# Doubt in proof that product of $3$ consecutive integers is divisible by $6$.

A common proof, but have a doubt in why only one conclusion is drawn from the analysis of the proof, as given below:
Prove that the product of $3$ consecutive integers is divisible by 3.
Let P$=n(n+1)(n+2)$, then P is divisible by both $2,3$. Hence, for suitable integer $k, m$, have that $2k=3m$. But as $(2,3)=1$, and $2\mid 3m$.
Therefore, $2 \mid m$. Thus $m=2s,$ and P$=3(2s)=6s$.

My doubt: have $2k = 3m$, so if $(2,3)=1$, it is the value of $k,m$ that makes the equality possible of the l.h.s & r.h.s values. so, why not say that $k \mid m$, or even better $k \mid 3$.

• $k|m$ or $k|3$ can't be concluded because it could be $k=9$ and $m=6$. – user525761 Feb 6 '18 at 14:09
• Basically you don't know that $k$ and/or $m$ are prime. – Joffan Feb 6 '18 at 14:14
• @Joffan Please elaborate, else vet: if either or both of $k,m$ are prime, then not possible to have that $k\mid 3$, or $m\mid 2$. But, there is one further rider on $m$, it cannot be odd (as $2$(=even)*$k$ = $3$(=odd)$*m \implies$ even*$k$= odd*$m$). So, doubt is only regarding $k$, an odd quantity. Hence, can very well say that $2\mid m$, as this is a special case with $m$ being even, and $2$ being the smallest even. I request you to take more complex cases: as $8k=3m$, now nothing can be said as $m$ can be $2$ or $16$. – jitender Feb 6 '18 at 14:25
• $k$ and $m$ are convenient "fillers" - you don't really need to know much about them to reach the proof result. The key is that $2$ and $3$ are coprime so $2\mid 3m \implies 2\mid m$ - or for the same result, $3 \mid 2k \implies 3\mid k$. In fact it's fairly easy to see, from $2k=3m$, that the only case for $k$ to be prime is $k=3, m=2$ – Joffan Feb 6 '18 at 14:35
• You don't have $k|3$ but you can conclude $3|k$ in the same way that you can conclude $2|m$ – Mark Bennet Feb 6 '18 at 14:41

We can conclude from $2 \mid 3m$ that $2 \mid m$ because we know that no prime factors of $2$ appear in the prime factorization of $3$ (i.e., $(2,3)=1$) so they must all appear in the prime factorization of $m$.
If we try to apply the same logic to say that $k \mid 3m$ therefore $k \mid m$, it doesn't go through; some prime factors of $k$ might appear in the prime factorization of $3$, so they might not all be in the prime factorization of $m$. Similarly, we can't say $k \mid 3$ because some prime factors of $k$ might appear in the prime factorization of $m$, so they might not all be in the prime factorization of $3$.