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For example the lcm(7,15) = 105 and if we multiply 7 by 5 i.e., here our k is 5 then we see that lcm(35,15) is also 105. Hence we se that our k=5 is a factor of 15

How to prove this ?

Aman Gandhi
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    Maybe you know a formula that relates $\text{lcm}$ and $\text{gcd},$. – dxiv Jul 07 '21 at 06:15
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    Yes i tried that and was able to prove it that way..but i was wondering if there may be any different approach to prove it – Aman Gandhi Jul 07 '21 at 06:20
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    I suggest you edit the question and show what you tried, so that others don't repeat your work. This could also help the question survive the close votes for lack of context that started posting. – dxiv Jul 07 '21 at 06:25
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    Yes i would edit it...but it would be useful if you can give your proof using gcd and lcm formula. I would help me to learn how to write proofs – Aman Gandhi Jul 07 '21 at 06:28
  • I am not very comfortable in writing the proof in my own words. All i did was some scratch work – Aman Gandhi Jul 07 '21 at 06:30
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    ${\rm lcm}(\color{#c00}{kx},y)!=!{\rm lcm}(x,y)\mid xy,\Rightarrow, \color{#c00}{kx}\mid xy,\Rightarrow, k\mid y,,$ by the lcm universal property. $\ \ $ – Bill Dubuque Jul 07 '21 at 08:17
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    @AmanGandhi You get $,k \gcd(x,y)=\gcd(kx,y) \implies k \mid \gcd(kx,y) \mid y,$. – dxiv Jul 07 '21 at 17:27

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Edit(on noticing the increasing downvotes): I do understand that my answer seems a bit foolish, but please, if you did notice any errors, I'd just like to learn from you and note them down. Note that I had mentioned in this post that I am a beginner - I mean, I had learnt Euclid's Division Lemma, rational and irrational numbers (currently learning complex numbers), GCD-LCM relation for two numbers, etc. and a few were learnt by me a bit by myself, but that hasn't made me any better and I've got still a lot more to learn - so if you downvote, please do let me know how I can correct my answers or how I can do the problem/proof-writing right. That will always be a good lesson for me to learn from(as well as for the OP; from how quickly the OP had marked my reply as answer, it seems to me that the OP blindly believed me).

If you want to write proofs in number theory on your own, check this out:"Art Of Proofs", chapter in "Intermediate Number Theory" by Justin Stevens.

Okay, so now to the proof. Please note that I am not an expert in proof writing (neither have I touched the PDF in the ab ove link :D) and would like to know from you whether there's anything wrong with it.

We're given, $$lcm(x,y) = m\rightarrow(1)$$ and $$lcm(kx,y)=m\rightarrow(2)$$ From $(1)$ we can see that $$\exists a,b : ax = by = m \rightarrow(3)$$
Similarly, from $(2)$ as well we can see that $$\exists a',b' : a'kx = b'y = m\rightarrow(4)$$.
Equating $a'kx = by$ from the above inferences, we see that $k \mid by$.

Now, using the GCD-LCM formula in both the above equations $(1) \& (2)$, we can see that $$m\times \gcd(x,y) = xy$$ and $$m\times \gcd(kx,y) = kxy = k\times\gcd(x,y)\times m \\ \implies \gcd(kx,y) = k\gcd(x,y)$$.

Since $\gcd(kx,y) \mid y,\space k\gcd(x,y) \mid y$. In your question, it is a special case of this as $\gcd(7,15) = 1$. So if we assume $\gcd(x,y) = 1,\space \gcd(kx,y) = k$. I guess you can do the rest on your own, catching up from here.

I know I must have made serious errors worthy of strict scrutiny (I am in a hurry to have my supper), so please do let me know if I have gone wrong.

Spectre
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  • Riding my answer on yours due to similarity... $$\gcd(kx,y),|y\\frac{|kxy|}{\operatorname{lcm}(kx,y)}|y\\frac{|k||xy|}{\operatorname{lcm}(x,y)}|y\k\gcd(x,y),|y\k|y.$$ @amangandhi – ryang Jul 07 '21 at 17:42
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    Please refrain from answering poor quality, problem statement questions with no context. Search math.meta.se for "Enforcement of Quality Standards". – amWhy Jul 07 '21 at 23:32
  • @amWhy so should this answer be deleted? – Spectre Jul 08 '21 at 01:18
  • @amWhy: ??????? – Spectre Jul 08 '21 at 03:34
  • @AmanGandhi, please do verify before you jump into a conclusion on accepting this answer. I am also a beginner and I would like to have my answer reviewed – Spectre Jul 08 '21 at 05:19