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if m,n are positive integers such that; $3m+n=3 lcm(m,n)+gcd(m,n)$ prove that $n|m$

if $m=nk$ the result follows directly as $\gcd(m,n)=n$ and $lcm(m,n)=m$.

let $\gcd(m,n)=d$ then $$3m+n=\frac{mn}{d}+d$$

But i dont see any thing helpful from here.

Also i think the following inequalities may be helpful :

$\gcd(m,n)\le \min(m,n)$ and

$lcm(m,n)|\le mn$

Albus Dumbledore
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  • As here we can reduce to the case $,m,n,$ coprime where it becomes $\ 3m+n = 3mn+1,,$ which becomes $\ (3m-1)(n-1)= 0\ $ by completing the product, with obvious solutions. – Bill Dubuque Oct 14 '20 at 08:11
  • @BillDubuque thank you for the reference but i personally don't think this is a duplicate.Agreed the combination of sources you mention help,but again i think this is far from a duplicate – Albus Dumbledore Oct 14 '20 at 08:15
  • As I explained above, it follows immediately by applying the general method I describe in the first link, then completing the product. There is nothing at all novel in this instance. We already have many problems like this. – Bill Dubuque Oct 14 '20 at 08:19
  • @BillDubuque sorry ,i didn't know there were so many similar problems.I merely mentioned it as i have never seen questions being flagged duplicate in this fashion maybe because i am not as expereinced as you with this site. Thank you for mentioning this. – Albus Dumbledore Oct 14 '20 at 08:23
  • See here and here for more on abstract dupes. In short: in order for the site to flourish as a Q&A site it is necessary to stop if from being flooded with hundreds of minor variation of problems, e.g. homework questions, because that greatly complicates searching for the best answers - which is one of the primary purposes of the site. Generally it is best to delete dupes which add nothing novel, but "gamification" often gets in the way of that. – Bill Dubuque Oct 14 '20 at 08:28
  • @BillDubuque so should i delete the question? – Albus Dumbledore Oct 14 '20 at 09:01

1 Answers1

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let $d=\gcd(m,n)$. Let $m=dk,n=dl$, then dividing both sides of the equation $3m+n=3\cdot\mathrm{lcm}(m,n)+\gcd(m,n)$ by $d$ we get $$3k+l=3kl+1\iff (3k-1)(l-1)=0$$ which is true if either $k=\frac13$ or $l=1$. Now $k$ is an integer and cannoot be $\frac13$. So we must have $l=1$, or that $n=d$.

QED
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