Prime number lemma proof needed

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I would like to prove that the maximum divisor of a number $n$ (excluding $n$ itself), is always divisible by the minimum prime factor of $n$ if and only if $n$ itself is divisible by the square of it's minimum prime factor.




number-theory




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  • And what have you done towards accomplishing this? Can you prove it in one direction?
    – saulspatz
    Jul 26 at 19:27










  • Let $k<n $ be the maximum proper divisor of $n$. Then $kr=n$ for some $r>1$. Then - what can you say about $r$?
    – Joffan
    Jul 26 at 19:33











  • I intuitively believe it to be true and if a proof exists then I would prefer to see another more experienced person's proof rather than muck about with my own. In the meantime I will try on my own, but why would a waste times asking everyone else?
    – Adam
    Jul 26 at 19:36










  • $r$ would have to be the minimum proper divisor of $n$
    – Adam
    Jul 26 at 19:37











  • Yes (provided $n$ is not prime) - can you prove that $r$ must be prime?
    – Joffan
    Jul 26 at 19:39















up vote
0
down vote

favorite












I would like to prove that the maximum divisor of a number $n$ (excluding $n$ itself), is always divisible by the minimum prime factor of $n$ if and only if $n$ itself is divisible by the square of it's minimum prime factor.




number-theory




share|cite|improve this question





















  • And what have you done towards accomplishing this? Can you prove it in one direction?
    – saulspatz
    Jul 26 at 19:27










  • Let $k<n $ be the maximum proper divisor of $n$. Then $kr=n$ for some $r>1$. Then - what can you say about $r$?
    – Joffan
    Jul 26 at 19:33











  • I intuitively believe it to be true and if a proof exists then I would prefer to see another more experienced person's proof rather than muck about with my own. In the meantime I will try on my own, but why would a waste times asking everyone else?
    – Adam
    Jul 26 at 19:36










  • $r$ would have to be the minimum proper divisor of $n$
    – Adam
    Jul 26 at 19:37











  • Yes (provided $n$ is not prime) - can you prove that $r$ must be prime?
    – Joffan
    Jul 26 at 19:39













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I would like to prove that the maximum divisor of a number $n$ (excluding $n$ itself), is always divisible by the minimum prime factor of $n$ if and only if $n$ itself is divisible by the square of it's minimum prime factor.




number-theory




share|cite|improve this question













I would like to prove that the maximum divisor of a number $n$ (excluding $n$ itself), is always divisible by the minimum prime factor of $n$ if and only if $n$ itself is divisible by the square of it's minimum prime factor.





number-theory





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 26 at 19:50
























asked Jul 26 at 19:24









Adam

11411




11411











  • And what have you done towards accomplishing this? Can you prove it in one direction?
    – saulspatz
    Jul 26 at 19:27










  • Let $k<n $ be the maximum proper divisor of $n$. Then $kr=n$ for some $r>1$. Then - what can you say about $r$?
    – Joffan
    Jul 26 at 19:33











  • I intuitively believe it to be true and if a proof exists then I would prefer to see another more experienced person's proof rather than muck about with my own. In the meantime I will try on my own, but why would a waste times asking everyone else?
    – Adam
    Jul 26 at 19:36










  • $r$ would have to be the minimum proper divisor of $n$
    – Adam
    Jul 26 at 19:37











  • Yes (provided $n$ is not prime) - can you prove that $r$ must be prime?
    – Joffan
    Jul 26 at 19:39

















  • And what have you done towards accomplishing this? Can you prove it in one direction?
    – saulspatz
    Jul 26 at 19:27










  • Let $k<n $ be the maximum proper divisor of $n$. Then $kr=n$ for some $r>1$. Then - what can you say about $r$?
    – Joffan
    Jul 26 at 19:33











  • I intuitively believe it to be true and if a proof exists then I would prefer to see another more experienced person's proof rather than muck about with my own. In the meantime I will try on my own, but why would a waste times asking everyone else?
    – Adam
    Jul 26 at 19:36










  • $r$ would have to be the minimum proper divisor of $n$
    – Adam
    Jul 26 at 19:37











  • Yes (provided $n$ is not prime) - can you prove that $r$ must be prime?
    – Joffan
    Jul 26 at 19:39
















And what have you done towards accomplishing this? Can you prove it in one direction?
– saulspatz
Jul 26 at 19:27




And what have you done towards accomplishing this? Can you prove it in one direction?
– saulspatz
Jul 26 at 19:27












Let $k<n $ be the maximum proper divisor of $n$. Then $kr=n$ for some $r>1$. Then - what can you say about $r$?
– Joffan
Jul 26 at 19:33





Let $k<n $ be the maximum proper divisor of $n$. Then $kr=n$ for some $r>1$. Then - what can you say about $r$?
– Joffan
Jul 26 at 19:33













I intuitively believe it to be true and if a proof exists then I would prefer to see another more experienced person's proof rather than muck about with my own. In the meantime I will try on my own, but why would a waste times asking everyone else?
– Adam
Jul 26 at 19:36




I intuitively believe it to be true and if a proof exists then I would prefer to see another more experienced person's proof rather than muck about with my own. In the meantime I will try on my own, but why would a waste times asking everyone else?
– Adam
Jul 26 at 19:36












$r$ would have to be the minimum proper divisor of $n$
– Adam
Jul 26 at 19:37





$r$ would have to be the minimum proper divisor of $n$
– Adam
Jul 26 at 19:37













Yes (provided $n$ is not prime) - can you prove that $r$ must be prime?
– Joffan
Jul 26 at 19:39





Yes (provided $n$ is not prime) - can you prove that $r$ must be prime?
– Joffan
Jul 26 at 19:39











2 Answers
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Let $k$ be the maximum divisor of $n$. Then $n=kr$, where $r$ is the minimum prime factor. Now $rmid kiff k=sr$ for some $s iff n=sr^2iff r^2mid n$






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    $n=p^2r$ where $p$ is the smallest factor of $n$ iff the largest divisor of $n$ is $n/p=pr$.






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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      1
      down vote



      accepted










      Let $k$ be the maximum divisor of $n$. Then $n=kr$, where $r$ is the minimum prime factor. Now $rmid kiff k=sr$ for some $s iff n=sr^2iff r^2mid n$






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted










        Let $k$ be the maximum divisor of $n$. Then $n=kr$, where $r$ is the minimum prime factor. Now $rmid kiff k=sr$ for some $s iff n=sr^2iff r^2mid n$






        share|cite|improve this answer























          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Let $k$ be the maximum divisor of $n$. Then $n=kr$, where $r$ is the minimum prime factor. Now $rmid kiff k=sr$ for some $s iff n=sr^2iff r^2mid n$






          share|cite|improve this answer













          Let $k$ be the maximum divisor of $n$. Then $n=kr$, where $r$ is the minimum prime factor. Now $rmid kiff k=sr$ for some $s iff n=sr^2iff r^2mid n$







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 26 at 20:03









          Chris Custer

          5,2782622




          5,2782622




















              up vote
              2
              down vote













              $n=p^2r$ where $p$ is the smallest factor of $n$ iff the largest divisor of $n$ is $n/p=pr$.






              share|cite|improve this answer



























                up vote
                2
                down vote













                $n=p^2r$ where $p$ is the smallest factor of $n$ iff the largest divisor of $n$ is $n/p=pr$.






                share|cite|improve this answer

























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  $n=p^2r$ where $p$ is the smallest factor of $n$ iff the largest divisor of $n$ is $n/p=pr$.






                  share|cite|improve this answer















                  $n=p^2r$ where $p$ is the smallest factor of $n$ iff the largest divisor of $n$ is $n/p=pr$.







                  share|cite|improve this answer















                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jul 26 at 20:16









                  Chris Custer

                  5,2782622




                  5,2782622











                  answered Jul 26 at 20:04









                  Keith Backman

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                  39227






















                       

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