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I have three variables $x,y,z$, but I want to find the smallest positive $M$ for which $M*M*x*y$ is divisible by $2*z*M$ which is the same as modulo 0. Can I do this without needing to brute check all values of $M$?

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  • $\begingroup$ By $\ast$, do you mean multiplication? $\endgroup$ – Peter LeFanu Lumsdaine Dec 3 '12 at 18:56
  • $\begingroup$ @PeterLeFanuLumsdaine Yes $\endgroup$ – KaliMa Dec 3 '12 at 18:57
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$2Mz$ divides $M^2xy$ if and only if $2z$ divides $Mxy$. So the question is just: find the least positive $M$ such that $Mxy$ is divisible by $2z$.

But this can be rephrased as: find the least positive multiple of $xy$ that is also a multiple of $2z$. So we want $Mxy = \mathop{\mathrm{lcm}} (xy, 2z)$; so the solution is: $$M = \frac{\mathop{\mathrm{lcm}}(xy,2z)}{xy}.$$

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Hint $\rm\,\ 2zM\mid xyM^2\!\iff\! 2z\mid xy M\!\iff\! 2z\mid (xy,2z)M\!\iff\! \color{#C00}{2z/(xy,2z)}\mid M,\ \ \, $ (a,b) = gcd(a,b)

Or: $ $ said dually $\rm\!\iff\! xy,2z\mid xyM\!\iff\!\, [xy,2z]\mid xyM\!\iff\!\, \color{#0A0}{[xy,2z]/xy}\mid M,\ \ \ [a,b] = lcm(a,b)$

They're equivalent by the $\rm\,gcd * lcm\:$ formula $\rm\ ab = (a,b)[a,b] = gcd(a,b)\,lcm(a,b),\ $ so

$$\rm (xy,2z)[xy,2z]\, =\, xy\,2z\ \ \Rightarrow\ \color{#C00}{\dfrac{2z}{(xy,2z)}}\, =\, \color{#0A0}{\dfrac{[xy,2z]}{xy}}$$

Remark $\ $ You can find many more examples of such universal proofs of gcd and lcm properties in my prior posts. Learning these techniques makes the proofs mechanical, just like calculus makes calculating volumes mechanical.

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