reserve a,b,c,h for Integer;
reserve k,m,n for Nat;
reserve i,j,z for Integer;
reserve p for Prime;

theorem
  not ex x,y,z being positive Integer st 4*x*y-x-y = z^2
  proof
    given x,y,z being positive Integer such that
A1: 4*x*y-x-y = z^2;
    consider p,k being Nat such that
A2: p = 4*k+3 and
A3: p is prime and
A4: p divides 4*(x-1)+3 by NUMBER05:3;
    reconsider p as Prime by A3;
A5: p > 1 by INT_2:def 4;
A6: p-1 <> 1 by A2;
A7: Euler p = p-1 by EULER_1:20;
A8: (2*z)|^(p-1) = (2*z)|^(2*(2*k+1)) by A2
    .= (2*z)|^2|^(2*k+1) by NEWTON:9;
A9: (2*z)|^2 = (2*z)^2 by WSIERP_1:1;
A10: (4*x-1)*(4*y-1) = (2*z)^2+1 by A1;
    then
A11: p divides (2*z)^2+1 by A4,INT_2:2;
    (2*z)^2,-1 are_congruent_mod p by A4,A10,INT_2:2;
    then (2*z)^2|^(2*k+1),(-1)|^(2*k+1) are_congruent_mod p by GR_CY_3:34;
    then
A12: (2*z)|^2|^(2*k+1) mod p = (-1) mod p by A9,NAT_D:64
    .= p-1 by NUMBER10:5;
    2*z,p are_coprime
    proof
      let w be Prime;
      assume that
A13:  w divides 2*z and
A14:  w divides p;
      w > 1 by INT_2:def 4;
      then w = p by A14,INT_2:def 4;
      then p divides (2*z)^2 by A13,INT_2:2;
      then p divides (2*z)^2+1-(2*z)^2 by A11,INT_5:1;
      hence contradiction by A5,WSIERP_1:15;
    end;
    hence contradiction by A6,A7,A8,A12,INT_2:def 4,EULER_2:18;
  end;
