
theorem Th27:
  for a,n be Nat holds a-1 divides a|^n-1
proof
  let a,n be Nat;
A1: 2|^n-1 is Nat by NAT_1:21,PREPOWER:11;
  per cases;
  suppose a <= 1;
    then a = 0 or a = 1 by NAT_1:25;
    hence thesis by INT_2:12;
  end;
  suppose a > 1;
    then
A2: a >= 1+1 by NAT_1:13;
    per cases by A2,XXREAL_0:1;
    suppose
A3:   a = 2;
      then a|^n-1 mod (a-1) = 0 by A1,RADIX_2:1;
      hence thesis by A3,INT_1:62;
    end;
    suppose
A4:   a > 2;
      then
A5:   a-1 > 2-1 by XREAL_1:9;
A6:   a-1 is Nat by A4,NAT_1:20;
      a mod (a-1) = (a+(a-1)*(-1)) mod (a-1) by NAT_D:61
      .= 1 by A5,PEPIN:5;
      then
A7:   (a|^n) mod (a-1) = 1 by A5,A6,PEPIN:35;
      (a|^n-1) mod (a-1) = ((a|^n mod (a-1))-(1 mod (a-1))) mod (a-1)
      by INT_6:7
      .= (1-1) mod (a-1) by A5,A7,PEPIN:5
      .= 0 by A6,NAT_D:26;
      hence thesis by A5,INT_1:62;
    end;
  end;
end;
