# Question pertaining to the relationship between the GCD and LCM of 3 numbers.

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

$$(m,n)(m,k)(n,k)[m,n,k]^2=[m,n][m,k][n,k](m,n,k)^2$$

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

$$\frac{(m,n)(m,k)(n,k)[m,n,k]}{(m,n,k)}=mnk$$

And I suspect that this may be true as well:

$$\frac{[m,n][m,k][n,k](m,n,k)}{[m,n,k]}=mnk$$

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

So

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

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

• 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. – Naman Kumar Jan 14 at 15:33
• As far as I know, while $(m,n)[m,n]=mn$, this is not necessarily true for more than two numbers. – Naman Kumar Jan 14 at 15:42
• @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) – Yanko Jan 14 at 15:42
• I think your answer is correct: we can do this except we change $[m,n]$ only for the terms with two numbers. – Naman Kumar Jan 14 at 15:44
• @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. – Yanko Jan 14 at 15:46