theorem Th26:
  a mod p <> 0 implies Lege (a^2,p) = 1
proof
  assume a mod p <> 0; then
  not p divides a by INT_1:62; then
  not p divides a^2 by Th7; then
  a^2 mod p <> 0 by INT_1:62;
  hence thesis by Def3,Th9;
end;
