theorem Th14:
  a gcd p = 1 & (x^2 - a) mod p = 0 implies x,p are_coprime
proof
  assume that
A1: a gcd p = 1 and
A2: (x^2 - a) mod p = 0;
  assume not x,p are_coprime;
  then
A3: p divides x^2 by Lm2,INT_2:2;
  p divides (x^2 - a) by A2,Lm1;
  then p divides (x^2 - (x^2 - a)) by A3,Th1;
  then p gcd a = |.p.| by Th12
    .= p by ABSVALUE:def 1;
  hence contradiction by A1,INT_2:def 4;
end;
