2

I honestly have no idea how to approach this question. I have done easier questions eg. show $gcd(a,b)lcm(a,b)=ab$ by letting $a=Alcm(a,b)$ and $b=Blcm(a,b)$ and then using the fact that $A$ and $B$ are coprime. However this approach only works in two variables. Any help with the question in the title would be greatly appreciated, as well as the thought process behind the answer so that I can approach similar problems.

Ben Martin
  • 1,885
  • 1
  • 9
  • 17

1 Answers1

2

For any prime $p$, let $p_a$ be its exponent in the prime factorisation of $a$, and similarily define $p_b, p_c$. Thus, for instance, $a = 2^{2_a}\cdot 3^{3_a}\cdot 5^{5_a}\cdots$, where most of these factors are $1$.

Now consider the exponent of $p$ in the prime factorisations of the left and right hand sides. On the left-hand side it's $\max(p_a, \min(p_b, p_c))$ and on the right-hand side it's $\min(\max(p_a, p_b), \max(p_a, p_c))$. We see that if $p_a<p_b, p_c$, then we get $\min(p_b, p_c)$ on both sides, and otherwise we get $p_a$ on both sides.

In either case they're both equal, and since $p$ was arbitrary this means that the prime factorisation of $\operatorname{lcm}(a, \gcd(b, c))$ is the same as the prime factorisation of $\gcd(\operatorname{lcm}(a, b), \operatorname{lcm}(a, c))$, so the two numbers must be equal.

Arthur
  • 204,511
  • I really do wish I could find a proof without resorting to prime factorisation (it doesn't feel elegant at all), but I could only prove that the left side divides the right side and not the other way around. – Arthur Jun 29 '18 at 08:30
  • Not elegant but it gets the job done and I can use this strategy in other problems. Thank you – Ben Martin Jun 29 '18 at 08:38
  • @B.Martin See (the dual of) this answer for a proof using only gcd laws (vs. prime factorizations). It boils down to the identity $(a,b,c)(ab,bc,ac) = (a,b)(b,c)(a,c)\ \ $ – Bill Dubuque Aug 29 '20 at 15:38