Let 4 variables a,b,c,d be non-negative integers in the domain [0,255] when they are multiplied by 255. That is to say, they are fractions in the domain [0,1] of a denominator 255 with non-negative integer numerators in the domain [0,255]. Examples of valid values are 1/255, 2/255, 3/255, etc.

The variables are related in one equation. I want to prove that there are no solutions to this equation, by which I mean there are no valid values for the 4 variables that will satisfy the equation.

ac+(1-a)bd   1
---------- = -
a +(1-a)b    2

Now I'm going to redefine a,b,c,d to be non-negative integers in the domain [0,255]. The equation will still hold if I add the denominator 255 to the variables.

   a   c           a     b   d
  --- --- + ( 1 - --- ) --- ---
  255 255         255   255 255     1
--------------------------------- = -
   a               a     b          2
  ---     + ( 1 - --- ) ---
  255             255   255


   ac      (255-a)bd
  ----- +  ---------
  255^2      255^3       1
---------------------- = -
   a       (255-a)b      2
  ---   +  --------
  255        255^2


  255ac + (255-a)bd
  -----------------
        255^3           1
--------------------- = -
  255a  + (255-a)b      2
  ----------------
        255^2


255ac + (255-a)bd      255^2        1
----------------- --------------- = -
      255^3       255a + (255-a)b   2


    255ac + (255-a)bd    1
---------------------- = -
255(255a  + (255-a)b)    2

_____________________________
|                           |
| 255  ac +   (255-a)bd   1 |
| --------------------- = - |
| 255^2a + 255(255-a)b    2 |
|___________________________|

a,b,c,d are non-negative integers in the domain [0,255].  Is it possible to prove that there are no solutions to this equation?

One way to determine this is to test all (255^4=4228250625) possible combinations, however I'm looking for a more compelling proof.

Both the numerator and denominator will each evaluate to a non-negative integer value. That being said, a part of the set of possible evaluated fractions will look like this:

1   2   3   4    5    6    7    8    9   10
- , - , - , - , -- , -- , -- , -- , -- , -- ...
2   4   6   8   10   12   14   16   18   20

The denominator must evaluate to an even number.

Here are some of the rules of parity (even or odd) arithmetic:

Addition/subtraction:
      Even Odd
     __________
Even |Even Odd
Odd  |Odd  Even

Multiplication:
      Even Odd
     __________
Even |Even Even
Odd  |Even Odd

The denominator has only two variables a and b that I need to worry about. Let's consider the possible cases of parity and see which combinations result in an even number.

255^2a + 255(255-a)b
(Odd)a + (Odd)((Odd)-a)b

a: Even; b: Even
(Odd)(Even) + (Odd)((Odd)-(Even))(Even)
(Even) + (Odd)(Odd)(Even)
(Even) + (Odd)(Even)
(Even) + (Even)
(Even)

a: Odd; b: Even
(Odd)(Odd) + (Odd)((Odd)-(Odd))(Even)
(Odd) + (Odd)(Even)(Even)
(Odd) + (Even)(Even)
(Odd) + (Even)
(Odd)

a: Even; b: Odd
(Odd)(Even) + (Odd)((Odd)-(Even))(Odd)
(Even) + (Odd)(Odd)(Odd)
(Even) + (Odd)(Odd)
(Even) + (Odd)
(Odd)

a: Odd; b: Odd
(Odd)(Odd) + (Odd)((Odd)-(Odd))(Odd)
(Odd) + (Odd)(Even)(Odd)
(Odd) + (Even)(Odd)
(Odd) + (Even)
(Odd)

Therefore, the denominator is only even with both a and b are even. Let's see the parity of the numerator with a and b both being even.

255ac + (255-a)bd
(Odd)(Even)c + ((Odd)-(Even))(Even)d
(Even)c + (Odd)(Even)d
(Even)c + (Even)d
(Even) + (Even)
(Even)

Therefore, the numerator must be an even number as well, reducing the set of possible evaluated fractions to those with even numerators:

2   4    6    8   10
- , - , -- , -- , -- ...
4   8   12   16   20