Proof of a formula involving Euler's totient function: $varphi (mn) = varphi (m) varphi (n) cdot...












17












$begingroup$


The third formula on the wikipedia page for the Totient function states that $$varphi (mn) = varphi (m) varphi (n) cdot dfrac{d}{varphi (d)} $$
where $d = gcd(m,n)$.



How is this claim justified?



Would we have to use the Chinese Remainder Theorem, as they suggest for proving that $varphi$ is multiplicative?






elementary-number-theory totient-function







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    There might be a direct proof, but of course if you show that $varphi$ is multiplicative (using the Chinese Remainder Theorem) and that $varphi(p^a) = p^a - p^{a-1}$, then you get your result.
    $endgroup$
    – Joel Cohen
    Feb 29 '12 at 15:18






  • 1




    $begingroup$
    It'd be nice to relate this formula with the natural mapping $U_{mn}to U_m times U_n$ by proving that the kernel has size $d$ and the image has index $phi(d)$.
    $endgroup$
    – lhf
    Mar 13 '12 at 2:22










  • $begingroup$
    See also math.stackexchange.com/questions/119911/…. (Thanks @Dane!)
    $endgroup$
    – lhf
    Mar 14 '12 at 12:07
















17












$begingroup$


The third formula on the wikipedia page for the Totient function states that $$varphi (mn) = varphi (m) varphi (n) cdot dfrac{d}{varphi (d)} $$
where $d = gcd(m,n)$.



How is this claim justified?



Would we have to use the Chinese Remainder Theorem, as they suggest for proving that $varphi$ is multiplicative?






elementary-number-theory totient-function







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    There might be a direct proof, but of course if you show that $varphi$ is multiplicative (using the Chinese Remainder Theorem) and that $varphi(p^a) = p^a - p^{a-1}$, then you get your result.
    $endgroup$
    – Joel Cohen
    Feb 29 '12 at 15:18






  • 1




    $begingroup$
    It'd be nice to relate this formula with the natural mapping $U_{mn}to U_m times U_n$ by proving that the kernel has size $d$ and the image has index $phi(d)$.
    $endgroup$
    – lhf
    Mar 13 '12 at 2:22










  • $begingroup$
    See also math.stackexchange.com/questions/119911/…. (Thanks @Dane!)
    $endgroup$
    – lhf
    Mar 14 '12 at 12:07














17












17








17


9



$begingroup$


The third formula on the wikipedia page for the Totient function states that $$varphi (mn) = varphi (m) varphi (n) cdot dfrac{d}{varphi (d)} $$
where $d = gcd(m,n)$.



How is this claim justified?



Would we have to use the Chinese Remainder Theorem, as they suggest for proving that $varphi$ is multiplicative?






elementary-number-theory totient-function







share|cite|improve this question











$endgroup$




The third formula on the wikipedia page for the Totient function states that $$varphi (mn) = varphi (m) varphi (n) cdot dfrac{d}{varphi (d)} $$
where $d = gcd(m,n)$.



How is this claim justified?



Would we have to use the Chinese Remainder Theorem, as they suggest for proving that $varphi$ is multiplicative?






elementary-number-theory totient-function




elementary-number-theory totient-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 12 '15 at 9:30









Martin Sleziak

44.7k9117272




44.7k9117272










asked Feb 29 '12 at 15:02









The Chaz 2.0The Chaz 2.0

8,10863574




8,10863574








  • 1




    $begingroup$
    There might be a direct proof, but of course if you show that $varphi$ is multiplicative (using the Chinese Remainder Theorem) and that $varphi(p^a) = p^a - p^{a-1}$, then you get your result.
    $endgroup$
    – Joel Cohen
    Feb 29 '12 at 15:18






  • 1




    $begingroup$
    It'd be nice to relate this formula with the natural mapping $U_{mn}to U_m times U_n$ by proving that the kernel has size $d$ and the image has index $phi(d)$.
    $endgroup$
    – lhf
    Mar 13 '12 at 2:22










  • $begingroup$
    See also math.stackexchange.com/questions/119911/…. (Thanks @Dane!)
    $endgroup$
    – lhf
    Mar 14 '12 at 12:07














  • 1




    $begingroup$
    There might be a direct proof, but of course if you show that $varphi$ is multiplicative (using the Chinese Remainder Theorem) and that $varphi(p^a) = p^a - p^{a-1}$, then you get your result.
    $endgroup$
    – Joel Cohen
    Feb 29 '12 at 15:18






  • 1




    $begingroup$
    It'd be nice to relate this formula with the natural mapping $U_{mn}to U_m times U_n$ by proving that the kernel has size $d$ and the image has index $phi(d)$.
    $endgroup$
    – lhf
    Mar 13 '12 at 2:22










  • $begingroup$
    See also math.stackexchange.com/questions/119911/…. (Thanks @Dane!)
    $endgroup$
    – lhf
    Mar 14 '12 at 12:07








1




1




$begingroup$
There might be a direct proof, but of course if you show that $varphi$ is multiplicative (using the Chinese Remainder Theorem) and that $varphi(p^a) = p^a - p^{a-1}$, then you get your result.
$endgroup$
– Joel Cohen
Feb 29 '12 at 15:18




$begingroup$
There might be a direct proof, but of course if you show that $varphi$ is multiplicative (using the Chinese Remainder Theorem) and that $varphi(p^a) = p^a - p^{a-1}$, then you get your result.
$endgroup$
– Joel Cohen
Feb 29 '12 at 15:18




1




1




$begingroup$
It'd be nice to relate this formula with the natural mapping $U_{mn}to U_m times U_n$ by proving that the kernel has size $d$ and the image has index $phi(d)$.
$endgroup$
– lhf
Mar 13 '12 at 2:22




$begingroup$
It'd be nice to relate this formula with the natural mapping $U_{mn}to U_m times U_n$ by proving that the kernel has size $d$ and the image has index $phi(d)$.
$endgroup$
– lhf
Mar 13 '12 at 2:22












$begingroup$
See also math.stackexchange.com/questions/119911/…. (Thanks @Dane!)
$endgroup$
– lhf
Mar 14 '12 at 12:07




$begingroup$
See also math.stackexchange.com/questions/119911/…. (Thanks @Dane!)
$endgroup$
– lhf
Mar 14 '12 at 12:07










2 Answers
2






active

oldest

votes


















27












$begingroup$

You can write $varphi(n) = n prod_{p mid n} left( 1 - frac 1p right)$.
Using this identity, we have



$$
varphi(mn)
= mn prod_{p mid mn} left( 1 - frac 1p right)
= mn frac{prod_{p mid m} left( 1 - frac 1p right) prod_{p mid n} left( 1 - frac 1p right)}{prod_{p mid d} left( 1 - frac 1p right)}
= varphi(m)varphi(n) frac{d}{varphi(d)}
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This should probably be restricted to prime $p$.
    $endgroup$
    – G. Bach
    Apr 25 '14 at 19:55



















5












$begingroup$

Hint $ $ A multiplicative function $rm:f(n):$ satisfies said identity if for all primes $rm:p:$



$$rm jle k Rightarrow f(p^{j+k}) = frac{f(p^j): f(p^k): p^j}{f(p^j)} = p^j f(p^k)$$



Indeed we have $rm phi(p^{j+k}) = p^{j+k}-p^{j+k-1} = p^j (p^k-p^{k-1}) = p^j phi(p^k)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks, Bill. I'll try to wrap my head around that. It seems useful.
    $endgroup$
    – The Chaz 2.0
    Feb 29 '12 at 16:44











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









27












$begingroup$

You can write $varphi(n) = n prod_{p mid n} left( 1 - frac 1p right)$.
Using this identity, we have



$$
varphi(mn)
= mn prod_{p mid mn} left( 1 - frac 1p right)
= mn frac{prod_{p mid m} left( 1 - frac 1p right) prod_{p mid n} left( 1 - frac 1p right)}{prod_{p mid d} left( 1 - frac 1p right)}
= varphi(m)varphi(n) frac{d}{varphi(d)}
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This should probably be restricted to prime $p$.
    $endgroup$
    – G. Bach
    Apr 25 '14 at 19:55
















27












$begingroup$

You can write $varphi(n) = n prod_{p mid n} left( 1 - frac 1p right)$.
Using this identity, we have



$$
varphi(mn)
= mn prod_{p mid mn} left( 1 - frac 1p right)
= mn frac{prod_{p mid m} left( 1 - frac 1p right) prod_{p mid n} left( 1 - frac 1p right)}{prod_{p mid d} left( 1 - frac 1p right)}
= varphi(m)varphi(n) frac{d}{varphi(d)}
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This should probably be restricted to prime $p$.
    $endgroup$
    – G. Bach
    Apr 25 '14 at 19:55














27












27








27





$begingroup$

You can write $varphi(n) = n prod_{p mid n} left( 1 - frac 1p right)$.
Using this identity, we have



$$
varphi(mn)
= mn prod_{p mid mn} left( 1 - frac 1p right)
= mn frac{prod_{p mid m} left( 1 - frac 1p right) prod_{p mid n} left( 1 - frac 1p right)}{prod_{p mid d} left( 1 - frac 1p right)}
= varphi(m)varphi(n) frac{d}{varphi(d)}
$$






share|cite|improve this answer









$endgroup$



You can write $varphi(n) = n prod_{p mid n} left( 1 - frac 1p right)$.
Using this identity, we have



$$
varphi(mn)
= mn prod_{p mid mn} left( 1 - frac 1p right)
= mn frac{prod_{p mid m} left( 1 - frac 1p right) prod_{p mid n} left( 1 - frac 1p right)}{prod_{p mid d} left( 1 - frac 1p right)}
= varphi(m)varphi(n) frac{d}{varphi(d)}
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Feb 29 '12 at 15:18









DaneDane

3,2041634




3,2041634












  • $begingroup$
    This should probably be restricted to prime $p$.
    $endgroup$
    – G. Bach
    Apr 25 '14 at 19:55


















  • $begingroup$
    This should probably be restricted to prime $p$.
    $endgroup$
    – G. Bach
    Apr 25 '14 at 19:55
















$begingroup$
This should probably be restricted to prime $p$.
$endgroup$
– G. Bach
Apr 25 '14 at 19:55




$begingroup$
This should probably be restricted to prime $p$.
$endgroup$
– G. Bach
Apr 25 '14 at 19:55











5












$begingroup$

Hint $ $ A multiplicative function $rm:f(n):$ satisfies said identity if for all primes $rm:p:$



$$rm jle k Rightarrow f(p^{j+k}) = frac{f(p^j): f(p^k): p^j}{f(p^j)} = p^j f(p^k)$$



Indeed we have $rm phi(p^{j+k}) = p^{j+k}-p^{j+k-1} = p^j (p^k-p^{k-1}) = p^j phi(p^k)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks, Bill. I'll try to wrap my head around that. It seems useful.
    $endgroup$
    – The Chaz 2.0
    Feb 29 '12 at 16:44
















5












$begingroup$

Hint $ $ A multiplicative function $rm:f(n):$ satisfies said identity if for all primes $rm:p:$



$$rm jle k Rightarrow f(p^{j+k}) = frac{f(p^j): f(p^k): p^j}{f(p^j)} = p^j f(p^k)$$



Indeed we have $rm phi(p^{j+k}) = p^{j+k}-p^{j+k-1} = p^j (p^k-p^{k-1}) = p^j phi(p^k)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks, Bill. I'll try to wrap my head around that. It seems useful.
    $endgroup$
    – The Chaz 2.0
    Feb 29 '12 at 16:44














5












5








5





$begingroup$

Hint $ $ A multiplicative function $rm:f(n):$ satisfies said identity if for all primes $rm:p:$



$$rm jle k Rightarrow f(p^{j+k}) = frac{f(p^j): f(p^k): p^j}{f(p^j)} = p^j f(p^k)$$



Indeed we have $rm phi(p^{j+k}) = p^{j+k}-p^{j+k-1} = p^j (p^k-p^{k-1}) = p^j phi(p^k)$






share|cite|improve this answer











$endgroup$



Hint $ $ A multiplicative function $rm:f(n):$ satisfies said identity if for all primes $rm:p:$



$$rm jle k Rightarrow f(p^{j+k}) = frac{f(p^j): f(p^k): p^j}{f(p^j)} = p^j f(p^k)$$



Indeed we have $rm phi(p^{j+k}) = p^{j+k}-p^{j+k-1} = p^j (p^k-p^{k-1}) = p^j phi(p^k)$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Feb 29 '12 at 16:02







user242

















answered Feb 29 '12 at 15:55









Bill DubuqueBill Dubuque

209k29191637




209k29191637












  • $begingroup$
    Thanks, Bill. I'll try to wrap my head around that. It seems useful.
    $endgroup$
    – The Chaz 2.0
    Feb 29 '12 at 16:44


















  • $begingroup$
    Thanks, Bill. I'll try to wrap my head around that. It seems useful.
    $endgroup$
    – The Chaz 2.0
    Feb 29 '12 at 16:44
















$begingroup$
Thanks, Bill. I'll try to wrap my head around that. It seems useful.
$endgroup$
– The Chaz 2.0
Feb 29 '12 at 16:44




$begingroup$
Thanks, Bill. I'll try to wrap my head around that. It seems useful.
$endgroup$
– The Chaz 2.0
Feb 29 '12 at 16:44


















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