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