Proving Eulers totient function is multiplicative
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There is a proof at the bottom of page 1, I understand it up until the point where it says "It follows from Theorem
4.7 of Rosen that the entries in such a row form a complete residue system
modulo n. Thus, exactly $phi(n)$ of them will be relatively prime to $n$, and
thus relatively prime to $mn$." I don't know where the theorem is, but would appreciate if someone could explain why an element would have to be in one of the $phi(n)$ rows such that it is coprime with $mn$.
http://gauss.math.luc.edu/greicius/Math201/Fall2012/Lectures/euler-phi.article.pdf
number-theory elementary-number-theory
edited Dec 19 '18 at 18:16
twnly
1,0361213
asked Dec 19 '18 at 17:55
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
$endgroup$
add a comment |
$begingroup$
There is a proof at the bottom of page 1, I understand it up until the point where it says "It follows from Theorem
4.7 of Rosen that the entries in such a row form a complete residue system
modulo n. Thus, exactly $phi(n)$ of them will be relatively prime to $n$, and
thus relatively prime to $mn$." I don't know where the theorem is, but would appreciate if someone could explain why an element would have to be in one of the $phi(n)$ rows such that it is coprime with $mn$.
http://gauss.math.luc.edu/greicius/Math201/Fall2012/Lectures/euler-phi.article.pdf
number-theory elementary-number-theory
edited Dec 19 '18 at 18:16
twnly
1,0361213
asked Dec 19 '18 at 17:55
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
$endgroup$
$begingroup$
I believe the theorem in question is: If $a$, $b$, $k$, and $m$ are integers such that $k>0$, $m>0$, and $aequiv bpmod m$, then $a^kequiv b^kpmod m$.
$endgroup$
– Math1000
Dec 19 '18 at 18:02
$begingroup$
For the title, see this duplicate.
$endgroup$
– Dietrich Burde
Dec 19 '18 at 19:43
add a comment |
$begingroup$
There is a proof at the bottom of page 1, I understand it up until the point where it says "It follows from Theorem
4.7 of Rosen that the entries in such a row form a complete residue system
modulo n. Thus, exactly $phi(n)$ of them will be relatively prime to $n$, and
thus relatively prime to $mn$." I don't know where the theorem is, but would appreciate if someone could explain why an element would have to be in one of the $phi(n)$ rows such that it is coprime with $mn$.
http://gauss.math.luc.edu/greicius/Math201/Fall2012/Lectures/euler-phi.article.pdf
number-theory elementary-number-theory
edited Dec 19 '18 at 18:16
twnly
1,0361213
asked Dec 19 '18 at 17:55
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
$endgroup$
There is a proof at the bottom of page 1, I understand it up until the point where it says "It follows from Theorem
4.7 of Rosen that the entries in such a row form a complete residue system
modulo n. Thus, exactly $phi(n)$ of them will be relatively prime to $n$, and
thus relatively prime to $mn$." I don't know where the theorem is, but would appreciate if someone could explain why an element would have to be in one of the $phi(n)$ rows such that it is coprime with $mn$.
http://gauss.math.luc.edu/greicius/Math201/Fall2012/Lectures/euler-phi.article.pdf
number-theory elementary-number-theory
number-theory elementary-number-theory
edited Dec 19 '18 at 18:16
twnly
1,0361213
asked Dec 19 '18 at 17:55
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
edited Dec 19 '18 at 18:16
twnly
1,0361213
asked Dec 19 '18 at 17:55
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
edited Dec 19 '18 at 18:16
twnly
1,0361213
edited Dec 19 '18 at 18:16
twnly
1,0361213
edited Dec 19 '18 at 18:16
twnly
1,0361213
1,0361213
asked Dec 19 '18 at 17:55
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
asked Dec 19 '18 at 17:55
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
asked Dec 19 '18 at 17:55
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
386
$begingroup$
I believe the theorem in question is: If $a$, $b$, $k$, and $m$ are integers such that $k>0$, $m>0$, and $aequiv bpmod m$, then $a^kequiv b^kpmod m$.
$endgroup$
– Math1000
Dec 19 '18 at 18:02
$begingroup$
For the title, see this duplicate.
$endgroup$
– Dietrich Burde
Dec 19 '18 at 19:43
add a comment |
$begingroup$
I believe the theorem in question is: If $a$, $b$, $k$, and $m$ are integers such that $k>0$, $m>0$, and $aequiv bpmod m$, then $a^kequiv b^kpmod m$.
$endgroup$
– Math1000
Dec 19 '18 at 18:02
$begingroup$
For the title, see this duplicate.
$endgroup$
– Dietrich Burde
Dec 19 '18 at 19:43
$begingroup$
I believe the theorem in question is: If $a$, $b$, $k$, and $m$ are integers such that $k>0$, $m>0$, and $aequiv bpmod m$, then $a^kequiv b^kpmod m$.
$endgroup$
– Math1000
Dec 19 '18 at 18:02
$begingroup$
I believe the theorem in question is: If $a$, $b$, $k$, and $m$ are integers such that $k>0$, $m>0$, and $aequiv bpmod m$, then $a^kequiv b^kpmod m$.
$endgroup$
– Math1000
Dec 19 '18 at 18:02
$begingroup$
For the title, see this duplicate.
$endgroup$
– Dietrich Burde
Dec 19 '18 at 19:43
$begingroup$
For the title, see this duplicate.
$endgroup$
– Dietrich Burde
Dec 19 '18 at 19:43
add a comment |
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$begingroup$
I believe the theorem in question is: If $a$, $b$, $k$, and $m$ are integers such that $k>0$, $m>0$, and $aequiv bpmod m$, then $a^kequiv b^kpmod m$.
$endgroup$
– Math1000
Dec 19 '18 at 18:02
$begingroup$
For the title, see this duplicate.
$endgroup$
– Dietrich Burde
Dec 19 '18 at 19:43