I am a high school student self-studying Number Theory and came across this question in the book Challenge and Thrill of Pre-College Mathematics (For reference, $(m,n)$ means $\gcd(m,n)$ and $[m,n]$ means $\text{LCM}(m,n)$):

If $m$, $n$, and $k$ are any three positive integers, prove that


I was able to derive this identity by trial and error and then prove it mathematically:


And I suspect that this may be true as well:


Which would prove the proposition. However, I am unable to prove this statement and am unsure of its truth.

Please offer a hint to how I would go about proving the second part, or in case if my assumption is incorrect than the correct method of proof.


We can use that $[m,n]\cdot(m,n)=m\cdot n\Rightarrow [m,n]=\frac{mn}{(m,n)}$.


$$\frac{[m,n][m,k][n,k](m,n,k)}{[m,n,k]}=\frac{mn\cdot mk\cdot nk \cdot(m,n,k)}{(m,n)(m,k)(n,k)[m,n,k]\cdot mnk} = m^2n^2k^2 \cdot A $$

Where $A$ is the inverse of the term you calculated to be $mnk$.

  • $\begingroup$ Wait... I think you flipped the terms around. How did $[m,n,k]$ come in the numerator and $(m,n,k)$ in the denominator? Suggested an edit. $\endgroup$ – Naman Kumar Jan 14 at 15:33
  • $\begingroup$ As far as I know, while $(m,n)[m,n]=mn$, this is not necessarily true for more than two numbers. $\endgroup$ – Naman Kumar Jan 14 at 15:42
  • $\begingroup$ @NamanKumar You're right I can't switch these two. Hold on I'll try to correct it (If I won't succeed in 5 minutes I'll delete the asnwer) $\endgroup$ – Yanko Jan 14 at 15:42
  • 1
    $\begingroup$ I think your answer is correct: we can do this except we change $[m,n]$ only for the terms with two numbers. $\endgroup$ – Naman Kumar Jan 14 at 15:44
  • $\begingroup$ @NamanKumar Yes right! I got confused because this $A$ is the inverse of $mnk$. So now it's correct :P I wish you luck with your self-study. $\endgroup$ – Yanko Jan 14 at 15:46

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