
theorem LM204K:
  for p being Prime, j, m, k be Nat st m = p|^k & not p divides j
  holds j gcd m = 1
  proof
    let p be Prime, j, m, k be Nat;
    assume
    AS: m = p|^k & not p divides j;
    assume
    A1: j gcd m <> 1;
    set q = j gcd m;
    q divides j by NAT_D:def 5; then
    A4: j=q*(j div q) by NAT_D:3;
    q divides m by NAT_D:def 5;
    then consider n be Nat such that
    A5: q =p|^(n+1) by A1, LM204K1, AS;
    j = (p|^n)*p*(j div q) by A4, A5, NEWTON:6
    .= ((p|^n)*(j div q))*p;
    hence contradiction by AS, INT_1:def 3;
  end;
