
theorem Th8:
  for p being Nat holds
  (ex n being Element of NAT st n divides p & 1 < n & n < p)
  implies ex n being Element of NAT st n divides p & 1 < n & n*n <= p
proof
  let p be Nat;
  given n be Element of NAT such that
A1: n divides p and
A2: 1<n and
A3: n<p;
  per cases;
  suppose
A4: n*n<=p;
    take n;
    thus thesis by A1,A2,A4;
  end;
  suppose
A5: n*n>p;
    consider k be Nat such that
A6: p=n*k by A1,NAT_D:def 3;
A7: k*k<=p
    proof
      assume k*k>p;
      then (k*k)*(n*n)>p*p by A5,XREAL_1:96;
      hence contradiction by A6;
    end;
    take k;
    n/n<n*k/n by A2,A3,A6,XREAL_1:74;
    then 1<k*n/n by A2,XCMPLX_1:60;
    hence thesis by A2,A6,A7,NAT_D:def 3,ORDINAL1:def 12,XCMPLX_1:89;
  end;
end;
