Prime number raised to a powerquestion on prime numberPrime number questionFind the prime-power decomposition of 999999999999If all pairs of addends that sum up to $N$ are coprime, then $N$ is prime.Is the square of a prime necessarily larger than the next prime?There are infinitely many odd numbers not expressible as the sum of a prime number and a power of $2$Is there any deep reason that 23456789 is prime?Is this number prime?Advise on determining prime numbers such that the product of their geometric series is equivalent to another prime number raised to a power.Is there a name for a multiplicative function $psi(p^n) = (p^n-1)(p^n-1-1)cdots(p-1)$?
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Prime number raised to a power
question on prime numberPrime number questionFind the prime-power decomposition of 999999999999If all pairs of addends that sum up to $N$ are coprime, then $N$ is prime.Is the square of a prime necessarily larger than the next prime?There are infinitely many odd numbers not expressible as the sum of a prime number and a power of $2$Is there any deep reason that 23456789 is prime?Is this number prime?Advise on determining prime numbers such that the product of their geometric series is equivalent to another prime number raised to a power.Is there a name for a multiplicative function $psi(p^n) = (p^n-1)(p^n-1-1)cdots(p-1)$?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Let $p$ be a prime number. If $n, a$ are two positive integers such that $p^n | a$, then $p|a$.
I've recently come across this result in a problem. They didn't give any proof for it, they just asserted that it is true. I am not too good at number theory, so please tell me whether this is some well-known result.
number-theory elementary-number-theory
asked 8 hours ago
Math GuyMath Guy
1807 bronze badges
$endgroup$
add a comment |
$begingroup$
Let $p$ be a prime number. If $n, a$ are two positive integers such that $p^n | a$, then $p|a$.
I've recently come across this result in a problem. They didn't give any proof for it, they just asserted that it is true. I am not too good at number theory, so please tell me whether this is some well-known result.
number-theory elementary-number-theory
asked 8 hours ago
Math GuyMath Guy
1807 bronze badges
$endgroup$
2
$begingroup$
$p^n|aiff a=p^nmiff dfrac ap=p^n-1minmathbb N$
$endgroup$
– Piquito
8 hours ago
add a comment |
$begingroup$
Let $p$ be a prime number. If $n, a$ are two positive integers such that $p^n | a$, then $p|a$.
I've recently come across this result in a problem. They didn't give any proof for it, they just asserted that it is true. I am not too good at number theory, so please tell me whether this is some well-known result.
number-theory elementary-number-theory
asked 8 hours ago
Math GuyMath Guy
1807 bronze badges
$endgroup$
Let $p$ be a prime number. If $n, a$ are two positive integers such that $p^n | a$, then $p|a$.
I've recently come across this result in a problem. They didn't give any proof for it, they just asserted that it is true. I am not too good at number theory, so please tell me whether this is some well-known result.
number-theory elementary-number-theory
number-theory elementary-number-theory
asked 8 hours ago
Math GuyMath Guy
1807 bronze badges
asked 8 hours ago
Math GuyMath Guy
1807 bronze badges
asked 8 hours ago
Math GuyMath Guy
1807 bronze badges
asked 8 hours ago
Math GuyMath Guy
1807 bronze badges
asked 8 hours ago
Math GuyMath Guy
1807 bronze badges
1807 bronze badges
2
$begingroup$
$p^n|aiff a=p^nmiff dfrac ap=p^n-1minmathbb N$
$endgroup$
– Piquito
8 hours ago
add a comment |
2
$begingroup$
$p^n|aiff a=p^nmiff dfrac ap=p^n-1minmathbb N$
$endgroup$
– Piquito
8 hours ago
2
2
$begingroup$
$p^n|aiff a=p^nmiff dfrac ap=p^n-1minmathbb N$
$endgroup$
– Piquito
8 hours ago
$begingroup$
$p^n|aiff a=p^nmiff dfrac ap=p^n-1minmathbb N$
$endgroup$
– Piquito
8 hours ago
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
We have $p^nmid a$ and obviously $pmid p^n$. Then $pmid a$ by transitivity.
answered 8 hours ago
Azif00Azif00
9479 bronze badges
$endgroup$
$begingroup$
Thank you! Which part uses specifically that $p$ is a prime number?
$endgroup$
– Math Guy
8 hours ago
2
$begingroup$
This does not depend on $p$ prime. It works for any $n$.
$endgroup$
– Antonios-Alexandros Robotis
8 hours ago
2
$begingroup$
Nowhere, we do not need prime numbers here.
$endgroup$
– Azif00
8 hours ago
add a comment |
$begingroup$
We have that $p mid p^n$ and that $p^n mid a$, such that $p mid a$. Maybe try to prove for yourself that divisibility is transitive.
Otherwise, let me prove it for you here (at least for you to compare):
Suppose that $a mid b$ and $b mid c$, that means we have $aa' = b$ and $bb' = c$ for some integers $a'$ and $b'$. We get $c = bb' = (aa')b' = a(a'b')$, such that $a mid c$.
answered 8 hours ago
ThorWittichThorWittich
1,8061 silver badge12 bronze badges
$endgroup$
add a comment |
$begingroup$
If $n = 1$, we have $p mid a$ and we are done.
For $n ge 2$:
$p^n mid a Longrightarrow exists c, ; a = cp^n = (cp^n - 1)p Longrightarrow p mid a. tag 1$
This result, being so simple, is very well-known.
We note that $p$ needn't be prime.
answered 8 hours ago
Robert LewisRobert Lewis
50.8k2 gold badges33 silver badges69 bronze badges
$endgroup$
add a comment |
$begingroup$
Maybe this is an easier way to look at this:
For integers $a,b,c$, if $b$ divides $c$, and $a$ divides $b$, then $a$ divides $c$.
answered 8 hours ago
Ruben du BurckRuben du Burck
9432 silver badges14 bronze badges
$endgroup$
add a comment |
$begingroup$
If $p^n | a$ then $a = p^nk$ for some integer $k$.
As $p | p^kk$, so we have $p|a$.
answered 7 hours ago
mlchristiansmlchristians
44414 bronze badges
$endgroup$
add a comment |
Your Answer
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have $p^nmid a$ and obviously $pmid p^n$. Then $pmid a$ by transitivity.
answered 8 hours ago
Azif00Azif00
9479 bronze badges
$endgroup$
$begingroup$
Thank you! Which part uses specifically that $p$ is a prime number?
$endgroup$
– Math Guy
8 hours ago
2
$begingroup$
This does not depend on $p$ prime. It works for any $n$.
$endgroup$
– Antonios-Alexandros Robotis
8 hours ago
2
$begingroup$
Nowhere, we do not need prime numbers here.
$endgroup$
– Azif00
8 hours ago
add a comment |
$begingroup$
We have $p^nmid a$ and obviously $pmid p^n$. Then $pmid a$ by transitivity.
answered 8 hours ago
Azif00Azif00
9479 bronze badges
$endgroup$
$begingroup$
Thank you! Which part uses specifically that $p$ is a prime number?
$endgroup$
– Math Guy
8 hours ago
2
$begingroup$
This does not depend on $p$ prime. It works for any $n$.
$endgroup$
– Antonios-Alexandros Robotis
8 hours ago
2
$begingroup$
Nowhere, we do not need prime numbers here.
$endgroup$
– Azif00
8 hours ago
add a comment |
$begingroup$
We have $p^nmid a$ and obviously $pmid p^n$. Then $pmid a$ by transitivity.
answered 8 hours ago
Azif00Azif00
9479 bronze badges
$endgroup$
We have $p^nmid a$ and obviously $pmid p^n$. Then $pmid a$ by transitivity.
answered 8 hours ago
Azif00Azif00
9479 bronze badges
answered 8 hours ago
Azif00Azif00
9479 bronze badges
answered 8 hours ago
Azif00Azif00
9479 bronze badges
answered 8 hours ago
Azif00Azif00
9479 bronze badges
9479 bronze badges
$begingroup$
Thank you! Which part uses specifically that $p$ is a prime number?
$endgroup$
– Math Guy
8 hours ago
2
$begingroup$
This does not depend on $p$ prime. It works for any $n$.
$endgroup$
– Antonios-Alexandros Robotis
8 hours ago
2
$begingroup$
Nowhere, we do not need prime numbers here.
$endgroup$
– Azif00
8 hours ago
add a comment |
$begingroup$
Thank you! Which part uses specifically that $p$ is a prime number?
$endgroup$
– Math Guy
8 hours ago
2
$begingroup$
This does not depend on $p$ prime. It works for any $n$.
$endgroup$
– Antonios-Alexandros Robotis
8 hours ago
2
$begingroup$
Nowhere, we do not need prime numbers here.
$endgroup$
– Azif00
8 hours ago
$begingroup$
Thank you! Which part uses specifically that $p$ is a prime number?
$endgroup$
– Math Guy
8 hours ago
$begingroup$
Thank you! Which part uses specifically that $p$ is a prime number?
$endgroup$
– Math Guy
8 hours ago
2
2
$begingroup$
This does not depend on $p$ prime. It works for any $n$.
$endgroup$
– Antonios-Alexandros Robotis
8 hours ago
$begingroup$
This does not depend on $p$ prime. It works for any $n$.
$endgroup$
– Antonios-Alexandros Robotis
8 hours ago
2
2
$begingroup$
Nowhere, we do not need prime numbers here.
$endgroup$
– Azif00
8 hours ago
$begingroup$
Nowhere, we do not need prime numbers here.
$endgroup$
– Azif00
8 hours ago
add a comment |
$begingroup$
We have that $p mid p^n$ and that $p^n mid a$, such that $p mid a$. Maybe try to prove for yourself that divisibility is transitive.
Otherwise, let me prove it for you here (at least for you to compare):
Suppose that $a mid b$ and $b mid c$, that means we have $aa' = b$ and $bb' = c$ for some integers $a'$ and $b'$. We get $c = bb' = (aa')b' = a(a'b')$, such that $a mid c$.
answered 8 hours ago
ThorWittichThorWittich
1,8061 silver badge12 bronze badges
$endgroup$
add a comment |
$begingroup$
We have that $p mid p^n$ and that $p^n mid a$, such that $p mid a$. Maybe try to prove for yourself that divisibility is transitive.
Otherwise, let me prove it for you here (at least for you to compare):
Suppose that $a mid b$ and $b mid c$, that means we have $aa' = b$ and $bb' = c$ for some integers $a'$ and $b'$. We get $c = bb' = (aa')b' = a(a'b')$, such that $a mid c$.
answered 8 hours ago
ThorWittichThorWittich
1,8061 silver badge12 bronze badges
$endgroup$
add a comment |
$begingroup$
We have that $p mid p^n$ and that $p^n mid a$, such that $p mid a$. Maybe try to prove for yourself that divisibility is transitive.
Otherwise, let me prove it for you here (at least for you to compare):
Suppose that $a mid b$ and $b mid c$, that means we have $aa' = b$ and $bb' = c$ for some integers $a'$ and $b'$. We get $c = bb' = (aa')b' = a(a'b')$, such that $a mid c$.
answered 8 hours ago
ThorWittichThorWittich
1,8061 silver badge12 bronze badges
$endgroup$
We have that $p mid p^n$ and that $p^n mid a$, such that $p mid a$. Maybe try to prove for yourself that divisibility is transitive.
Otherwise, let me prove it for you here (at least for you to compare):
Suppose that $a mid b$ and $b mid c$, that means we have $aa' = b$ and $bb' = c$ for some integers $a'$ and $b'$. We get $c = bb' = (aa')b' = a(a'b')$, such that $a mid c$.
answered 8 hours ago
ThorWittichThorWittich
1,8061 silver badge12 bronze badges
answered 8 hours ago
ThorWittichThorWittich
1,8061 silver badge12 bronze badges
answered 8 hours ago
ThorWittichThorWittich
1,8061 silver badge12 bronze badges
answered 8 hours ago
ThorWittichThorWittich
1,8061 silver badge12 bronze badges
1,8061 silver badge12 bronze badges
add a comment |
add a comment |
$begingroup$
If $n = 1$, we have $p mid a$ and we are done.
For $n ge 2$:
$p^n mid a Longrightarrow exists c, ; a = cp^n = (cp^n - 1)p Longrightarrow p mid a. tag 1$
This result, being so simple, is very well-known.
We note that $p$ needn't be prime.
answered 8 hours ago
Robert LewisRobert Lewis
50.8k2 gold badges33 silver badges69 bronze badges
$endgroup$
add a comment |
$begingroup$
If $n = 1$, we have $p mid a$ and we are done.
For $n ge 2$:
$p^n mid a Longrightarrow exists c, ; a = cp^n = (cp^n - 1)p Longrightarrow p mid a. tag 1$
This result, being so simple, is very well-known.
We note that $p$ needn't be prime.
answered 8 hours ago
Robert LewisRobert Lewis
50.8k2 gold badges33 silver badges69 bronze badges
$endgroup$
add a comment |
$begingroup$
If $n = 1$, we have $p mid a$ and we are done.
For $n ge 2$:
$p^n mid a Longrightarrow exists c, ; a = cp^n = (cp^n - 1)p Longrightarrow p mid a. tag 1$
This result, being so simple, is very well-known.
We note that $p$ needn't be prime.
answered 8 hours ago
Robert LewisRobert Lewis
50.8k2 gold badges33 silver badges69 bronze badges
$endgroup$
If $n = 1$, we have $p mid a$ and we are done.
For $n ge 2$:
$p^n mid a Longrightarrow exists c, ; a = cp^n = (cp^n - 1)p Longrightarrow p mid a. tag 1$
This result, being so simple, is very well-known.
We note that $p$ needn't be prime.
answered 8 hours ago
Robert LewisRobert Lewis
50.8k2 gold badges33 silver badges69 bronze badges
edited 8 hours ago
answered 8 hours ago
Robert LewisRobert Lewis
50.8k2 gold badges33 silver badges69 bronze badges
answered 8 hours ago
Robert LewisRobert Lewis
50.8k2 gold badges33 silver badges69 bronze badges
answered 8 hours ago
Robert LewisRobert Lewis
50.8k2 gold badges33 silver badges69 bronze badges
50.8k2 gold badges33 silver badges69 bronze badges
add a comment |
add a comment |
$begingroup$
Maybe this is an easier way to look at this:
For integers $a,b,c$, if $b$ divides $c$, and $a$ divides $b$, then $a$ divides $c$.
answered 8 hours ago
Ruben du BurckRuben du Burck
9432 silver badges14 bronze badges
$endgroup$
add a comment |
$begingroup$
Maybe this is an easier way to look at this:
For integers $a,b,c$, if $b$ divides $c$, and $a$ divides $b$, then $a$ divides $c$.
answered 8 hours ago
Ruben du BurckRuben du Burck
9432 silver badges14 bronze badges
$endgroup$
add a comment |
$begingroup$
Maybe this is an easier way to look at this:
For integers $a,b,c$, if $b$ divides $c$, and $a$ divides $b$, then $a$ divides $c$.
answered 8 hours ago
Ruben du BurckRuben du Burck
9432 silver badges14 bronze badges
$endgroup$
Maybe this is an easier way to look at this:
For integers $a,b,c$, if $b$ divides $c$, and $a$ divides $b$, then $a$ divides $c$.
answered 8 hours ago
Ruben du BurckRuben du Burck
9432 silver badges14 bronze badges
answered 8 hours ago
Ruben du BurckRuben du Burck
9432 silver badges14 bronze badges
answered 8 hours ago
Ruben du BurckRuben du Burck
9432 silver badges14 bronze badges
answered 8 hours ago
Ruben du BurckRuben du Burck
9432 silver badges14 bronze badges
9432 silver badges14 bronze badges
add a comment |
add a comment |
$begingroup$
If $p^n | a$ then $a = p^nk$ for some integer $k$.
As $p | p^kk$, so we have $p|a$.
answered 7 hours ago
mlchristiansmlchristians
44414 bronze badges
$endgroup$
add a comment |
$begingroup$
If $p^n | a$ then $a = p^nk$ for some integer $k$.
As $p | p^kk$, so we have $p|a$.
answered 7 hours ago
mlchristiansmlchristians
44414 bronze badges
$endgroup$
add a comment |
$begingroup$
If $p^n | a$ then $a = p^nk$ for some integer $k$.
As $p | p^kk$, so we have $p|a$.
answered 7 hours ago
mlchristiansmlchristians
44414 bronze badges
$endgroup$
If $p^n | a$ then $a = p^nk$ for some integer $k$.
As $p | p^kk$, so we have $p|a$.
answered 7 hours ago
mlchristiansmlchristians
44414 bronze badges
answered 7 hours ago
mlchristiansmlchristians
44414 bronze badges
answered 7 hours ago
mlchristiansmlchristians
44414 bronze badges
answered 7 hours ago
mlchristiansmlchristians
44414 bronze badges
44414 bronze badges
add a comment |
add a comment |
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2
$begingroup$
$p^n|aiff a=p^nmiff dfrac ap=p^n-1minmathbb N$
$endgroup$
– Piquito
8 hours ago