Euler's phi function and perfect numbers

Question

Here is the prompt

Prove that $\phi(P)=2^{k-1}(2^{k-1}-1)$, given that $P>6$ is an even perfect number.

---

This is what i did,

$P = 2^{p-1}(2^p-1)$, through Euler's theorem $\phi(P)=\phi(2^{p-1}(2^p-1))$, because they are multiplicative = phi(2^(p-1)) * phi(2^p - 1) Then taking it individually

phi(2^(p-1)) = phi(2^k), given that k = p-1 = 2^(k-1)(2-1) through some lemma = 2^(k-1) Now for phi(2^p - 1)

I'm stuck because i don't know what to do with the -1 how does phi(2^k+1) = 2^(k-1) - 1

Answer 1

For even perfect numbers, $2^p-1$ must be a Mersenne prime. This implies that $\phi(2^p-1)=(2^p-1)-1=2(2^{p-1}-1)$, so

$$ \phi(P)=\phi(2^{p-1}(2^p-1))=2^{p-2}2(2^{p-1}-1)=2^{p-1}(2^{p-1}-1) $$

Answer 2

$\phi(2^{p-1}(2^p-1))=\phi(2^{p-1})\cdot \phi(2^p-1)=2^{p-2}\cdot(2^p-2)$ using the assumption that $2^p-1$ is prime.

Answer 3

First : if $n$ is an even perfect number, then $n=2^m(2^{m+1}-1)$, and $q=2^{m+1}-1$ is prime:

Let $n$ be a even perfect number. Then $n=2^{m}q$ with $m\geqslant 1$ , $2\nmid q$ and $\sigma(n)=\sigma(2^{m}q)=2n$.

Now, take a natural number $k$, then $k=p_{1}^{\alpha_{1}}\cdots p_{l}^{\alpha_{l}}$ and

$\displaystyle \sigma (k)=(p_{1}^{\alpha_{1}}+\cdots +1)\cdots(p_{l}^{\alpha_{l}}+\cdots +1)=\sigma (p_{1}^{\alpha_{1}})\cdots\sigma (p_{l}^{\alpha_{l}}) \displaystyle =\sigma (p_{1}^{\alpha_{1}})[\sigma (p_{2}^{\alpha_{2}})\cdots\sigma (p_{l}^{\alpha_{l}})]=\sigma(p_{1}^{\alpha_{1}})[\sigma(p_{2}^{\alpha_{2}}\cdots p_{l}^{\alpha_{l}})]$

Then $\sigma(n)=\sigma(2^{m})\sigma(q)=2^{m+1}q$

$\displaystyle \frac{2^{m+1}}{\sigma(2^{m})}=\frac{\sigma(q)}{q}\displaystyle \frac{2^{m+1}}{2^{m+1}-1}=\frac{\sigma(q)}{q}$

Because $(2^{m},2^{m+1}-1)=1$ and $2^{m+1}-1\geqslant 3$ ,then

$\displaystyle \frac{2^{m+1}}{2^{m+1}-1}$ is a reduced fraction that is not an integer.

Then $q\geqslant 2$ y $q\nmid \sigma(q)$.

So, if q is prime, we have $q=(2^{m+1}-1)$ and the result follows.

Let $q=p_{1}^{\beta_{1}}\cdots p_{t}^{\beta_{t}}$ , then

$\displaystyle \frac{\sigma(q)}{q}=\frac{(p_{1}^{\beta_{1}}+\cdots +1)}{p_{1}^{\beta_{1}}}\cdots\frac{(p_{t}^{\beta_{t}}+\cdots +1)}{p_{t}^{\beta_{t}}}$

$\displaystyle\geqslant\frac{(p_{1}^{\beta_{1}}+1)}{p_{1}^{\beta_{1}}}\cdots\frac{(p_{t}^{\beta_{t}}+1)}{p_{t}^{\beta_{t}}}>\frac{q+1}{q}$

Then $q$ must be prime.

So $q=(2^{m+1}-1)$ and $n=2^{m}(2^{m+1}-1)$ with $(2^{m+1}-1)$ prime.

The result you want follows easily from this, because if $P$ is perfect, then $P=2^{k−1}(2^{k−1}-1)$ with $(2^{k−1}-1)$ a prime number.

So:

$\phi(P)=\phi(2^{k−1})\phi(2^{k}-1)=2^{k−2}\phi(2^{k}−1)=2^{k−2}2(2^{k−1}−1)=2^{k−1}(2^{k−1}−1)$