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This question already has an answer here:

$\sqrt {3} \in Q$. Then, $\sqrt{3} = \frac ab$ with the lowest term for $a,b \in Z$.

Then, $3b^2=a^2$, which implies that $a^2$ is divisible by 3.

That is, $a$ is also divisible by 3 (by fundamental theorem of arithmetic).

I don't understand here $a^2$ divisible by 3 implies $a$ divisible by 3.

Could you explain it?

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marked as duplicate by user296602, user223391, Mohammad Riazi-Kermani, Claude Leibovici, Paramanand Singh real-analysis Mar 6 '18 at 8:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Euclid's lemma. $\endgroup$ – K B Dave Mar 6 '18 at 0:13
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    $\begingroup$ Can $a^2$ be a multiple of three without $a$ being a multiple of three? Consider that $$(3n\pm 1)^2 = 3(3n^2\pm 2n)\color{red}{+1}.$$ $\endgroup$ – Jack D'Aurizio Mar 6 '18 at 0:17
  • $\begingroup$ The linked question has 2 instead of 3, but the proof is virtually identical. $\endgroup$ – user296602 Mar 6 '18 at 0:17
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Working with integers, recall that $cd$ divisible by a prime $k$ implies $c$ is divisible by $k$ or $d$ is divisible by $k$. $a^2$ divisible by 3 implies $a \cdot a$ divisible by 3. So, $a$ is divisible by $3$ or $a$ is divisible by $3$ (This statement occurs by taking the left $a$ and the right $a$; if you don't understand this, in the example mentioned let $c = a$ and $d=a$).

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    $\begingroup$ cd divisible by k implies c is divisible by k or d is divisible by k Only if $k$ is a prime.Otherwise take $c=d, k=c^2$ for example. $\endgroup$ – dxiv Mar 6 '18 at 0:27
  • $\begingroup$ It's simple. Thanks. $\endgroup$ – Sihyun Kim Mar 6 '18 at 0:29
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    $\begingroup$ @SihyunKim Be careful! This answer is wrong. $\endgroup$ – user296602 Mar 6 '18 at 0:33
  • $\begingroup$ why is it wrong?? $\endgroup$ – Sihyun Kim Mar 6 '18 at 0:39
  • $\begingroup$ What am I missing here? Last time I checked, 3 was still prime. $\endgroup$ – MJD Mar 6 '18 at 0:45
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Consider the prime factorization of $a$.

Suppose on the contrary that $3$ is not a factor of $a$. Then squaring $a$ will also not make $3$ appears in the prime factorization of $a^2$. Hence $3$ will not be a factor of $a^2$.

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