Mixing
Prove $phi(p^e)not|p^et-1$
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Let p be a prime, m an odd positive integer such that $ p^e||m$, where $ e>1 $ is an integer. I am trying to show that
$$ phi(p^e)=p^e-p^{e-1}nmid p^et-1=m-1 $$
Is it enough to just say that the left-hand side will be some multiple of p whereas the right hand side will be some multiple of p minus one so the right-hand side cannot divide the left?
elementary-number-theory divisibility
asked 2 days ago
mjoseph
477
add a comment |
up vote
0
down vote
favorite
Let p be a prime, m an odd positive integer such that $ p^e||m$, where $ e>1 $ is an integer. I am trying to show that
$$ phi(p^e)=p^e-p^{e-1}nmid p^et-1=m-1 $$
Is it enough to just say that the left-hand side will be some multiple of p whereas the right hand side will be some multiple of p minus one so the right-hand side cannot divide the left?
elementary-number-theory divisibility
asked 2 days ago
mjoseph
477
Yes, that is enough.
– lulu
2 days ago
how would I write that though
– mjoseph
2 days ago
1
What you wrote is very nearly formal. If you want to say more, I'd remark that (clearly) if $a,|,b$ then any divisor of $a$ divides $b$. Now, $e>1$ implies that $p,|,p^e-p^{e-1}$. Thus, if $p^e-p^{e-1}$ were a divisor of $m-1$, $p$ would also have to be a divisor of $m-1$. But $p,|,m implies p, nmid m-1$ and we are done.
– lulu
2 days ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let p be a prime, m an odd positive integer such that $ p^e||m$, where $ e>1 $ is an integer. I am trying to show that
$$ phi(p^e)=p^e-p^{e-1}nmid p^et-1=m-1 $$
Is it enough to just say that the left-hand side will be some multiple of p whereas the right hand side will be some multiple of p minus one so the right-hand side cannot divide the left?
elementary-number-theory divisibility
asked 2 days ago
mjoseph
477
Let p be a prime, m an odd positive integer such that $ p^e||m$, where $ e>1 $ is an integer. I am trying to show that
$$ phi(p^e)=p^e-p^{e-1}nmid p^et-1=m-1 $$
Is it enough to just say that the left-hand side will be some multiple of p whereas the right hand side will be some multiple of p minus one so the right-hand side cannot divide the left?
elementary-number-theory divisibility
elementary-number-theory divisibility
asked 2 days ago
mjoseph
477
asked 2 days ago
mjoseph
477
edited 2 days ago
asked 2 days ago
mjoseph
477
asked 2 days ago
mjoseph
477
asked 2 days ago
mjoseph
477
477
Yes, that is enough.
– lulu
2 days ago
how would I write that though
– mjoseph
2 days ago
1
What you wrote is very nearly formal. If you want to say more, I'd remark that (clearly) if $a,|,b$ then any divisor of $a$ divides $b$. Now, $e>1$ implies that $p,|,p^e-p^{e-1}$. Thus, if $p^e-p^{e-1}$ were a divisor of $m-1$, $p$ would also have to be a divisor of $m-1$. But $p,|,m implies p, nmid m-1$ and we are done.
– lulu
2 days ago
add a comment |
Yes, that is enough.
– lulu
2 days ago
how would I write that though
– mjoseph
2 days ago
1
What you wrote is very nearly formal. If you want to say more, I'd remark that (clearly) if $a,|,b$ then any divisor of $a$ divides $b$. Now, $e>1$ implies that $p,|,p^e-p^{e-1}$. Thus, if $p^e-p^{e-1}$ were a divisor of $m-1$, $p$ would also have to be a divisor of $m-1$. But $p,|,m implies p, nmid m-1$ and we are done.
– lulu
2 days ago
Yes, that is enough.
– lulu
2 days ago
Yes, that is enough.
– lulu
2 days ago
how would I write that though
– mjoseph
2 days ago
how would I write that though
– mjoseph
2 days ago
1
1
What you wrote is very nearly formal. If you want to say more, I'd remark that (clearly) if $a,|,b$ then any divisor of $a$ divides $b$. Now, $e>1$ implies that $p,|,p^e-p^{e-1}$. Thus, if $p^e-p^{e-1}$ were a divisor of $m-1$, $p$ would also have to be a divisor of $m-1$. But $p,|,m implies p, nmid m-1$ and we are done.
– lulu
2 days ago
What you wrote is very nearly formal. If you want to say more, I'd remark that (clearly) if $a,|,b$ then any divisor of $a$ divides $b$. Now, $e>1$ implies that $p,|,p^e-p^{e-1}$. Thus, if $p^e-p^{e-1}$ were a divisor of $m-1$, $p$ would also have to be a divisor of $m-1$. But $p,|,m implies p, nmid m-1$ and we are done.
– lulu
2 days ago
add a comment |
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Yes, that is enough.
– lulu
2 days ago
how would I write that though
– mjoseph
2 days ago
1
What you wrote is very nearly formal. If you want to say more, I'd remark that (clearly) if $a,|,b$ then any divisor of $a$ divides $b$. Now, $e>1$ implies that $p,|,p^e-p^{e-1}$. Thus, if $p^e-p^{e-1}$ were a divisor of $m-1$, $p$ would also have to be a divisor of $m-1$. But $p,|,m implies p, nmid m-1$ and we are done.
– lulu
2 days ago