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Version 11 |
| \PMlinkescapeword{forces} |
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| \PMlinkescapeword{perfect} |
\PMlinkescapeword{perfect} |
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| An integer $n$ is called \emph{perfect} if it is the sum of all positive divisors of $n$ less than $n$ itself. It is not known if there are any odd perfect numbers, but all even perfect numbers have been classified according to the following lemma: |
An integer $n$ is called \emph{perfect} if it is the sum of all positive divisors of $n$ less than $n$ itself. It is not known if there are any odd perfect numbers, but all even perfect numbers have been classified according to the following lemma: |
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| \begin{lemma*} |
\begin{lemma*} |
| An even number is perfect if and only if it equals $2^{k-1}(2^k-1)$ for some integer $k>1$ and $2^k-1$ is prime.\\ |
An even number is perfect if and only if it equals $2^{k-1}(2^k-1)$ for some integer $k>1$ and $2^k-1$ is prime.\\ |
| \end{lemma*} |
\end{lemma*} |
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| \begin{proof} |
\begin{proof} |
| Let $\sigma$ denote the sum of divisors function. Recall that this function is multiplicative. |
Necessity: Let $p=2^k-1$ be prime, let $n=2^{k-1}p$, and define $\sigma(a)$ as the sum of all positive divisors of the integer $a$. Since $\sigma$ is multiplicative (meaning $\sigma(ab)=\sigma(a)\sigma(b)$ when $\gcd(a,b)=1$), we have: |
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| Necessity: Let $p=2^k-1$ be prime and $n=2^{k-1}p$. We have that |
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| \begin{eqnarray*} |
\begin{eqnarray*} |
| \sigma(n) & = & \sigma(2^{k-1}p)\\ |
\sigma(n) & = & \sigma(2^{k-1}p)\\ |
| & = & \sigma(2^{k-1}) \sigma(p)\\ |
& = & \sigma(2^{k-1}) \sigma(p)\\ |
| & = & (2^k-1)(p+1)\\ |
& = & (2^k-1)(p+1)\\ |
| & = & (2^k-1)(2^k)\\ |
& = & (2^k-1)(2^k)\\ |
| & = & 2n, |
& = & 2n, |
| \end{eqnarray*} |
\end{eqnarray*} |
| which shows $n$ is perfect. |
which shows $n$ is perfect. |
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Sufficiency: Assume $n$ is an even perfect number. Write $n=2^{k-1}m$ for some odd $m$ and $k \geq 2$. Then we have $\gcd(2^{k-1},m)=1$. Thus,
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Sufficiency: Assume $n$ is an even perfect number. Write $n = 2^{k-1}m$ for some odd $m$ and $k \geq 2$. Then we have $\text{gcd}(2^{k-1},m)=1$, so
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| \[ \sigma (n) = \sigma (2^{k-1}m) = \sigma (2^{k-1}) \sigma (m) = (2^k-1)\sigma (m). \] |
\[ \sigma (n) = \sigma (2^{k-1}m) = \sigma (2^{k-1}) \sigma (m) = (2^k-1)\sigma (m). \] |
| Since $n$ is perfect, then by definition $\sigma(n)=2n$ which in our case means $\sigma(n)=2n=2^km$. Piecing together the two formulas for $\sigma(n)$ yields |
But if $n$ is perfect, then by definition $\sigma(n)=2n$ which in our case means $\sigma(n)=2n=2^km$. Piecing together our two formulas for $\sigma(n)$, we get |
| \[ 2^km=(2^k-1)\sigma(m). \] |
\[ 2^km=(2^k-1)\sigma(m). \] |
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Therefore, $(2^k-1)|2^km$, which forces $(2^k-1)|m$. Write $m=(2^k-1)M$. From above, we have:
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So $(2^k-1)|2^km$, which \PMlinkescapetext{forces} $(2^k-1)|m$. Write $m=(2^k-1)M$. So from above we have:
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| \begin{eqnarray*} |
\begin{eqnarray*} |
| 2^km & = & (2^k-1)\sigma(m) \\ |
2^km & = & (2^k-1)\sigma(m) \\ |
| 2^k(2^k-1)M & = & (2^k-1)\sigma(m) \\ |
2^k(2^k-1)M & = & (2^k-1)\sigma(m) \\ |
| 2^kM & = \sigma(m) |
2^kM & = \sigma(m) |
| \end{eqnarray*} |
\end{eqnarray*} |
| Since $m|m$ by definition and $M|m$ by assumption, we have |
Since $m|m$ by definition and $M|m$ by assumption, we have |
| \[ 2^kM=\sigma(m)\geq m+M=2^kM, \] |
\[ 2^kM=\sigma(m)\geq m+M=2^kM, \] |
| which forces $\sigma(m)=m+M$. Thus $m$ has only two positive divisors, $m$ and $M$. Hence $m$ must be prime, $M=1$, and $m=(2^k-1)M=2^k-1$, from which the result follows. |
which forces $\sigma(m)=m+M$. Thus $m$ has only two positive divisors, $m$ and $M$. Hence $m$ must be prime, $M=1$, and $m=(2^k-1)M=2^k-1$, from which the result follows. |
| \end{proof} |
\end{proof} |
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| The lemma can be used to produce examples of (even) perfect numbers: |
The lemma can be used to produce examples of (even) perfect numbers: |
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| \begin{itemize} |
\begin{itemize} |
| \item If $k=2$, then $2^k-1=2^2-1=3$, which is prime. According to the lemma, $2^{k-1}(2^k-1)=2^{2-1} \cdot 3=6$ is perfect. Indeed, $1+2+3=6$. |
\item If $k=2$, then $2^k-1=2^2-1=3$, which is prime. According to the lemma, $2^{k-1}(2^k-1)=2^{2-1} \cdot 3=6$ is perfect. Indeed, $1+2+3=6$. |
| \item If $k=3$, then $2^k-1=2^3-1=7$, which is prime. According to the lemma, $2^{k-1}(2^k-1)=2^{3-1} \cdot 7=28$ is perfect. Indeed, $1+2+4+7+14=28$. |
\item If $k=3$, then $2^k-1=2^3-1=7$, which is prime. According to the lemma, $2^{k-1}(2^k-1)=2^{3-1} \cdot 7=28$ is perfect. Indeed, $1+2+4+7+14=28$. |
| \item If $k=5$, then $2^k-1=2^5-1=31$, which is prime. According to the lemma, $2^{k-1}(2^k-1)=2^{5-1} \cdot 31=496$ is perfect. Indeed, $1+2+4+8+16+31+62+124+248=496$. |
\item If $k=5$, then $2^k-1=2^5-1=31$, which is prime. According to the lemma, $2^{k-1}(2^k-1)=2^{5-1} \cdot 31=496$ is perfect. Indeed, $1+2+4+8+16+31+62+124+248=496$. |
| \end{itemize} |
\end{itemize} |
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| Note that $k=4$ yields that $2^k-1=2^4-1=15$, which is not prime. |
Note that $k=4$ yields that $2^k-1=2^4-1=15$, which is not prime. |
