reserve a,b,d,n,k,i,j,x,s for Nat;

theorem Th17:
  ex k be Nat st 17 divides k = (10|^(16*n+9)-7)/3
proof
A1: 10|^(1+1) = 10*10|^1 by NEWTON:6;
  100 = 17*5 + 15;
  then 10|^2 mod 17 = 15 & 15*15 = 13*17 +4 by A1,NUMBER02:16;
  then (10|^2*10|^2) mod 17 = (15*15) mod 17 = 4 & 10|^2*10|^2 = 10|^(2+2) &
    16= 17*0+16 by NUMBER02:16,NAT_D:67,NEWTON:8;
  then
A2: (10|^4*10|^4) mod 17 = (4*4) mod 17 = 16 & 16*16 = 15*17+1 &
  10|^4*10|^4 = 10|^(4+4) by NEWTON:8,NUMBER02:16,NAT_D:67;
  then(10|^8*10|^8) mod 17 = (16*16) mod 17 = 1 &
  10|^8*10|^8 = 10|^(8+8) by NEWTON:8,NUMBER02:16,NAT_D:67;
  then
A3:(10|^16) |^n mod 17 = 1|^n mod 17 = 1 by GR_CY_3:30;
  10 mod 17 = 10 & 16*10 = 17*9 + 7 by NAT_D:24;
  then
A4:(10|^8*10) mod 17 = (16*10) mod 17 = 7 & 10|^8*10 = 10|^(8+1)
  by A2,NEWTON:6,NUMBER02:16,NAT_D:67;
  10|^(16*n+9) = 10|^(16*n)* (10|^9) by NEWTON:8
  .= (10|^16) |^n * (10|^9) by NEWTON:9;
  then 10|^(16*n+9) mod 17 = 1*7 mod 17 & 7 mod 17 = 7
    by A3,A4,NAT_D:67;
  then
A5: (10|^(16*n+9) - 7) mod 17 = (7-7) mod 17 = 0 by INT_6:7;
  value(<%1%>^((16*n+8)-->3),10) = (10|^(16*n+8+1)-7)/3 by Th16;
  then reconsider k = (10|^(16*n+9)-7)/3 as Nat;
  take k;
  3*k*6 mod 17 = ((6 mod 17) * 0) mod 17 = 0 & 3*6*k = 3*k*6
    by A5,NAT_D:67;
  then 0 = (18 mod 17)*(k mod 17) mod 17 & (17*1+1) mod 17 = 1
    by NUMBER02:16,NAT_D:67;
  then k mod 17 = 0;
  hence thesis by PEPIN:6;
end;
