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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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