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 even & n >= 1 & p/b>0 &p/a>0 &a*b
<>0 implies x=0 & y=n-root(p/b) or x=0 & y=-n-root(p/b) or x=n-root(p/a) & y=0
  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 even & n >= 1 and
A4: p/b>0 and
A5: p/a>0 and
A6: a*b<>0;
  n>=1 by A3;
  then
A7: n>0;
  per cases by A2;
  suppose
A8: x=0;
    then a*0 to_power n+b*y|^ n=p by A1;
    then a*0 +b*y|^ n=p by A7,POWER:def 2;
    then y|^ n=p/b by A6,XCMPLX_1:89;
    hence thesis by A3,A4,A8,Th4;
  end;
  suppose
A9: y=0;
    then a*x|^ n+b*0 to_power n=p by A1;
    then a*x|^ n+b*0=p by A7,POWER:def 2;
    then x|^ n=p/a by A6,XCMPLX_1:89;
    hence thesis by A3,A5,A9,Th4;
  end;
end;
