theorem PSQ:
  for p be prime Nat holds p divides (p+(k+1))*(p-(k+1)) implies k+1 >= p
  proof
    let p be prime Nat;
    p divides p*p; then
A1: p divides p|^2 by NEWTON:81;
A2: p|^2 = (p|^2 - (k+1)|^2)+(k+1)|^2;
    assume p divides (p+(k+1))*(p-(k+1)); then
    p divides p|^2 - (k+1)|^2 by NEWTON01:1;then
    p divides (k+1)|^2 by A1,A2,INT_2:1;
    hence thesis by NAT_D:7,NAT_3:5;
  end;
