Question

If 4 divides a^2-3b^2, then at least one of the integers a and b is even.

Answer 1

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

$a=2k+1,~~b=2l+1$ (for some integers k and l)

So, let's compute what $a^2-3b^2$ would be in this case:

$a^2-3b^2=(2k+1)^2-3(2l+1)^2=4k^2+4k+1-3(4l^2+4l+1)$

$= 4k^2+4k+1-12l^2-12l-3=4k^2+4k-12l^2-12l-2$

$=4(k^2+k+3l^2-3l)-2$

which notice wouldn't be divisible by 4. This shows then that since $a^2-3b^2$ is divisible by 4, at least one of the integers a and b is even.