How can I show that if $p^3$ divides $n^2$ then $p^2$ divides n? Can I generalise it?

Question

I found this question in a textbook and I thought about using FTA but I couldn't prove it 'rigourously'. I'm also interested if there's a generalisation of this, like the powers being different than 2 or 3.

(n being a natural number and p is prime.)

Answer 1

Suppose $n = p^a b$ where $p \not\mid b$.

If $p^3 | n^2$ then, since $n^2 = p^{2a}b^2$, $2a \ge 3$. Therefore $a \ge 2$, so $p^2 | n$.

More generally, suppose $p^m | n^k$. Since $n^k = p^{ak}b^k$, $m \le ak$ so $a \ge \frac{m}{k} $ so, since $a$ is an integer, $a \ge \lceil \frac{m}{k} \rceil $.

This is the case $m=3, k=2$.

Answer 2

Let the highest exponent of $p$ that divides $n$ be $a$ where integer $n\ge0$

$p^3|n^2\implies2a\ge3\implies a\ge1.5\implies a\ge2$ as $a$ is an integer

Answer 3

$\begin{align}{\rm By }\ \color{#c00}{\rm E}={\rm Euclid's\ Lemma\!:\ }\ \ \ \ \ \ \ \ \ &p\mid \color{#0a0}{p^{\large 3}\!\mid n^{\large 2}} \overset{\color{#c00}{\rm E}}\Rightarrow\, p\mid n,\ \ \text{so cancelling }\ \color{#0a0}{p^{\large 2}}\\ \overset{\large\color{#0a0}{(\ \ )/p^2}}\Longrightarrow \ &\color{#0a0}{p\mid(n/p)^{\large 2}}\overset{\color{#c00}{\rm E}}\Rightarrow\,p\mid n/p\,\overset{\times\, p}\Rightarrow\, p^{\large 2}\mid n \end{align}$

To generalize it - rather that using Euclid's Lemma to "peel off" one prime at a time as above - it is more efficient to simply compare powers of primes in their unique prime factorizations.