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I need help with this excercise.

Let $p$ be a prime, and $n$ and $a$ positive integers. Then $p^a$ exactly divides $n$ if $p^a\mid n$, but $p^{a+1} \not \mid n$; we then write $p^a\parallel n$. Prove each.

If $p^a\parallel n$ and $p^b\parallel m$, then $p^{a+b}\parallel nm$.

I know that, if $p^a\parallel n$ and $p^b\parallel m$, then $p^a\mid n$ and $p^b\mid m$. Exist integers $X,Y$ such that

$$n=p^aX \hspace{1cm} \text{ and } \hspace{1cm} m=p^bY $$ then, $mn= p^{a+b}(XY)$.

Is it necessary to prove that $p \not \mid X$ and $p \not \mid Y$? If so, how can I prove it?

Bill Dubuque
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I think yes it is necessary to show that $p \nmid X$ and $p \nmid Y$ because u need to show $p^{(a+b)} || nm $ not just $p^{(a+b)} | nm $ and to show that u can simply use $p^{a+1} \nmid p^{a}X$ and $p^{b+1} \nmid p^{b}Y$.

Rishav Gupta
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  • As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community Mar 03 '22 at 16:28
  • Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Mar 03 '22 at 17:04
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Since we know that $p^a | n$ and $p^b | m$, there exist integers $N$ and $M$ such that $p^aN = n$ and $p^bM= m$. If for a contradiction $p^{(a+b)+1} | mn$, then there exists an integer $X$ such that $p^{(a+b)+1}X = mn$. But then we can replace $m$ and $n$ on the right hand side to obtain $$p^{(a+b)+1}X = (p^aN)(p^bM) = p^{(a+b)}MN$$ Dividing both sides by $p^{a+b}$, we obtain $pX = MN$, so $p | MN$. By Euclid's Lemma, $p$ must divide at least one of $M, N$. Suppose without loss of generality that $p |N$. Then, there exists an integer $k$ so that $pk = N \implies n = p^aN = p^a(pk) = p^{a+1}k$. Thus, $p^{a+1} | n$, a contradiction to the fact $p^a || n$.

Michael
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  • Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Mar 03 '22 at 17:04
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    I'm going to be honest - with the way the OP asked their question, I'd be willing to bet that they are unlikely to understand number theory at the level described in the duplicate post. Being able to phrase the same question at a more elementary level is likely to help more users. I doubt too many undergraduates taking number theory for their first time are going to be looking for posts titled "Properties of $p$-adic valuations"... – Michael Mar 03 '22 at 17:32
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    That's not an "honest" description of the linked answers. They employ no valuation theory. They are very elementary - at the same level of this question. Searches that locate that post needn't involve its title. No doubt there are also plenty of other dupes. – Bill Dubuque Mar 03 '22 at 18:00