If $2a^2 = b^2$ then $2$ is a common divisor of $a$ and $b$?

Question

The question is:

Prove the statement or disprove it using a counterexample. If $2a^2 = b^2$, where $a,b\in \mathbb Z$, then $2$ is a common divisor of $a$ and $b$?

The only thing that works in this case is when $a=b=0$. And $2$ is a common divisor of $0$ . I'm not quite sure if that's a valid proof.

Answer 1

Hint: if $b^{2} = 2a^{2}$, then $b$ must be even (why?). Hence, $b = 2k$ for some $k \in \mathbb{Z}$. Thus, $2a^{2} = (2k)^{2}$ and...

Answer 2

You can also see this from the fact that $2$ is prime, and for integers primes $p | ab$ implies $p|a$ or $p|b$. Then $2|b^2$ implies $2|b$. But since $b^2$ is a perfect square, it must contain an even power of $2$ in its prime factorization. This implies $a^2$ must contain an odd number of copies of 2, so $a$ must be even. Then both a,b are even, and 2 is a common divisor. The case here for 2 same can be generalized to primes.