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Prime number lemma proof needed
Clash Royale CLAN TAG#URR8PPP
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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
asked Jul 26 at 19:24
Adam
11411
 |Â
show 7 more comments
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
asked Jul 26 at 19:24
Adam
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
 |Â
show 7 more comments
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
asked Jul 26 at 19:24
Adam
11411
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
asked Jul 26 at 19:24
Adam
11411
edited Jul 26 at 19:50
asked Jul 26 at 19:24
Adam
11411
asked Jul 26 at 19:24
Adam
11411
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
 |Â
show 7 more comments
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
 |Â
show 7 more comments
2 Answers
2
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 $s iff n=sr^2iff r^2mid n$
answered Jul 26 at 20:03
Chris Custer
5,2782622
add a comment |Â
up vote
2
down vote
$n=p^2r$ where $p$ is the smallest factor of $n$ iff the largest divisor of $n$ is $n/p=pr$.
edited Jul 26 at 20:16
Chris Custer
5,2782622
answered Jul 26 at 20:04
Keith Backman
39227
add a comment |Â
2 Answers
2
active
oldest
votes
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 $s iff n=sr^2iff r^2mid n$
answered Jul 26 at 20:03
Chris Custer
5,2782622
add a comment |Â
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 $s iff n=sr^2iff r^2mid n$
answered Jul 26 at 20:03
Chris Custer
5,2782622
add a comment |Â
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 $s iff n=sr^2iff r^2mid n$
answered Jul 26 at 20:03
Chris Custer
5,2782622
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 $s iff n=sr^2iff r^2mid n$
answered Jul 26 at 20:03
Chris Custer
5,2782622
answered Jul 26 at 20:03
Chris Custer
5,2782622
answered Jul 26 at 20:03
Chris Custer
5,2782622
answered Jul 26 at 20:03
Chris Custer
5,2782622
5,2782622
add a comment |Â
add a comment |Â
up vote
2
down vote
$n=p^2r$ where $p$ is the smallest factor of $n$ iff the largest divisor of $n$ is $n/p=pr$.
edited Jul 26 at 20:16
Chris Custer
5,2782622
answered Jul 26 at 20:04
Keith Backman
39227
add a comment |Â
up vote
2
down vote
$n=p^2r$ where $p$ is the smallest factor of $n$ iff the largest divisor of $n$ is $n/p=pr$.
edited Jul 26 at 20:16
Chris Custer
5,2782622
answered Jul 26 at 20:04
Keith Backman
39227
add a comment |Â
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 $n/p=pr$.
edited Jul 26 at 20:16
Chris Custer
5,2782622
answered Jul 26 at 20:04
Keith Backman
39227
$n=p^2r$ where $p$ is the smallest factor of $n$ iff the largest divisor of $n$ is $n/p=pr$.
edited Jul 26 at 20:16
Chris Custer
5,2782622
answered Jul 26 at 20:04
Keith Backman
39227
edited Jul 26 at 20:16
Chris Custer
5,2782622
edited Jul 26 at 20:16
Chris Custer
5,2782622
edited Jul 26 at 20:16
Chris Custer
5,2782622
5,2782622
answered Jul 26 at 20:04
Keith Backman
39227
answered Jul 26 at 20:04
Keith Backman
39227
answered Jul 26 at 20:04
Keith Backman
39227
39227
add a comment |Â
add a comment |Â
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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