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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.

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  • $\begingroup$ Your statement is not a proof, but an example of when the statement is true. If you want to prove your statement is true, you must show it is true for all integers $a$ and $b$ where the hypothesis holds. If you want to disprove it, you only need show there is an $a$ and $b$ where it is not true. $\endgroup$ – Alex Wertheim Dec 7 '14 at 3:46
  • $\begingroup$ $2a^2=b^2$ for $a, b \in \mathbb Z \implies a=b=0$, so... $\endgroup$ – Macavity Dec 7 '14 at 4:00
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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...

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  • $\begingroup$ My pleasure! :) $\endgroup$ – Alex Wertheim Dec 7 '14 at 3:49
  • $\begingroup$ @AWertheim: Please see my edit. $\endgroup$ – Passing By Dec 7 '14 at 4:01
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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.

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  • $\begingroup$ How did you deduce $2|a^{2}$? Also, why does it follow that since $b^{2}$ is a perfect square, it must contain an odd number of copies of $2$? Did you mean $2|b^{2}$ implies $2|b$, which means that $2^{2} \mid b^{2}$? $\endgroup$ – Alex Wertheim Dec 7 '14 at 3:54
  • $\begingroup$ Sorry, I should have said $2|b^2$, let me edit. $\endgroup$ – Passing By Dec 7 '14 at 3:56

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