Step-by-step explanation:
The proof can be done by contradiction. Suppose both a, and b weren't even. So that a, and b are both odd. This means they both look like
[tex]a=2k+1,~~b=2l+1[/tex] (for some integers k and l)
So, let's compute what [tex]a^2-3b^2[/tex] would be in this case:
[tex]a^2-3b^2=(2k+1)^2-3(2l+1)^2=4k^2+4k+1-3(4l^2+4l+1)[/tex]
[tex]= 4k^2+4k+1-12l^2-12l-3=4k^2+4k-12l^2-12l-2 [/tex]
[tex]=4(k^2+k+3l^2-3l)-2[/tex]
which notice wouldn't be divisible by 4. This shows then that since [tex]a^2-3b^2[/tex] is divisible by 4, at least one of the integers a and b is even.
