| Version 18 |
Version 17 |
| \PMlinkescapeword{forces} |
\PMlinkescapeword{forces} |
| \PMlinkescapeword{formulas} |
\PMlinkescapeword{formulas} |
| \PMlinkescapeword{perfect} |
\PMlinkescapeword{perfect} |
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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. |
Let $\sigma$ denote the sum of divisors function. Recall that this function is multiplicative. |
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| Necessity: Let $p=2^k-1$ be prime and $n=2^{k-1}p$. We have that |
Necessity: Let $p=2^k-1$ be prime and $n=2^{k-1}p$. We have that |
| \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 that $n$ is perfect. |
which shows that $n$ 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 some $k>1$. 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 $\gcd(2^{k-1},m)=1$. Thus,
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| \[ \sigma(n)=\sigma(2^{k-1}m)=\sigma(2^{k-1})\sigma(m)=(2^k-1)\sigma(m). \] |
\[ \sigma(n)=\sigma(2^{k-1}m)=\sigma(2^{k-1})\sigma(m)=(2^k-1)\sigma(m). \] |
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Since $n$ is perfect, $\sigma(n)=2n$ by definition. Therefore, $\sigma(n)=2n=2^km$. Piecing together the two formulas for $\sigma(n)$ yields
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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
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| \[ 2^km=(2^k-1)\sigma(m). \] |
\[ 2^km=(2^k-1)\sigma(m). \] |
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Thus, $(2^k-1)\mid 2^km$, which forces $(2^k-1)\mid m$. Write $m=(2^k-1)M$. Note that $1\le M<m$. From above, we have:
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Therefore, $(2^k-1)\mid 2^km$, which forces $(2^k-1)\mid m$. Write $m=(2^k-1)M$. Note that $1\le M<m$. 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 $m\mid m$ by definition of \PMlinkname{divides}{Divides} and $M\mid m$ by assumption, we have |
Since $m\mid m$ by definition of \PMlinkname{divides}{Divides} and $M\mid m$ by assumption, we have |
| \[ 2^kM=\sigma(m)\geq m+M=2^kM, \] |
\[ 2^kM=\sigma(m)\geq m+M=2^kM, \] |
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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.
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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.
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| \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. |
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| The sequence of known perfect numbers appears in the OEIS as sequence \PMlinkexternal{A000396}{http://www.research.att.com/~njas/sequences/?q=A000396}. |
The sequence of known perfect numbers appears in the OEIS as sequence \PMlinkexternal{A000396}{http://www.research.att.com/~njas/sequences/?q=A000396}. |
