reserve a,b,c,k,m,n for Nat;
reserve p for Prime;

theorem Th35:
  not ex m,n being positive Nat st m^2 - n^2 = 4
  proof
    given m,n being positive Nat such that
A1: m^2 - n^2 = 4;
    now
      assume m^2 <= n^2;
      then m^2 - n^2 <= n^2 - n^2 by XREAL_1:9;
      then m^2 - n^2 <= 0;
      hence contradiction by A1;
    end;
    then
A2: m > n by SQUARE_1:15;
    per cases;
    suppose m is even;
      then consider a being Nat such that
A3:   m = 2*a;
      per cases;
      suppose n is even;
        then consider b being Nat such that
A4:     n = 2*b;
        4*a*a - 4*b*b = 4 by A1,A3,A4;
        then
A5:     a^2-b^2 = 1;
        a is positive & b is positive by A3,A4;
        hence thesis by A5,Th34;
      end;
      suppose n is odd;
        then consider b being Nat such that
A6:     n = 2*b+1 by ABIAN:9;
        4*a*a - (4*b*b+4*b+1) = 4 by A1,A3,A6;
        then a*a-b*b-b = 5/4;
        hence contradiction by Lm2;
      end;
    end;
    suppose m is odd;
      then consider a being Nat such that
A7:   m = 2*a+1 by ABIAN:9;
      per cases;
      suppose n is even;
        then consider b being Nat such that
A8:     n = 2*b;
        4*a*a+4*a+1 - 4*b*b = 4 by A1,A7,A8;
        then a*a+a-b*b = 3/4;
        hence contradiction by NAT_D:33;
      end;
      suppose n is odd;
        then consider b being Nat such that
A9:     n = 2*b+1 by ABIAN:9;
A10:    4*a*a+4*a+1 - (4*b*b+4*b+1) = 4 by A1,A7,A9;
        2*b+1-1 < 2*a+1-1 by A2,A7,A9,XREAL_1:9;
        then
A11:    b < a by XREAL_1:64;
        then b^2 < a^2 by XREAL_1:96;
        then a^2 - b^2 > b^2 - b^2 by XREAL_1:9;
        then
A12:    a^2 - b^2 >= 0+1 by INT_1:7;
        a - b > b - b by A11,XREAL_1:9;
        then a-b >= 0+1 by INT_1:7;
        then a^2 - b^2 + (a-b) >= 1+1 by A12,XREAL_1:7;
        hence contradiction by A10;
      end;
    end;
  end;
