
theorem :: Problem 135
  for p being Prime st
    p+2 is Prime & p+6 is Prime & p+8 is Prime & p+12 is Prime & p+14 is Prime
      holds
   p = 5
  proof
    let p be Prime;
    assume
A1: p+2 is Prime & p+6 is Prime & p+8 is Prime & p+12 is Prime & p+14 is Prime;
    assume
FF: p <> 5;
    consider k being Nat such that
ZZ: p = 5*k or p = 5*k+1 or p = 5*k+2 or p = 5*k+3 or p = 5*k+4
      by NUMBER02:25;
    per cases by ZZ;
    suppose
S1:   p = 5 * k;
S2:   k <> 0 by S1;
      k <> 1 by FF,S1; then
      reconsider k as non trivial non zero Nat by S2,NAT_2:def 1;
      5 * k is composite by XPRIMES1:5;
      hence thesis by S1;
    end;
    suppose
      p = 5 * k + 1; then
      p + 14 = 5 * (k + 3);
      hence thesis by A1,XPRIMES1:5;
    end;
    suppose
      p = 5 * k + 2; then
      p + 8 = 5 * (k + 2);
      hence thesis by A1,XPRIMES1:5;
    end;
    suppose
      p = 5 * k + 3; then
      p + 12 = 5 * (k + 3);
      hence thesis by A1,XPRIMES1:5;
    end;
    suppose
S1:   p = 5 * k + 4;
      p + 6 = 5 * (k + 2) by S1;
      hence thesis by A1,XPRIMES1:5;
    end;
  end;
