2023/012) prove that there is no integer solution to $x^2 + y^2 = 3z^2$

We shall prove it by contradiction.

Let the smallest solution of the same be $(a,b,c)$ such that $a^2 + b^2 = 3c^2\cdots(1)$

Working in mod 3 we have $x^2 \equiv 0 \pmod 3\cdots(2)$ or  $x^2 \equiv 1  \pmod 3\cdots(3)$

in (1) RHS is divisible by 3 so LHS is also divisible by 3

For this to be true using (2) and (3) we must have 

$a^2 \equiv 0 \pmod 3$ 

and $b^2 \equiv 0 \pmod 3$ 

So a and b are multiples of 3 that is $a=3p$ and $b= 3q$ for some p and q

Putting in (1)

$3p^2 + 3q^2 = c^2$ or $c^2$ is multiple of 3 or c is multiple of 3

So hence $\frac{a}{3}, \frac{p}{3}, \frac{c}{3}$ is a smaller solution

Which is a contradiction

So equation does not have integer solution 

This method is known as proof by infinite descent