- **enzocreti**
(*https://www.mersenneforum.org/forumdisplay.php?f=156*)

- - **Numbers of the form 41s+r**
(*https://www.mersenneforum.org/showthread.php?t=24012*)

Numbers of the form 41s+rLet be k an integer
I search the values of k such that: 1- k is a multiple of 43 2-k is of the form 41s+r where r can be one of these numbers: 1, 10, 16, 18, 37. s in the integers obviously Somebody can give me a routine for SAGE for exampel to find them out? |

41*21 = 1 (mod 43) so 41^(-1) = 21 (mod 43):
41s+1: --------- 41s+1 = 0 (mod 43) => 41s = -1 (mod 43) => s=(-1)*41^(-1) = (-1)*21 = -21 = 22 (mod 43). so s=22+43n and the numbers are then: 41*(43n+22)+1 = [U]1763n+903[/U] 41s+10: --------- 41s+10 = 0 (mod 43) => 41s = 33 (mod 43) => s=33*21 = 5 (mod 43) so s=5+43n and the numbers are then: 41*(43n+5)+10 = [U]1763n+215[/U] 41s+16: --------- 41s+16 = 0 (mod 43) => 41s = 27 (mod 43) => s=27*21 = 8 (mod 43) so s=8+43n and the numbers are then: 41*(43n+8)+16 = [U]1763n+344[/U] 41s+18: --------- 41s+18 = 0 (mod 43) => 41s = 25 (mod 43) => s=25*21 = 9 (mod 43) so s=9+43n and the numbers are then: 41*(43n+9)+18 = [U]1763n+387[/U] 41s+37: --------- 41s+37 = 0 (mod 43) => 41s = 6 (mod 43) => s=6*21 = 40 (mod 43) so s=40+43n and the numbers are then: 41*(43n+40)+37 = [U]1763n+1677[/U] So 1763n + m, where m is 215,344,387,903,1677 |

congruent to 6 or 7 mod 13And what if I search for values of k such that
k is multiple of 43 k is of the form 41s+r with r=1,10,16,18,37 then if k is even, k have to be congruent to 6 mod 13 if k is odd, k have to be congruent to 7 mod 13 ? |

[QUOTE=enzocreti;508421]And what if I search for values of k such that
k is multiple of 43 k is of the form 41s+r with r=1,10,16,18,37 then if k is even, k have to be congruent to 6 mod 13 if k is odd, k have to be congruent to 7 mod 13 ?[/QUOTE] :picard: The Pari-GP function chinese() [named in honor of the Chinese Remainder Theorem, or CRT] is made for such things. The conditions 43|k, k == 1 (mod 41), 2|k, and k == 6 (mod 13) give ? chinese([Mod(0,43),Mod(1,41),Mod(0,2),Mod(6,13)]) %1 = Mod(23822, 45838) while 43|k, k == 1 (mod 41), k == 1 (mod 2), and k == 7 (mod 13) give ? chinese([Mod(0,43),Mod(1,41),Mod(1,2),Mod(7,13)]) %2 = Mod(32637, 45838) and similarly for the others. |

[QUOTE=enzocreti;508421]And what if I search for values of k such that
k is multiple of 43 k is of the form 41s+r with r=1,10,16,18,37 then if k is even, k have to be congruent to 6 mod 13 if k is odd, k have to be congruent to 7 mod 13 ?[/QUOTE] 41s+1: 22919n+903, 22919n+9718 41s+10: 22919n+215, 22919n+14319 41s+16: 22919n+344, 22919n+9159 41s+18: 22919n+10965, 22919n+19780 41s+37: 22919n+8729, 22919n+17544 |

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