides 92020 23005 is congruent to 1 mod 71 curious that there are pg(k) primes with k multiple of the reverse of 645=546 645=(18^21)*21 92020=215*428 Both 215 and 428 are congruent to 2 mod 71 The inverse of 215 in the ring F426 is 107 curious that pg(7) is prime , 7 is prime...the inverse of 7 in the ring F426 is 61...(61*71)/426=1 541456 is a multiple of 787 in the ring F426 the inverse is 367 367*787=288828 288828=6^2*(8*10031) (359*891)/426=75=(8*1003+1)/107 8*1003 is 1 mod 71 pg(359) is prime and the inverse mod 426 of 359 is 89 remark: in the ring F426 a^2 is congruent to 1 mod 426, for a=143 (143*1431)/426=288 lcm(426,288)=143^21 maybe there is a link to the fact that 541456429*10^3143^2+13=92020 in the ring Z428 (215*2151)/428=108 (211*711)/428=35 215 69660 92020 and 541456 are congruent to + or  (108 or 35) mod 323 pg(3371) is probable prime with 3371 prime pg(331259) is proabble prime with 331259 prime 3371=59+46*72 331259=59+46*10^2*72 I would conjecture that there are infinitely many prime pg(k) with k a prime of the form 59+72s. I note that 46*10^2+1=4601 which divides 92020 this is equivalent to primes of the form (72b13), as Peter Scholze pointed out I note that 331259 equiv (congruent) 999 mod (33711) 3371 and 331259 leave also the same resiude 13 mod 46 so they are primes of the form 59+1656x 1656 is a multiple of 46 naturally I think it is not a chance that 4601 divides 92020 72*460113 is 331259 so 3371 and 331259 are primes of the form 72*k13 with k congruent to 1 mod 46 because 331259=239239+92020 follows that (239239+13) is a multiple of 428 and (239239+13)/428=the famous 559 3371 and 331259 are primes of the form 59+3312x 331259=(4601*41)*13+4601*20 where 4601*20=92020 because 59 and 3312 are coprime we expect infinitely many primes of the form 59+3312x so there could be infinitely many pg(59+3312x) primes with 59+3312x prime I note that 3371, 331259, 92020 are all congruent to 5 mod 11 so 3371 and 331259 are primes of the form 589+792s curious that 589*792+1=683^2 putting all together we have that 3371 and 331259 are primes of the form 14845+18216s. infact 3371 and 331259 are 13 mod 46, 13 mod 72 and 5 mod 11 3371 and 331259 leave the same remainder 82 (mod 23*11) so they are also primes of the form 23*11*s+82 3371=82+253*(13) 331259=82+253*(13+6^4) 3371 and 331259 are primes of the form 59+414s where 414 is 1 mod 59 and 0 mod 23 4601 which divides 92020 is 1 mod 23 and 1 mod 59 so 3371 and 331259 are primes congruent to 59 mod (46*72) 3371 and 331259 are congruent to 13 mod 46 (337113)/46=73 which is congruent to 1 mod 72 (33125913)/46=19*379 which is congruent to 1 mod 72 19*379 is congruent to 51^2 mod 46 51^2 is congruent to 1 mod 26 4601 which divides 92020 is congruent to 1 mod 26 19*379+1 is a multiple of 26 4602 is a multiple of 26 and 59 19*379+14602+1=51^2 51^2 is congruent to 1 mod 26 51^2 is congruent to 2 mod 23 4601 is congruent to 1 mod 200 4601 divides 92020 19*379 is congruent to 1 mod 200 we can say 3371 and 331259 are congruent to 13 mod 46 3371 and 331259 are congruent to 13 mod ((46s+1)*72) 3371 infact is congruent to 13 mod (47*72) 331259 is congruent to 13 mod 4601*72 where 4601=46s+1 92020 is 0 mod 4601 72 is the least integer such that 4601*x is congruent to 1 mod 337 3371 is 1 mod 337 69660 (divisible by 215*3) is congruent to 215*3 mod 4601 4601 is congruent to 1 mod 59 22360 is congruent to 1 mod 59 4601 divides 92020 69660+22360=92020 774 divides 69660, 428 divides 92020 331259 is congruent to 13 mod (774*428) 774*428 is congruent to 1 mod 337 3371 is 1 mod 337 91*100 is congruent to 3371 mod 337 331259 is 12 mod 337 and 13 mod (428*774) 331259 and 3371 are primes of the form 59+3312s 3312 is congruent to 2 mod 331 92020=428*215 is 2 mod 331 428*774 is congruent to 59 mod 331 curious that pg(1323) is prime and 1323 is congruent to 1 mod 331 69660 is (6^4+4) mod 230 so 69660 is 200 mod 230 the inverse mod 230 of 43 is 107 so 5*324 is 10 mod 230 5*94 is 10 mod 230 5*47*2=5*94 the inverse mod 230 of 47 is 93 so 10 is congruent to 930 mod 230 92020 is congruent to 20 mod 920 43*107*72 is congruent to 1 mod 337 43*107*72 is congruent to 3371 mod 337 43*107*721=331271, the number is 3371 with a 12 sandwiched 33(12)71 the difference between 331271 and 3371 is 1093*300 where 1093 is a Wieferich prime 300 is the least integer such that 1093x is congruent to 1 mod 337 331259 is congruent to 12 mod 331271 and to 13 mod (43*107*72) 3371 is congruent to 1001 mod 1093 331259 is congruent to 100112 mod 1093 working mod (1093) 3371 is 1001 331259=3371+327888k working mod (1093) 331259=100112s 3371 is congruent to 12*28 mod 337 the inverse mod 337 of 28 is 325 3371*325 is congruent to 12 mod 337 so 3371*325 is congruent to 331259 mod 337 and so 331259 is congruent to 12 mod 337 Se know that 69660 is congruent to 6^6 mod 71 23004 (23005 divides 92020) has the form 288*3^x18^2 Also 69660 should have the form 288*3^x18^2 6^6 is 648 mod (71*324) (69660+648)/(288*3^2324)=31 69660 is divisible by 540=288*3324 given 69660 is congruent to 2^3*3^4 mod 213 after some passage : 111 is congruent to 2^7*(2^7+1) mod 213 this implies 324 is congruent to 2^7*(2^7+1) mod 213 and 69660=324*215 so is congruent to 2*(2^7+1)*2^7 mod 213 (2^7+1)=129 is a divisor of 69660 I notice that 128*129=2^7*(2^7+1) is congruent to 1 mod 337 so 3371 is congruent to (128*129+2) mod 337 I notice that 128*129+2 is a multiple of 359 and pg(359) is prime I notice that 128*129 is congruent to 7^3 mod 3371 3371 is congruent to 336*128*129 mod 337 336*128*1293371=5544661=10*(2^101)*271*2 331259 is congruent to 5544649 congruent to 325 mod 337 so 331259 is congruent to 11*83*6073 mod 337 and so 331259 is congruent to 239*7 mod 337 if we multiply by 143 143*331259 is congruent to 239239 mod 337 331259=92020+239239 because 239239 is (2^5+1) mod 337, then 33125992020 is conruent to (2^5+1) mod 337 92020 is 10 mod 9203 (this strange prime is 1 mod 43*107) 331259 is 9203 mod 23004=215*1071=71*... 331259 is 7^2 mod 9203 71*6^6+1 is congruent to 2131*1399 mod 9203 pg(2131) is prime 71*6^6 is congruent to 8699 mod 9203 71*6^6 is congruent to 8699 mod (9203*359) pg(359) is prime 331259 and 71*6^6 are both congruent to 72 mod 331 or (259) mod 331 (71*6^6+72)/331(331259+72)/331=9007=10^43*331 71*6^4 is congruent to 2 mod 331 and 92020 is congruent to 2 mod 331 in particular (71*6^6+72) is divisible by (920202=92018, which is multiple of 331) (331*1391)/2=23004=71*... 23005 divides 92020 139*331 is 1 mod (43*107) curious that 331259 is congruent to (9203*2^2+1) mod (331*139) (331*139215*107)+6^6=69660 331259 is congruent to 2000 mod 9007=10^43*331 71*6^6 also is 2000 mod 9007 curiously 92020 is congruent to (84^2+1) mod 9007 after a post on mathexchange I had a proof that: 6^(6+35j) for some j, has the form 648+23004s 69660 has the form 648+23004s 69660 is divisible by (6^56^2) 6^5+1=7777 6^5 is congruent to (6^2+1) mod 71 i didn't realized that 331259 is congruent to 9203 mod (71*324) 331259 and 9203 are primes 331259 and 9203 are both congruent to 44 mod 71 (3312599203)/71=4665 6^6=46656 4665 has in common with 46656 the first four digits 6^6 is congruent to 44^2 mod 2236 I remember that 69660=(44+1)*(441)*2 and 69660 is congruent to 6^6 mod 23004 331259=(2236*2+193)*71+44 193 is 1936=44^2 with a six truncated mod 2236 331259 is 331 69660=(44^21)*36=(193*10+6)*36 (6^61) is congruent to 1935 mod 2236 1935=(44^21) divides 69660 I looked at the factorization of 215, 69660, 92020, 541456 (multiple of 43 leading to a pg prime) 215=43*5 for example every prime in the factorization of one of these numbers is a quadratic residue mod 71. This is quite surprising. for example 43 is a quadratic residue mod 71 also 5 but also 787 which divides 541456 331259 is congruent to (4667) mod 6^6 4667 is a multiple of 359 (pg(359) is prime) 69660 is congruent to 345 mod 359 as 6^6 is congruent to 345 mod 359 69660+6^6 infact is 359*324=359*18^2 consider 71x is congruent to 1 mod 215 359x is congruent to 1 mod 215 x=109+215n 71*(215+109)=23004 71*109=7739 7739+1 divides 69660 23004+1 divides 92020 curious that (359*1091)/215=182 and 541456 is congruent to 182 mod 2131, pg(2131) is prime 182^2+1=33125 and pg(331259) is prime 359*(109+215x)1 for x=2 we have 193500 1935 divides 69660 (359*(109+215x)1)/10^x is an integer for x=2 69660 is congruent to 6^6 mod (71*324) and to 6^6 mod (359*324) pg(359) is prime the difference between 359 and 71 is the famous 288=phi(323) I think it's not chance that 92020 is congruent to 288 mod 323 288=2*6^6/324 in the ring F116316 69660 is congruent to 6^6 mod (116316) in the ring F116316 (116316=359*324) : 331259 is congruent to 133^2 mod (116316) pg(331259) is prime Last fiddled with by enzocreti on 20210416 at 09:16 Reason: 3 
