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If $\gcd(a,b) = 1$ and $a,b\mid x$ then $ab\mid x$.

My attempt: $\,a,b\mid x\Rightarrow ma = x = nb\,$ for integers $\,m,n,\,$ so

\begin{align*} x^2 &= (ma)(mb)\\\ x^2 &= (m^2)(ab)\\\ x &= \sqrt{m^2ab}\\\ x &= m\sqrt{a}\sqrt{b} \end{align*} Let $m$ be $k\sqrt{a}\sqrt{b}$. Then \begin{align*} x &= kab\Longrightarrow ab\mid x \end{align*}

Is this correct, if not can someone point me in the right direction?

Bill Dubuque
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fmunshi
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  • You know that $x=ma$ for some $m$; and you know that $x=kb$ for some $k$. You do not know that $m=k$. So you cannot conclude that $x^2=m^2ab$ (which would imply that $ab$ is a square). – Arturo Magidin Dec 03 '10 at 15:50

4 Answers4

11

You are given that $a$ divides $x$. Therefore, you can write $x = ma$ where m is an integer. You are also given that $b$ divides $x$. This implies that $b$ divides $ma$. But $b$ and $a$ are coprime. Therefore $b$ must divide $m$. So you can write $m = kb$ where $k$ is another integer. Therfore, you have $x = kab$.

svenkatr
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  • Thanks a lot for your help, I can't believe I missed such a simple substitution for x, I was assuming the proof to be more complicated than it actually was. – fmunshi Oct 29 '10 at 05:06
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$\gcd(a,b)=1$ so there are integers $s$ and $t$ such that $sa+tb=1$, whence $sax+tbx=x$. But $a$ divides $x$ so $x=ax'$ for some integer $x'$. As $b$ divides $x$, $x=bx''$ for some integer $x''$. Substituting on the left-hand side of the preceding equation yields $sabx''+tabx'=x$. Factor out the common $ab$ on the left to get $ab(ax''+tx'')=x$, so that $ab$ divides $x$.

vbm
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Chris Leary
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2

No. This is not correct.
1. Integers $m_1$ and $m_2$ in $x=m_1 a$ and $x=m_2 b$ can be different.
2. In the beginning of your proof all numbers are integers. However when you write "let m be k*sqrt(a)*sqrt(b)", you don't know, that k is integer too.

Right direction: use, the fact, that each number $y$ can be uniquely represented as $p_1^{\alpha_1} p_2^{\alpha_2}\dots p_n^{\alpha_n}$, where $p_i$ are primes and $\alpha_i$ are integers.

Fiktor
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  • I've already understood the question from svenkatr's response, although I'm curious as to how you were suggesting to solve the problem, do you mind elaborating on your "right direction" section? – fmunshi Oct 29 '10 at 05:07
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By Euclid's Lemma: $\rm\, \gcd(\color{#C00}{a,b})\! =\! 1,\,$ $\rm\ \overbrace{\color{#c00}{a\mid b}\,(x/b)}^{\!\!\large {\rm by}\,\ a,b\,\mid\, x}\ \Rightarrow\ a\mid x/b\, \overset{\!\!\times\, b}\Longrightarrow\ ab\mid x$

Or $\rm\, \ b,a\:|\:x\ \Rightarrow\ ab\: |\: ax,bx\ \Rightarrow\ ab\ |\ gcd(ax,bx) = x\gcd(\color{#c00}{a,b}) = x\, $ by gcd Distributive Law.

This is the special case $\rm\ gcd(a,b) = 1\ $ of $\rm\ gcd(a,b)\ lcm(a,b)\ =\ ab\ $ which has a similar proof that has an elegant view in terms of cofactor reflection.

See also the LCM Universal Property $\ a,b\mid x\iff {\rm lcm}(a,b)\mid x$

By induction this generalizes to lcm = product for pair coprimes (see also here), i.e.

$$ a_1,\cdots a_k\mid x\,\Rightarrow\, a_1\cdots a_k \mid x\, \ \ {\bf if}\,\ \,a_i\, \text{ are pair-coprime}\qquad$$

Bill Dubuque
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