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Integers $a$ and $b$ are co-prime and $3\cdot b^2=a^2$.

$3\cdot b^2=a^2$, implies $a^2$ is divisible by 3 since, $3b^2$ is divisible by 3.

Is $a$ divisible by 3?

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    $\begingroup$ you don't even need $a$ and $b$ to be coprime to deduce. Since $3\mid a^2$, it automatically divides $a$. By the way, I think you're trying to prove that $\sqrt{3}$ is irrational. $\endgroup$
    – TBTD
    May 24 '19 at 14:41
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Yes, $a$ must be divisible by $3$. If the integer $a=3k+1$ or $3k+2$ for arbitrary $k$, (which are the cases in which $a$ is not divisible by $3$) then the remainder is $9k^2+6k+1\equiv1\pmod3$ or $9k^2+12k+4\equiv1\pmod3$.

Therefore, $a^2|3$ cannot occur if $a$ is not divisible by $3$, and thus, $a$ must be divisible by $3$.

However, the equation $3a^2=b^2$ cannot be solved for coprime integers, so if you consider that, then it is impossible for $a|3$ since there are no $a$ that fit the equation.

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Yep; in general, if $p|ab$ then $p|a$ or $p|b$ ($p$ is a prime). In your case, $3|a^2$ so $3|a$ or $3|a$; thus $a$ is divisible by $3$

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Hint $:$ Let $p$ be a prime number and $a,b$ be positive integers such that $p \mid ab.$ Then either $p \mid a$ or $p \mid b.$

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THEOREM: In general, if $p$ is prime and $p|n^2$, then $p|n$, where $n$$\in$$Z$.

The proof is based on the uniqueness of prime factorisation.

In your case, $3$ is a prime and $a$ is an integer. Since $3|a^2$, this implies that $3|a$

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