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I want to prove the following lemma:

Let $x$ be the LCM of three consecutive integers. Prove $x$ is always even.

This is my current thought process: We can express $x$ as LCM$(n,n+1,n+2)$ where n is an integer. Now, we can divide our proof on a case-by-case basis, by considering the parity of $n$. The first case is if $n$ is odd. Note that each consecutive pair of terms is co-prime. Also, because $n$ is odd, $n$ and $n+2$ are also co-prime because consecutive odd integers are co-prime. Hence, all three integers are co-prime and the LCM of these integers is $(n)(n+1)(n+2)$. The product of two odd numbers and an even number is an even number; hence, if $n$ is odd, $x$ must be even.

My question is whether or not there is a way to similarly generalize the case where $n$ is even? Is there a nice, closed-form expression for the LCM of three consecutive integers where the first and the last are even? If we can prove this, then the proof of the lemma would naturally follow.

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  • $\begingroup$ what about two consecutive integers? One has to be even and it must divide the (least) common multiple. Isn't this sufficient? $\endgroup$ – Mirko May 19 '17 at 22:20
  • $\begingroup$ Actually, you can prove that $x$ is a multiple of $6$. $\endgroup$ – steven gregory May 20 '17 at 0:19
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gt6989b's suggestion is the easiest way to prove the lemma.

That said, yes, there is a closed-form expression. If $n$ is even, then $n$ and $n+2$ share (only) the prime factor $2$. Moreover, one of those numbers is singly even; that is, one of the consecutive integers $n/2, (n+2)/2$ is odd. So $\mathrm{lcm}(n,n+2)=n(n+2)/2$ and $\mathrm{lcm}(n,n+1,n+2)=n(n+1)(n+2)/2$.

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HINT You don't need a closed expression for the LCM, just a definition. If $p$ is a prime and $p | m$ then $p | \mathrm{lcm}\{m,n\}$. So it suffices to show one of the three integers is even...

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