reserve x,y,a,b,c,p,q for Real;
reserve m,n for Element of NAT;

theorem
  a*x|^ n+b*y|^ n=p & x*y=0 & n is odd & a*b<>0 implies x=0 & y
  =n-root(p/b) or x=n-root(p/a) & y=0
proof
  assume that
A1: a*x|^ n+b*y|^ n=p and
A2: x*y=0 and
A3: n is odd and
A4: a*b<>0;
  consider m being Nat such that
A5: n=2*m+1 by A3,ABIAN:9;
A6: n>0 by A5;
  now
    per cases by A2;
    suppose
A7:   x=0;
      then a*0 to_power n+b*y|^ n=p by A1;
      then a*0 +b*y|^ n=p by A6,POWER:def 2;
      then y|^ n=p/b by A4,XCMPLX_1:89;
      hence thesis by A3,A7,POWER:4;
    end;
    suppose
A8:   y=0;
      then a*x|^ n+b*0 to_power n=p by A1;
      then a*x|^ n+b*0=p by A6,POWER:def 2;
      then x|^ n=p/a by A4,XCMPLX_1:89;
      hence thesis by A3,A8,POWER:4;
    end;
  end;
  hence thesis;
end;
