equality of LCM Let $a,b,c,d$ be natural numbers. I want to understand the necessary and sufficient condition on $a,b,c,d$ in order to have $[a,b] = [c,d]$ where $[m,n]$ denote the least common multiple of $m$ and $n$. 
Any help would be appreciated.
 A: I think the most direct and thorough way to put it:
Le $a = \prod p_{a,i}^{k_{a,i}}$ be the prime factorization of $a$ and $b = \prod p_{b,i}^{k_{b,i}}$ etc.
Then we must have $\{p_{a,i}\}\cup \{p_{b,i}\} = \{p_{c,i}\}\cup \{p_{d,i}\}$.  Or in other words, collectively $a$ and $b$ have the same prime divisors and $c$ and $d$ do collectively.
And and for each $i$ then $\max(k_{a,i},k_{b,i}) = \max(k_{c,i},k_{d,i})$. That is to say, the maximum power of each prime factor in $a$ or $b$ is the same as the maximum power for the prime factor in $c$ or $d$.
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Another possible answer:  $[a,b] = \frac {ab}{\gcd(a,b)}=a'b\gcd(a,b)$ where $a = a'\gcd(a,b)$ etc. So we need  $ a'b'\gcd(a,b) = c'd'\gcd(c,d)$.  That is an accurate answer to your question, but IMO, I think it begs the question as we must now determine, well, when will $ a'b'\gcd(a,b) = c'd'\gcd(c,d)$?  Which basically comes down to examing the prime factors of $a,b,c,d$ and comparing powers.
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Ex:  $72$ and $56$ have prime factors if $2,3$ and $2,7$, and max powers of $2^3, 3^2, 7^1$.  And $63, 8$ have prime factors of $3,7,2$ and max powers of $3^2,7^1, 2^3$.  So $[72,56] = [63,8] = 2^3*3^2*7 = 504$.
