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I want to prove that for any $a,b\in\mathbb{Z}^+$ there are relatively prime $m,n\in\mathbb{Z}$ such that $mn=\text{lcm}(a,b)$.

Since $\gcd(a,b)\text{lcm}(a,b)=ab$, if either $$\gcd\left(\frac{a}{\gcd(a,b)},b\right)=1$$ or $$\gcd\left(\frac{b}{\gcd(a,b)},a\right)=1$$ we can put $m=\frac{a}{\gcd(a,b)}$ and $n=b$. I'm stuck on showing that one of these pairs are relatively prime.

Xena
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    Isn't $m= \mathrm{lcm}(a,b), n=1$ enough? This holds for every number $N=N \cdot 1$, and $N,1$ are coprime numbers. Now, do the same thing for $N=\mathrm{lcm}(a,b)$. – Crostul Dec 22 '16 at 10:49
  • Oh, sure :) I think my instructor meant $a,b,n,m\neq 1$ – Xena Dec 22 '16 at 10:53
  • If you require $a,b,m,n \neq 1$ this is hopeless. For example, take $a=b=4$. This number can be written only as $4=2.2=1.4$. The first two are not coprime, the other two have $1$. – Crostul Dec 22 '16 at 10:54
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    Can you clarify the question? If we can take $m$ or $n$ to be $1$ there's nothing to prove. If we can't then the claim is false. – lulu Dec 22 '16 at 14:13

2 Answers2

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You can always set $m = lcm(a,b)$ and $n = 1$ and your claim holds. If $lcm(a,b)$ has only one prime factor this is the only way to do it (this can happen as you don't require $a$ and $b$ to be relatively prime: for example, with $a=3$ and $b=9$, being $lcm(a,b) = 9$, or even worse, $a = b = 2$ which gives $lcm(a,b) = 2$: this can only be expressed as a product of two (positive) integers as $2\cdot1$).

If an integer has two or more distinct prime factors you can always express it as a product of two coprime factors which are greater than one, "splitting" the prime factorization into two different numbers. Moreover, if $a$ and $b$ are relatively prime then $lcm(a,b) = ab$ and choosing $m = a$, $n = b$ we have what you want.

chelivery
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Your second claim is incorrect. Note that for $a = 2^2 \cdot 3 = 12$ and $b = 2 \cdot 3^2 = 18$ we have $\operatorname{gcd}(a, b) = 6$, but neither $12/6 = 2$ and $18$, nor $18/6 = 3$ and $12$ are coprime.

Dominik
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