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Will really appreciate some guidance.

Let $p$ be a prime and $n$ be a positive number.

Then $p^a$ exactly divides $n$ if $p^a|n$, but $p^{a+1} \not \! | \; n$. We then write $p^a\|n$ if $a$ is the largest component of $p$ such that $p^a|a$.

Prove that if $p^a\|n$ and $p^b\|m$ then $p^{a+b}\|mn$.

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2 Answers 2

up vote 3 down vote accepted

Write $n=p^a u$ and $m = p^b v$ where neither of $u,v$ is divisible by $p$.

Then $$m n = p^{a+b} u v,$$ and from this your conclusion follows immediately.

EDIT: The OP wanted more details. If $p^{a+b+1}$ were to divide $mn$, then we would have $mn=p^{a+b+1}w$ where $w$ is an integer (perhaps itself divisible by $p$.) In that case, $$mn=p^{a+b}uv=p^{a+b+1}w,$$ so that on cancelling $p^{a+b}$ we would have $$uv=pw,$$ which would imply that $p$ divides one of $u,v$, whereas we assumed neither of $u,v$ were divisible by $p$ above.

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Thank you but I thought that from this I need to show that $p^{a+b+1} is not divisible by n and this is where I stuck –  Mary Nov 19 '12 at 14:57
    
I'll put a bit more explanation in the above answer. But you really should say you want to show $p^{a+b+1}$ does not divide $mn$, rather than that it is not divisible by $n$. –  coffeemath Nov 19 '12 at 15:17
1  
@coffeemath But that's the only nontrivial part of the problem (and also the essence of the matter), so it is reasonable to presume that's where the OP was stuck. –  Bill Dubuque Nov 19 '12 at 15:50
    
In my first version of my answer (before what is now in the "EDIT" part) I thought the "exactly divisible by" $p^{a+b}$ was clear since neither of $u,v$ are divisible by $p$. By his comment it seems to him it was not clear. –  coffeemath Nov 19 '12 at 16:39

Hint $\rm\,\ p^{a+b+1}\mid (jp^a)(kp^b)\:\Rightarrow\:p\mid jk\:\Rightarrow\: p\mid j\ \ or\ \ p\mid k,\ $ by $\rm\:p\:$ prime

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