541456
 

541456 is 215 mod 307...




541456 and 215 are 10 mod 41...

51456 and 541456
 

41*43-307 is 1456


51456 and 541456 are congruent to 10^m mod 41

541456
 

541456=(307*3)^2-307*10^3+215


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


541456 is 215 mod 307


541456 and 215 have the same residue mod 41

307*3^2 is congruent to 10^3 mod (41*43)


541456 and 215 are multiples of 43 and have the same residue ten mod 41

so the above expression can be rewritten also as:


541456-215=307*(41*43+10^3)-307*10^3


541456*10 and 215*10 have the same residue 1 mod 307


last observation

541456=(307*3)^2-307*10^3+215where 10^3 is exactly the residue of 307*3^2 mod (1763=41*43). 2 is the least integer such that 307*3^m is confruent to 10^3 mod 1763

...541456...
 

observing that 41*43-307=1456
it follows that:


(540000)+41*43=(307*3)^2-307*10^3+522


(540000) is 522 mod (41*43)


and 522=215+307


540000-522=306*41*43=(307-1)*41*43

541456
 

541456=(307*3)^2-307*10^3+215




I observe that (307*3)^2 and 307*10^3 have the same residue (238) mod (1763=41*43)




(307*3)^2=1763*481+238
(307*10^3)=1763*174+238


481-174 again is 307


215=540000+41*43-307*42^2 or it is the same 215=540000+41*43-307*(41*43+1)


because 307*7+1=2150 it follows that


215=(7*3^2*307+9)/90 or it is the same 215=((7*(1763+10^3)+9)/90


so 541456=(307*3)^2-307*10^3+(7*3^2*307+9)/90
or is the same


541456=(307*3)^2-307*10^3+((999+42^2)*7+9)/90


1456=7*13*2^4


so 541456=540000+7*13*2^4


541456/2^4 is 344 mod 1763


so 541456/2^4 is of the form 1763s+344 so 1763s+344 is a multiple of 43


541456/7/13=238.,,,
307*10^3 is 238 mod 1763


7*13*2^4=1456=41*43-307

2^4 is the highest power of 2 dividing 541456. 541456/2^4 is again a number of the form 1763s+r with r=344
541456=(((307*9-10^3)*10+7)*307+1)/10
215=(307*7+1)/10


215=(307*7+1)/10
541456=(17637*307+1)/10


so 215 and 541456 are numbers of the form (307*A+1)/10 where A is a number 7 mod 1763


541456=1763*307+(307^4-1)/17017/10^3-307 where 17017=7*11*13*17.


17017 divides all the numbers of the form 307^(4+4k)-1
(307^4-1)/17017/10^3=522=215+307




(307^4-1)/10^3/17017 is congruent to 215 mod 307


215 divides numbers of the form (307^(4+60k)-1)/10^3/17017-307




(307^4-1)/10^3-85085*43=522*10^4+4219 where 4219 is a prime


the following identity:

541(prime)*10^3+3*2^3*19=307*31*43+1763*(2^2*19-1)

this is 541*10^3+456=541456=307*31*43+1763*(2^2*19-1) where 1763*(2^2*19-1) is 215 mod 307

and 307*31*43 is 215 mod 1763




215=(307-1)*42^2+6^3+41*43-307*(41*43+1)




so 541456=306*(42^2-1)+215+41*43


or 541456=307*(42^2-1)+215


50^2*(215+1)+41*43-307=541456


50^2*215 is 1548=43*6^2 mod 1763

pg(51456) and pg(541456)
 

pg(51456) and pg(541456) are probable primes




51456 and 541456 are both 10 mod 41.


51456=50000+41*43-307
541456=540000+41*43-307


the difference 540000-50000=490000 is a perfect square






both end with digits 1456=41*43-307


540000=11020*49+20
50000=1020*49+20


pg(215) and pg(541456) are prp


2150 and 5414560 are 1 mod 307 and 387 mod 1763=41*43


so 2150 and 5414560 have the same residue mod 307 and mod 1763 (215 and 541456 have the same residue mod 41)


the general solution to the equation

10x is congruent to 1 mod 307 is 307n+215

the general solution to the system

10x is congruent to 1 mod 307
x is 0 mod 43 is

x=13201n+215




541456=1763*308-6^2*43=6^3+42^2*307-308




the congruence system:


215+307x is congruent to 0 mod 43
215+307x is congruent to 10 mod 41


has the solution x=1763n


so pg(215) and pg(215+1763*307=541456) are both prp


2150 and 5414560 have the same residue 387 mod 1763


215+307x is congruent to 0 mod 43
215+307x is congruent to 1 mod 41 has the solution x=1763n+387


pg(92020) is prp


92020 is 1000 mod 41 and 92020 is 0 mod 43.


the solution to the system


215+307x is congruent to 0 mod 43
215+307x is congruent to 1000 mod 41 has the solution


x=1763n+1505


92020 mod 1505 is 215


1763=(3^2*307-10^3)


so x=(3^2*307-10^3)*n




10^(p-2) is congruent to 215 mod p


some solutions are p=2,5 and p=307




pg(215), pg(69660), pg(92020), pg(541456) are prp with 215, 69660, 92020, 541456 multiple of 43


215 is 215 mod 307
69660 is 278 mod 307
92020 is 227 mod 307
541456 is 215 mod 307


so the possible residues mod 307 are 215, 278 and 227


215+278+227 is -10^3 mod (215)


215+278+227=720
720+10^3=1720=215*8


and 1763 is 43 mod 1720






the system



x is congruent to 0 mod 43
x-10 is congruent to 0 mod 41
10x is congruent to 1 mod 307 is equivalent to


x is congruent to 0 mod 43
x-10 is congruent to 0 mod 41
x-215 is congruent to 0 mod 307


infact the values of x for both systems are:


x=215,541456, 1082697, 1623938,...


x=215+1763*307k


pg(215) and pg(541456) are prp

but pg(1082697) is NOT prp


anyway it coulkd be that there is another pg(215+1763*307k) prp


Finally 541456=215+10^3+41*43+307*(42^2-10)
Or more elegantly 541456=215+10^3+(42^2-1)+307*(42^2-10)




because 307*(42^2-10) is congruent to 763 mod 1763


the solution to

541456=215+10^3+1763x+763 is x=306