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 $k \mid 3$.
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 $3$.