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












1












$begingroup$


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.






number-theory elementary-number-theory greatest-common-divisor







share|cite|improve this question









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    1












    $begingroup$


    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.






    number-theory elementary-number-theory greatest-common-divisor







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      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.






      number-theory elementary-number-theory greatest-common-divisor







      share|cite|improve this question









      $endgroup$




      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.






      number-theory elementary-number-theory greatest-common-divisor




      number-theory elementary-number-theory greatest-common-divisor






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 14 at 15:08









      Naman KumarNaman Kumar

      8812




      8812






















          1 Answer
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          0












          $begingroup$

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



          So



          $$frac{[m,n][m,k][n,k](m,n,k)}{[m,n,k]}=frac{mncdot mkcdot nk cdot(m,n,k)}{(m,n)(m,k)(n,k)[m,n,k]cdot mnk} = m^2n^2k^2 cdot A $$



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






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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













          Your Answer





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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

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          0












          $begingroup$

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



          So



          $$frac{[m,n][m,k][n,k](m,n,k)}{[m,n,k]}=frac{mncdot mkcdot nk cdot(m,n,k)}{(m,n)(m,k)(n,k)[m,n,k]cdot mnk} = m^2n^2k^2 cdot A $$



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






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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


















          0












          $begingroup$

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



          So



          $$frac{[m,n][m,k][n,k](m,n,k)}{[m,n,k]}=frac{mncdot mkcdot nk cdot(m,n,k)}{(m,n)(m,k)(n,k)[m,n,k]cdot mnk} = m^2n^2k^2 cdot A $$



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






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          0












          0








          0





          $begingroup$

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



          So



          $$frac{[m,n][m,k][n,k](m,n,k)}{[m,n,k]}=frac{mncdot mkcdot nk cdot(m,n,k)}{(m,n)(m,k)(n,k)[m,n,k]cdot mnk} = m^2n^2k^2 cdot A $$



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






          share|cite|improve this answer











          $endgroup$



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



          So



          $$frac{[m,n][m,k][n,k](m,n,k)}{[m,n,k]}=frac{mncdot mkcdot nk cdot(m,n,k)}{(m,n)(m,k)(n,k)[m,n,k]cdot mnk} = m^2n^2k^2 cdot A $$



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







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 14 at 15:45

























          answered Jan 14 at 15:17









          YankoYanko

          6,5971529




          6,5971529












          • $begingroup$
            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




















          • $begingroup$
            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


















          $begingroup$
          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






          $begingroup$
          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




          $begingroup$
          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




          $begingroup$
          @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




          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




          $begingroup$
          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






          $begingroup$
          @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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