Prove that if $m,n,l \in \Bbb N$ and $\operatorname{lcm}(m+l,m)=\operatorname{lcm}(n+l,n)$, then $m=n$ is the only satisfactory solution.

Question

I've already solved this question but I have done so by equating the lower bounds of the lcm function.

WLOG x>=y then x<=lcm(x,y)<=xy

m+l <= [m+l,m] <= m^2+ml and n+l <= [n+l,n] <= n^2+nl , since neither can be lower than the other (easily proved by contradiction) m=l

I wanted to know if this is actually allowed or if there is a counterexample to this method (since it seems so much more easier than forming equations).

Thanks

Answer 1

$m$ is one lower bound of $lcm(m+l,m) = lcm(n+l,m)$ but we don't know that it is the greatest lower bound.

$n$ is another lower bound of $lcm(m+l, m) = lcm(n+l, m)$ but we don't know that it is the greatest lower bound. And we don't know that it is the same lower bound.

We can not use that $m\le lcm(m+l,m)=lcm(n+l,m)$ and $n\le lcm(m+l,m)=lcm(n+1,m)$ to conclude $m= n$.

After all. If $lcm(m+l,m)=lcm(n+1,m) = K$ then $1,2,3,4,..........., K-2, K-1, K \le lcm(m+l,m)=lcm(n+1,m)$ and $m,n \in \{1,2,3,4,.....,K\}$ as are many other numbers. That's no reason to assume that all the numbers are equal to each other.

Answer 2

Your argument does not make sense. You have two bounds on the $\operatorname{lcm}$, but then assert $m=l$(which should be $m=n$). If $m=12,n=13,l=5$, for example, the bounds on the common $\operatorname{lcm}$ are $17 \le \operatorname{lcm} \le 204$ and $18 \le \operatorname{lcm} \le 221$. There is a lot of overlap. Nothing you have said prevents the $\operatorname{lcm}$ from being something like $100$ or $150$. There can't be a counterexample because the underlying claim is true, but the proposed proof is inadequate.