Prove or reject this statement:
If $ a \mid bc $ then $\displaystyle \frac{a}{\gcd(a,b)} \mid c$
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Martin Sleziak
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Mahdi Khosravi
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3 Answers
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Hint: If $a\mid bc$ and $(a,b)=1$ then $a\mid c$. Use this to show:
$$ \displaystyle a \mid bc \implies \frac{a}{\gcd(a,b)} \mid \frac{b}{\gcd(a,b)}\times c \implies \frac{a}{\gcd(a,b)}\mid c $$ Because $\gcd\left(\frac{a}{\gcd(a,b)},\frac{b}{\gcd(a,b)}\right)=1$.
Anne Bauval
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Arash
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Didn't you add the extra condition that $(a,b)=1$ to the question? – Numbersandsoon Sep 29 '13 at 04:16
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There is no assumption about $\gcd(a,b)$,this works for general $a$ and $b$. – Arash Sep 29 '13 at 21:02
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HINT:
Let the highest power of prime $p$ that divides $a,b,c$ be $A,B,C$ respectively
As $a|bc, A\le B+C\iff A-B\le C\ \ \ \ (1) $
Now, the highest power$(D)$ of prime $p$ that divides $\displaystyle \frac a{(a,b)}=A-$min$(A,B)$
If $A\ge B,D=A-B\le C$ by $(1)$
Else $D=A-A=0$ which is always $\le C$
lab bhattacharjee
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@MahdiKhosravi, it proves that for all prime divisor of $c,$ the proposition is true, hence is the proposition – lab bhattacharjee Sep 28 '13 at 18:40
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@MahdiKhosravi, the sum of powers of any prime $p$ in $b,c$ must be $\ge$ the power that prime in $a$ – lab bhattacharjee Sep 28 '13 at 18:56
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$$ a|bc \implies a = \frac{\gcd(a,c) \gcd(a,b)}{\gcd(\gcd(b,c),a)} \ \frac{a}{\gcd(a,b)} = \frac{\gcd(a,c)}{\gcd(\gcd(b,c),a)} \implies \gcd(\gcd(b,c),a)|\gcd(a,c) ; & ; \gcd(a,c)|c$$
– S L Sep 28 '13 at 19:30