If p divides a^2, then p divides a

Let $p$ be a prime and $a$ an integer. If $p$ divides $a^2$ , then $p$ divides $a$

Proof. Given that

p

divides

a^2

, so that means that:

\begin{align*} p \mid a^2. \end{align*}

Each integer can be written as a unique prime factorization. Therefore:

\begin{align*} a = \prod_{i = 1}^{n} p_i^{m_i}. \end{align*}

This means that:

\begin{align*} p \mid (\prod_{i = 1}^{n} p_i^{m_i})^2 \iff p \mid \prod_{i = 1}^{n} p_i^{2m_i} \end{align*}

and

p = p_i

for some

i \in \{1,2,\ldots,n\}

p

is prime. Therefore:

\begin{align*} p \mid \prod_{i = 1}^{n} p_i^{m_i} \iff p \mid a, \end{align*}

which concludes the proof.