theorem
for a,p be Integer holds
p|^(2*n+k) divides a|^2 implies p|^n divides a
proof
  let a,p be Integer;
  assume
  A1: p|^(2*n+k) divides a|^2;
  p|^(2*n) divides p|^(2*n+k) by NAT_1:11,NEWTON89; then
  p|^(2*n) divides a|^2 by A1,INT_2:9; then
  A2: (p|^n)|^2 divides a|^2 by NEWTON:9;
  |.p|^n.||^2 = |.(p|^n)|^2.| by TAYLOR_2:1
  .= (p|^n)|^2 gcd a|^2 by A2,NEWTON02:3
  .= ((p|^n) gcd a)|^2 by NEWTON027;
  hence thesis by POW1,NEWTON02:3;
end;
