reserve a,b,c,d,x,j,k,l,m,n for Nat,
  p,q,t,z,u,v for Integer,
  a1,b1,c1,d1 for Complex;

theorem
  a>=b & c|^n - b|^n = a|^n implies (c-b) gcd (a|^n) = c-b &
    (c-a) gcd b|^n = c-a
  proof
    assume
    A1: a>=b & c|^n - b|^n = a|^n;
    A1a: c|^n = b|^n + a|^n by A1; then
    A1c: c - b >= a - b by Th6,XREAL_1:9;
    a - b >= b - b by A1,XREAL_1:9; then
    A1f: c - b in NAT by A1c,INT_1:3;
    c - a >= a - a by A1a,Th6,XREAL_1:9; then
    A1i: c - a in NAT by INT_1:3;
    c-a divides c|^n - a|^n by Th32;
    hence thesis by A1,A1i,A1f,Th32,NEWTON:49;
  end;
