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So i know that $\phi(ab) = \phi(a)\phi(b)$ if $gcd(a,b)=1$ and if p is prime then $\phi(p^e) = p^e - p^{e-1}$.

I was thinking of splitting $n^e$ into product of things so that I can use the multiplicative property but im not sure how i can ensure the split's gcd is still 1.

Bill Dubuque
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LTM
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3 Answers3

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Show the value of $\frac{\phi(n)}{n}$ depends only on the distinct prime factors of $n.$

So if $m_1,m_2$ have the same set of prime factors, then $$\frac{\phi(m_1)}{m_1}=\frac{\phi(m_2)}{m_2}.$$

$n$ and $n^e$ have the same prime factors.

So: $$\frac{\phi(n)}{n}=\frac{\phi(n^e)}{n^e}$$ Multiply both sides by $n^e$ to get your result.

Thomas Andrews
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HINT: The set of positive integers $M < n^e$ relatively prime to $n^e$ are precisely those integers $M$ of the following form: $M = jn + k_M$, where $j \in \{0,1,2, \ldots, n^{e-1}-1\}$, and $k_M \in \{1,2,\ldots, n-1\}$ satisfying $(k_M,n)=1$. So $\phi(n)$ choices for $k_M$, and $n^{e-1}$ choices for $j$....

If $n$ is prime, then $\phi(n)=n-1$, and $n^{e-1}(n-1)=n^e-n^{e-1}$.

Mike
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  • How did you come up with $M=jn + k_M$? – LTM Mar 29 '22 at 19:53
  • Well, first think $\pmod n$. Then you do agree that $M$ can be written $M=jn+k_n$, where $j$ is a nonnegative integer $k_n = M \pmod n$, right? The condition that $M<p^e$ gives $j$ an integer between $0$ and $p^{e-1}$. – Mike Mar 29 '22 at 20:00
  • @LTM Well, first think $\pmod n$ before thinking $\pmod {n^e}$. Then you do agree that $M$ can be written $M=jn+k_n$, where $j$ is a nonnegative integer $k_n = M \pmod n$, right? The further condition that $M<p^e$ gives $j$ an integer between $0$ and $p^{e-1}$. – Mike Mar 29 '22 at 20:01
  • *excuse me, $k_M$, and not $k_n$. – Mike Mar 29 '22 at 22:53
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Here's a different way to exploit multiplicativity. As both sides of the equation are multiplicative functions sending $n=1$ to $1$, you need only check equality in the case $n=p^k$ with $p\in\Bbb P,\,k\ge1$, which is trivial.

J.G.
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