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I have been trying to evaluate this GCD with different manners:

  1. For instance: $(x \wedge m)(x \wedge n) = \prod_{p \in \mathbb{P}} p^{\min\{ v_p(x), v_p(m) \}} \times \prod_{p \in \mathbb{P}} p^{\min \{ v_p(x), v_p(n) \}}$

I am not sure how to show that $\min\{ v_p(x), v_p(m) \} + \min\{v_p(x), v_p(n) \} = v_p(x)$.

  1. Also, I have been trying to rewrite $x$ as a divisor of $mn$ and extract some common factors without success.

  2. Euclid algorithm didn't work.

What is the evident way to show this without lengthy proof using decomposition in prime numbers of $x, m, n$?

EDIT2: The proof posted earlier was false.

Notations : $\wedge$ is the GCD. $\mathbb{P}$ is the set of prime numbers.

1 Answers 1

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Since $m\land n=1$, for each $p$ you have either $v_p(m)=0$ or $v_p(n)=0$. Since $x\mid mn$, $v_p(x)\le v_p(mn)=v_p(m)+v_p(n)$.

If $v_p(m)=0$, then $$ \min\{ v_p(x), v_p(m) \} + \min\{v_p(x), v_p(n) \} = 0+\min\{v_p(x), v_p(n) \}= \min\{v_p(x), v_p(mn) \}=v_p(x) $$ Similarly if $v_p(n)=0$.

Alternative proof. Since $m\land n=1$, also $(x\land m)\land(x\land n)=1$, $$ (x\land m)(x\land n)=(x\land m)\lor(x\land n) $$ By distributivity, $$ (x\land m)\lor(x\land n)=x\land(m\lor n)=x\land (mn)=x $$

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    Thank you, I wonder if a proof exists without prime factor decompositions? Using only GCD operations? Otherwise, nice! – 2017-01-09
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    @Raito Isn't this a proof with prime factor decompositions? I added an alternative proof – 2017-01-09
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    Sorry, I meant "without", I'll edit my post. – 2017-01-09
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    @Raito Already done. ;-) – 2017-01-09
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    Could you just detail or explain quickly why $m \wedge n = 1 \implies (x \wedge m) \wedge (x \wedge n) = 1$, it does not seems that immediate to me. Other than this step, this is really brilliant. (I understand why it would imply it, but I don't really see why it is immediate.) – 2017-01-09
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    @Raito If two numbers are coprime, their lcm is the same as their product. – 2017-01-09
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    I agree, but isn't this the second line : $(x \wedge m)(x \wedge n) = (x \wedge m) \vee (x \wedge n)$? I didn't understand the step about $(x \wedge m) \wedge (x \wedge n) = 1$ since $m \wedge n = 1$? – 2017-01-09
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    @Raito $\land$ is associative and commutative. Or, if you prefer, divisors of coprime numbers are coprime. – 2017-01-09