reserve X for set;
reserve a,b,c,k,m,n for Nat;
reserve i,j for Integer;
reserve r,s for Real;
reserve p,p1,p2,p3 for Prime;

theorem
  n <> 0 implies not ex x,y being positive Nat st x*(x+1) = p|^(2*n)*y*(y+1)
  proof
    assume
A1: n <> 0;
    given x,y being positive Nat such that
A2: x*(x+1) = p|^(2*n)*y*(y+1);
A3: x+0 < x+1 by XREAL_1:8;
A4: x,x+1 are_coprime;
    p|^(2*n) divides p|^(2*n)*(y*(y+1));
    then p|^(2*n) divides x or p|^(2*n) divides x+1 by A2,A4,Th107;
    then x+0 >= p|^(2*n) or x+1 >= p|^(2*n) by NAT_D:7;
    then x+1 >= p|^(2*n) by A3,XXREAL_0:2;
    then
A5: 2*(x+1) >= 2*p|^(2*n) by XREAL_1:64;
A6: p|^(2*n) = p|^n|^2 by NEWTON:9;
A7: p|^n|^2 = (p|^n)^2 by NEWTON:81;
A8: (p|^n*(2*y+1)+(2*x+1)) * (p|^n*(2*y+1)-(2*x+1))
     = (p|^n*p|^n*(2*y+1))*(2*y+1) - (4*p|^n|^2*y*(y+1)+1) by A2,A6
    .= (p|^n)^2-1 by A7;
    then
A9: p|^n*(2*y+1)-(2*x+1) >= 0;
A10: p|^(2*n)-1 = (p|^n*(2*y+1)+(2*x+1)) * (p|^n*(2*y+1)-(2*x+1))
    by A6,A8,NEWTON:81;
    now
      assume
A11:  p|^n*(2*y+1)-(2*x+1) < 1;
      (p|^n*(2*y+1)-(2*x+1)) in NAT by A9,INT_1:3;
      then p|^n*(2*y+1)-(2*x+1) = 0 by A11,NAT_1:14;
      hence contradiction by A1,A10;
    end;
    then
A12: (p|^n*(2*y+1)-(2*x+1))*(p|^n*(2*y+1)+(2*x+1)) >= 1*(p|^n*(2*y+1)+(2*x+1))
    by XREAL_1:64;
    p|^n*(2*y+1)+(2*x+1) > 0+(2*x+1) by XREAL_1:6;
    then p|^(2*n)-1 > 2*x+1 by A10,A12,XXREAL_0:2;
    then p|^(2*n)-1+1 > 2*x+1+1 by XREAL_1:8;
    then 1*p|^(2*n) > 2*p|^(2*n) by A5,XXREAL_0:2;
    hence contradiction by XREAL_1:64;
  end;
