# Finding all pairs of integers satisfying gcd(a,b) = 6 and lcm(a,b)=540 [duplicate]

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.

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

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 $$b=12$$.
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 $$AA^2\implies A<10\iff1\le A\le9$$