theorem Th70:
  for p be prime Nat st p = 2*n+1 holds
  p divides a or p divides a|^n-1 or p divides a|^n+1
  proof
    let p be prime Nat such that
    A1: p = 2*n+1;
    a|^(2*n+1) - a = a|^(2*n)*a - a by NEWTON:6
    .= a*(a|^(2*n) - 1)
    .= a*((a|^n)|^2 - 1|^2) by NEWTON:9
    .= a*((a|^n - 1)*(a|^n + 1)) by NEWTON01:1; then
    p divides a or p divides ((a|^n - 1)*(a|^n + 1)) by A1,Th58,INT_5:7;
    hence thesis by INT_5:7;
  end;
