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I'm trying to find positive integers $a$ and $b$ so that $\gcd(a,b) = 968$ and $\operatorname{lcm}(a,b)= 746105360$.

I know that $$\operatorname{lcm}(a,b) = \displaystyle \frac{a \cdot b}{\gcd(a,b)}$$

So if I substitute the values I want I have

$$746105360= \displaystyle \frac{a \cdot b}{968}$$

At this point my thought was to try to find another relationship between $a$ and $b$ so that I would have a system of equations that I could solve, but at this point I seem to be stuck.

Thank you

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    $\begingroup$ Prime factorization of each number would be especially useful. $\endgroup$ – Simply Beautiful Art Feb 22 '17 at 1:25
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    $\begingroup$ You just want one such pair $a$, $b$, or all? One such pair is simply $a=968$, $b=746105360$. $\endgroup$ – alex.jordan Feb 22 '17 at 1:34
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All that you need is : $$a=a'.d\\b=b'.d\\d=(a,b)=968\\(a',b')=1\\ lcm(a,b)=\dfrac{a.b}{gcd(a,b)}=\dfrac{a'd.b'd}{d}=a'b'd=770770\\$$ $$770770=770(1001)=7.11.2.5.(1001)=7.11.2.5.13.7.11=11^2.7^2.2.5.13 $$ so $$\begin{cases}a='1 & b' = 11^2.7^2.2.5.13 &\to {\begin{cases}a=a'd=1.968\\b=b'd=968. 11^2.7^2.2.5.13\end{cases}} \\ a'=11^2 & b'=7^2.2.5.13 \\a'=2 & b'=11^2.7^2.5.13 \\a'=5 &b'=11^2.7^2.2.13\\ \vdots\end{cases}$$

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  • $\begingroup$ Small error where you say $a'b'd=770770$, should be just $a'b'=770770$ $\endgroup$ – Joffan Feb 22 '17 at 1:45

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