Is it not Vieta's

Question

How many ordered pairs of positive integers $(m,n)$ satisfy $\gcd(m,n) = 2$ and $\mathop{\text{lcm}}[m,n] = 108$?

I think that it's all even pairs of factors of 216. I think that the answer is 10. Since gcd = 2, and lcm = 108. m * n = 216. Can someone please help me find all possible pairs of (m,n)?

Answer 1

See, if the $\gcd$ is $2$ and the $\operatorname{lcm}$ is $108$, then we know the product of the numbers is $216$. So one way, would be to just pairs of factors which fit this criteria.

$1,2,3,4,6,8,9,12,18,24,27,36,54,72,108,216$ are the factors. The $n$th number from the left and $n$th number from the right multiply to $216$. You can see, after some effort, that $(2,108)$, $(4,54)$ are the only pairs of numbers which satisfy the criteria. So there are only two such ordered pairs.

How can you reduce the amount of effort above? Well, if the $\operatorname{lcm}$ is $108$, then one of the numbers has to be a multiple of $27$(since $27$ is the largest prime power dividing $108$). Furthermore, since the $\gcd$ is $2$, such a number is a multiple of $2$, therefore a multiple of $27 \times 2 = 54$. This leaves very few numbers in the list above, so you need not check each one individually.

Answer 2

Let $\dfrac mM=\dfrac nN=(m,n)=2$

$\implies(M,N)=1\ \ \ \ (1)$

$[M,N]=54=MN$

Now $54=1\cdot54,2\cdot27,3\cdot18,6\cdot9$

Now use $(1),$ and we need ordered pairs