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This question already has an answer here:

Given that $$a\cdot b=gcd(a,b)\cdot lcm(a,b)$$

How can we find all the integer solutions $(a,b)$ if $gcd(a,b)=6$ and $lcm(a,b)=540$?

The first thing I did was factorizing using the fundamental theorem of arithmetic.

$$a\cdot b=gcd(a,b)\cdot lcm(a,b)=6\cdot540=2^3\cdot3^4\cdot5$$

I also know that $$gcd(a,b)=p_1^{min(a_1,b_1)}\cdot p_2^{min(a_2,b_2)}\dots p_n^{min(a_n,b_n)}$$ $$lcm(a,b)=p_1^{max(a_1,b_1)}\cdot p_2^{max(a_2,b_2)}\dots p_n^{max(a_n,b_n)}$$ where $a_1\dots a_n$ and $b_1\dots b_n$ are the exponents in the factorization of $a$ and $b$ respectively.

From there I am not sure where to go to obtain all the possible integers $a$ and $b$.

There is already a similar question on Mathematics SE, however the accepted solution is simply a hint towards the correct procedure to do it, while this question already acknowledged that hint.

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marked as duplicate by Sil, John Omielan, callculus, José Carlos Santos, A. Pongrácz Apr 27 at 22:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Since $a\times b=6\times540$, then $\dfrac a6\times\dfrac b6=\dfrac{540}6=90$. But $\gcd\left(\dfrac a6,\dfrac b6\right)=1$. So, find all pairs $(m,n)$ of coprime numbers such that $m\times n=90=2\times3^2\times5$. For each such pair, take $a=6m$ and $b=6n$. If, for instance, $m=45$ and $n=2$, then you get $a=270$ and $b=12$.

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HInt:

WLOG $$\dfrac aA=\dfrac bB=6\implies(A,B)=1$$

lcm$(a,b)=6AB=540\implies AB=90=2\cdot3^2\cdot5$ with $(A,B)=1$

WLOG $A<B\implies90= AB>A^2\implies A<10\iff1\le A\le9$

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