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I'm having a bit of trouble with this type of problem:

$∀a, b ∈ Z^+, gcd(a, b) = gcd(ab, lcm(a, b))$

How would you expand the RHS to eventually equal to the LHS or vice versa? And ultimately what is the correct way to prove this statement?

Owen Jordan
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    Well, the lcm of $a$ and $b$ always divides $ab$... so what is the gcd of two numbers,when one divides the other? (This statement is false, by the way). – Arturo Magidin Apr 07 '19 at 14:09
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    If $a=2, b=3$ then $\gcd(a,b)=1$ but $\gcd(6, 6)=6$. – lulu Apr 07 '19 at 14:10
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    It's always a good idea to work examples before trying to prove or disprove general statements. Here, the first things you try (like $(a,b)=(1,2)$) lead to counterexamples. Sometimes, of course, your examples confirm the general result but even in those cases it is useful as you may see steps in the particular example which you can generalize. – lulu Apr 07 '19 at 14:12
  • It is true that $\gcd(a,b) = \gcd(a!+!b,{\rm lcm}(a,b)),,$ so double check for typos in the statement. – Bill Dubuque Apr 07 '19 at 14:59
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Since $\operatorname{lcm}(a,b)$ is necessarily a factor of $ab,$ then $$\operatorname{gcd}\bigl(ab,\operatorname{lcm}(a,b)\bigr)=\operatorname{lcm}(a,b).$$ Thus, the claim amounts to saying that $$\operatorname{gcd}(a,b)=\operatorname{lcm}(a,b)$$ for all positive integers $a,b$. This will only be true when $a=b,$ however.

I highly recommend that, in the future, you work several examples to get a feel for general statements you're supposed to prove or disprove. You may find a counterexample (which, of course, disproves the general statement), or you may get a good feel for why it's true and how to prove it.

Cameron Buie
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