2
$\begingroup$

Question: If $p^{2k+1} | m^2$, show $p^{k+1} | m$.

Answer that was provided: If $p^k$ is the largest power of $p$ that divides $m$, then $p^{2k}$ is the largest power of $p$ that divides $m^2$. Hence if a power of $p$ larger than $2k$ can divide $m$, then $p^{k+1}$ surely also divides m.

Here is my question about the proof provided to me. If $p^{2k}$ is already considered to be the largest power of $p$ that divides $m^2$, how can we suggest there is an even larger power of $p$ that divides $m^2$? I guess I'm wondering if someone could add more detail to the above proof so that it is clearer to me. I would really appreciate that.

Thank you.

  • 1
    The two sentences in the answer should both be read as if preceded by "for any $k \in \Bbb{N}$". So the first sentence isn't constraining the $k$ in the second sentence. It would have been clearer to use $j$ rather than $k$ in the first sentence. – 2017-01-29
  • 0
    Thank you for the clarification! – 2017-01-29

1 Answers 1

3

If $p^k$ is the largest power of p that divides m, then $p^{2k}$ is the largest power of p that divvies $m^2$.

The confusion here lies with (re)using the same variable name for $k$.

The statement would have been easier to follow if it said: "let $p^a$ be the largest power of $p$ that divides $m$, then $p^{2a}$ is the largest power of $p$ that divides $m^2$".

Hence if a power of p larger than $2k$ can divide m, then $p^{k+1}$ surely also divides m.

Rephrased using the above: we know that $p^{2k+1}$ divides $m^2$, therefore $2k+1 \le 2a$. It follows that $a \ge k + \frac{1}{2}$ and, since both $a,k$ are integers, $a \ge k+1$, so in the end $p^{k+1} \mid p^a$ must divide $m$.

  • 1
    Thank you for the clarification! I think you answered another one of my questions previously, so I really appreciate all the help you're giving me. :) – 2017-01-29
  • 0
    @Cici Glad it helped. I may also point that yours was a well asked question, which invites answers on MSE. – 2017-01-29