Is there a proof that if $d \mid mn$, where $m$ and $n$ are coprime, then $d=d_1d_2$ where $d_1 \mid m$ and $d_2 \mid n$, where the $d_i$ are comprime?
I was working on Project Euler and came across this fact but the link to the proof had expired.
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Any condition on $d_i$'s? – May 21 '15 at 04:51
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Yes, they are coprime. See this link – Demetri Pananos May 21 '15 at 04:56
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1@DemetriP If $d_i$ must be coprime, then you need a condition on $m$ and $n$. Else take $d=4$, $m=n=2$. – Erick Wong May 21 '15 at 04:57
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6|6.5, 6=3.2 but neither 3|5 nor 2|5. – Amey Deshpande May 21 '15 at 04:57
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My mistake, m and n are also comprime. – Demetri Pananos May 21 '15 at 04:57
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@AmeyDeshpande In your example, pick $6=6\cdot1$ instead of $3\cdot 2$. The question doesn't say it holds for all possible $d_1$ and $d_2$. – Erick Wong May 21 '15 at 04:58
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1This is actually true without any coprimality condition (for instance, $d_1 = (d,m)$ and $d_2 = d/d_1$ works), which gives a stronger statement than the one in the question (if $m$ and $n$ are coprime then any divisors of $m$ and $n$ are necessarily coprime). However, adding coprimality additionally makes the choice of $d_1$, $d_2$ unique. – Erick Wong May 21 '15 at 05:11
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I'm not familiar with the notation $(d,m)$. What does it mean? – Demetri Pananos May 21 '15 at 05:25
2 Answers
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$\color{#c00}{cd = mn}\,\Rightarrow\, (d,m)(d,n) = ((d,m)d,(d,\color{#c00}m)\color{#c00}n) = (d,\color{#0a0}{m,n},\color{#c00}c)\color{#c00}d = d,\,$ by $\color{#0a0}{(m,n)=1}$
Hence $\ d = (d,m)(d,n)\ $ where $\ (d,m)\mid m,\,\ (d,n)\mid n\ \ \ $ QED
Or $\,\ d\mid mn\iff d\mid dn,mn \color{#c0f}\iff d\mid (dn,mn)\overset{\color{#0a0}{\rm D\,L}} = (d,m)n \iff d/(d,m)\mid n$
Therefore $\ d\, =\, (d,m) \dfrac{d}{(d,m)} \ $ where $\ (d,m)\mid m\ $ and $\ \dfrac{d}{(d,m)}\mid n\ \ \ $ QED
Above we used $\, d\mid j,k\color{#c0f}\iff d\mid (j,k),\,$ the universal property of the gcd, and we also used $\,\rm\color{#0a0}{DL} =$ the gcd Distributive Law, both here and in the first proof.
Bill Dubuque
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I'm not familiar with the notation $(d,m)$. What does it mean? – Demetri Pananos May 21 '15 at 05:27
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@DemetriP $\ (x,y) := \gcd(x,y),$ is standard notation in number theory. $\ \ $ – Bill Dubuque May 21 '15 at 05:28
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The case where $m = n = 1$ is trivial.
Since $d \mid mn,$ we have $d = p_{1}^{a_{1}}\cdots p_{k}^{a_{k}}$ for some integer $k \geq 1$, some integers $a_{1}, \dots, a_{k} \geq 0,$ and some distinct primes $p_{1}, \dots, p_{k}$ such that $p_{j} \mid m$ or $p_{j} \mid n$ for all $1 \leq j \leq k.$ Suppose that $p_{1}^{a_{1}}, \dots, p_{i}^{a_{i}} \mid m$ and $p_{i+1}^{a_{i+1}}, \dots, p_{k}^{a_{k}} \mid n$ for some $1 \leq i \leq k.$ Let $d_{1} := p_{1}^{a_{1}}\cdots p_{i}^{a_{i}}$ and $d_{2} := p_{i+1}^{a_{i+1}}\cdots p_{k}^{a_{k}}$. Then we see that $(d_{1}, d_{2}) = 1,$ that $d = d_{1}d_{2},$ that $d_{1} \mid m,$ and that $d_{2} \mid n.$
Yes
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@OP Notice $,d_1 = (d,m),$ and $,d_2 = (d,n),$ as in my answer. The hypothesis $,(m,n)=1,$ is not needed, see the 2nd proof there. – Bill Dubuque May 21 '15 at 05:36
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@BillDubuque: Okay, thanks. If it was you that downvoted my answer, did you downvote it simply because of a redundant hypothesis? – Yes May 21 '15 at 06:15
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I did not downvote any answers here so I don't know the answer to that. – Bill Dubuque May 21 '15 at 12:54