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# Let $p$ be a prime number. If $p$ divides $a^2$, then $p$ divides $a$, where $a$ is a positive integer.

**To do:**To prove $p$ divides $a$, if $p$ divides $a^2$. where $a$ is a positive integer.

**Solution:**

Let, $a=p_1.p_2.p_3.p_4.p_5.....\ p_n$

Where, $p_1,\ p_2,\ p_3,\ ....,\ p_n$ are prime numbers which are not distinct.

$\Rightarrow a^2=( p_1.p_2.p_3.p_4.p_5.....p_n). ( p_1.p_2.p_3.p_4.p_5......p_n)$

It is given that $p$ divides $a^2$

As known that every composite number can be expressed as product of unique prime numbers.

This means that $p$ is one of the numbers from $( p_1.p_2.p_3.p_4.p_5....p_n)$.

We have $a=( p_1.p_2.p_3.p_4.p_5....p_n)$ and $p$ is one of the numbers from $( p_1.p_2.p_3.p_4.p_5....p_n)$.

It means that $p$ also divides $a$.

Hence, it has been proved that if $p$ divides $a^2$, then it also divides $a$.

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