SOLUTION: Prove that there are no positive integers x, y, z such that x^2+y^2 = 3z^2 .

Consider, one after one, these steps.


1.  Assume that both x and y are multiple of 3, and get a contradiction.  (It is easy)

       Hence, both x and y can not be multiple of 3 simultaneously.


2.  Assume that x is multiple of 3, and get a contradiction.  (It is easy)


3.  Assume that y is multiple of 3, and get a contradiction.  (It is easy)

       Hence, nor x neither y can be multiple of 3 even separately.


4.  Hence,  x  must have the remainder 1 or 2 when divided by 3: x = 3a + r,
       where  r = 1 or 2.


5.  Same for y:  y  must have a remainder 1 or 2 when divided by 3:  y = 3b + s,
       where  s = 1 or 2.

6.  Hence,      when squared,  has a reminder 1 when divided by 3.
    Similarly,  when squared,  has a reminder 1 when divided by 3.

    It implies that  gives the remainder 2 when divided by 3.

    But the right side of the equation is a multiple of 3. 
    Contradiction.

So, there is no way for the given equation to be valid.