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I searched for questions on perfect numbers. Most of the questions are abbreviated, but it looks like Tharsis found a workaround:

What's going on here?

Answer 1

Test answer with links

let $N$ be an odd perfect number. It is not difficult to show.

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut sed pharetra erat, in lacinia nunc. Curabitur eget dictum massa. Sed ac nunc eu elit fermentum condimentum nec sed dui. Duis vel semper enim, ut molestie leo. Donec viverra velit enim. Nulla congue fermentum aliquam. Sed a arcu elit. Nulla sollicitudin eget urna a ultricies. Donec varius blandit augue, a mattis ipsum bibendum sed. Nullam auctor luctus risus sed eleifend. Proin pulvinar nisi pulvinar, tristique velit vitae, adipiscing leo. Nunc nisl orci, pulvinar nec tellus a, cursus consequat nunc.

Nullam semper velit quis massa fringilla rutrum. Nam et metus nec lorem mollis bibendum. Fusce sem justo, congue pharetra ultrices sit amet, pulvinar id leo. Donec posuere in lorem at rhoncus. Vestibulum mollis vestibulum posuere. Nullam dapibus euismod nulla, sed mattis enim lobortis commodo. Etiam luctus tellus ut tellus cursus, sed sollicitudin augue vehicula. Nunc condimentum, nibh eget consectetur ornare, elit metus tempus sapien, non fringilla sem nulla ut purus. Nulla id tellus porttitor, tristique sem sit amet, imperdiet orci. Morbi vehicula lobortis velit in facilisis. Integer a sollicitudin eros.

let $N$ be an odd perfect number. It is not difficult to show.

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut sed pharetra erat, in lacinia nunc. Curabitur eget dictum massa. Sed ac nunc eu elit fermentum condimentum nec sed dui. Duis vel semper enim, ut molestie leo. Donec viverra velit enim. Nulla congue fermentum aliquam. Sed a arcu elit. Nulla sollicitudin eget urna a ultricies. Donec varius blandit augue, a mattis ipsum bibendum sed. Nullam auctor luctus risus sed eleifend. Proin pulvinar nisi pulvinar, tristique velit vitae, adipiscing leo. Nunc nisl orci, pulvinar nec tellus a, cursus consequat nunc.

Nullam semper velit quis massa fringilla rutrum. Nam et metus nec lorem mollis bibendum. Fusce sem justo, congue pharetra ultrices sit amet, pulvinar id leo. Donec posuere in lorem at rhoncus. Vestibulum mollis vestibulum posuere. Nullam dapibus euismod nulla, sed mattis enim lobortis commodo. Etiam luctus tellus ut tellus cursus, sed sollicitudin augue vehicula. Nunc condimentum, nibh eget consectetur ornare, elit metus tempus sapien, non fringilla sem nulla ut purus. Nulla id tellus porttitor, tristique sem sit amet, imperdiet orci. Morbi vehicula lobortis velit in facilisis. Integer a sollicitudin eros.

let $N$ be an odd perfect number. It is not difficult to show.

Answer 2

Let $\omega(n)$ denote the number of distinct prime factors of a positive integer $n$, and let $N$ be an odd perfect number. It is not difficult to show that $\omega(N)\ge3$. In fact, Nocco already proved this in 1863. Showing that $\omega(N)\neq3$ requires a little more effort, but the proof is still fairly short. I will present some version of the proof below; the original one is due to Peirce in 1830 (which, interestingly, seems to have been published before Nocco's proof).

The two key ideas here are the geometric series and the factor chain method. Let $N=\displaystyle\prod_{i=1}^k p_i^{e_i}$ be the prime factorization of $N$. Then it follows that

$$\sigma(N)=\prod_{i=1}^k \sum_{j=0}^{e_i} p_i^j=\prod_{i=1}^k \frac{p_i^{e_i+1}-1}{p_i-1}.$$

This formula is useful, because each factor is necessarily greater than one when $e_i$ is positive, which is the case for all prime factors $p_i$. Hence $\sigma(N)\ge\sigma\left(\displaystyle\prod_{i\in{I}} p_i^{e_i}\right)$, where $I\subseteq\{1,2,\ldots,k\}$.

The second idea is perhaps even more fundamental, because we have yet to use the property of perfectness. Since $\sigma(N)=2N$, every divisor of $\sigma(N)$ must also divide $2N$; the other way around is not as interesting. In particular, every factor $\frac{p_i^{e_i+1}-1}{p_i-1}$ must divide $2N$. This result is really powerful; for example, observing that $2^1||2N$ yields Euler's well-known form of an odd perfect number after some algebra.

With these tools in our toolbox, the proof is almost immediate. Let $p^\alpha$, $q^\beta$, and $r^\gamma$ be three odd prime powers with $p<q<r$ and assume that $N=p^\alpha q^\beta r^\gamma$ is an odd perfect number. By the same token as above, we see that

$$\begin{align*}\sigma(N)=\sigma(p^\alpha q^\beta r^\gamma) &= \frac{p^{\alpha+1}-1}{p-1} \cdot \frac{q^{\beta+1}-1}{q-1} \cdot \frac{r^{\gamma+1}-1}{r-1} < \frac{p^\alpha p}{p-1} \cdot \frac{q^\beta q}{q-1} \cdot \frac{r^\gamma r}{r-1}\\\implies& \frac{\sigma(N)}{N} = \frac{\sigma(p^\alpha q^\beta r^\gamma)}{p^\alpha q^\beta r^\gamma} < \frac{p}{p-1} \cdot \frac{q}{q-1} \cdot \frac{r}{r-1}.\end{align*}$$

Suppose that $p=5$. Then $\frac{\sigma(N)}{N}<\frac{5}{4}\cdot\frac{7}{6}\cdot\frac{11}{10}=\frac{77}{48}<2$, which is impossible. Hence $p=3$.

Suppose that $q=7$. Then $\frac{\sigma(N)}{N}<\frac{3}{2}\cdot\frac{7}{6}\cdot\frac{11}{10}=\frac{77}{40}<2$, which is impossible. Hence $q=5$.

Suppose that $r=17$. Then $\frac{\sigma(N)}{N}<\frac{3}{2}\cdot\frac{5}{4}\cdot\frac{17}{16}=\frac{255}{128}<2$, which is impossible. Hence $r \le 13$.

In consequence, we have three possible cases to consider: $r=7$, $r=11$, and $r=13$.

In the first case, we have $N=3^\alpha 5^\beta 7^\gamma$. By the factor chain method, $3^1||N$ implies $4|2N$ and $7^1||N$ implies $8|2N$, neither of which is possible. Hence $\alpha\ge2$ and $\gamma\ge2$. It follows that

$$\frac{\sigma(N)}{N} \ge \frac{\sigma(3^2\cdot5\cdot7^2)}{3^2\cdot5\cdot7^2} = \left(1 + \frac{1}{3} + \frac{1}{3^2}\right)\left(1 + \frac{1}{5}\right)\left(1 + \frac{1}{7} + \frac{1}{7^2}\right)=\frac{494}{245}>2,$$

which is impossible, but we already knew that $105=3\cdot5\cdot7$ cannot divide an odd perfect number.

In the second case, we have $N=3^\alpha 5^\beta 11^\gamma$. Suppose that $\beta=1$. It follows that

$$\frac{\sigma(N)}{N} = \frac{\sigma(3^\alpha\cdot5\cdot11^\gamma)}{3^\alpha\cdot5\cdot11^\gamma} < \frac{3}{2} \cdot \frac{6}{5} \cdot \frac{11}{10} = \frac{99}{50} < 2,$$

which is impossible (could we know that beforehand?). Hence $\beta\ge2$. By the factor chain method, $\alpha\ge4$ and $\gamma\ge4$. It follows that

$$\frac{\sigma(N)}{N} \ge \frac{\sigma(3^4\cdot5^2\cdot11^4)}{3^4\cdot5^2\cdot11^4} = \left(1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4}\right)\left(1 + \frac{1}{5} + \frac{1}{5^2}\right)\left(1 + \frac{1}{11} + \frac{1}{11^2} + \frac{1}{11^3} + \frac{1}{11^4}\right)=\frac{99851}{49005}>2,$$

which is impossible, but we already knew that $825=3\cdot5^2\cdot11$ cannot divide an odd perfect number.

In the third case, we have $N=3^\alpha 5^\beta 13^\gamma$. $\beta=1$ implies deficiency, so $\beta\ge2$. Now, $\alpha=2$ also implies deficiency, so $\alpha\ge4$. By the factor chain method, we must have $\gamma\ge2$, which this time implies abundance. Hence the assumption leads to a contradiction; furthermore, $8775=3^3\cdot5^2\cdot13$ cannot divide an odd perfect number. We are finally done.

Unfortunately, proving that $\omega(N)\neq4$ is not this easy by hand. We can similarly show that $p=3$, but then we already have two possible choices for $q$. Of course, one could just write a program and loop through all the possibilities, but Sylvester did it successfully in 1888, unlikely by the aid of computers. Hence I wonder if this algorithm can be significantly improved, or if there are other more efficient techniques that achieve the same goal.

Answer 3

Let $\omega(n)$ denote the number of distinct prime factors of a positive integer $n$, and let $N$ be an odd perfect number. It is not difficult to show that $\omega(N)\ge3$. In fact, Nocco already proved this in 1863. Showing that $\omega(N)\neq3$ requires a little more effort, but the proof is still fairly short. I will present some version of the proof below; the original one is due to Peirce in 1830 (which, interestingly, seems to have been published before Nocco's proof).

The two key ideas here are the geometric series and the factor chain method. Let $N=\displaystyle\prod_{i=1}^k p_i^{e_i}$ be the prime factorization of $N$. Then it follows that

$$\sigma(N)=\prod_{i=1}^k \sum_{j=0}^{e_i} p_i^j=\prod_{i=1}^k \frac{p_i^{e_i+1}-1}{p_i-1}.$$

This formula is useful, because each factor is necessarily greater than one when $e_i$ is positive, which is the case for all prime factors $p_i$. Hence $\sigma(N)\ge\sigma\left(\displaystyle\prod_{i\in{I}} p_i^{e_i}\right)$, where $I\subseteq\{1,2,\ldots,k\}$.

The second idea is perhaps even more fundamental, because we have yet to use the property of perfectness. Since $\sigma(N)=2N$, every divisor of $\sigma(N)$ must also divide $2N$; the other way around is not as interesting. In particular, every factor $\frac{p_i^{e_i+1}-1}{p_i-1}$ must divide $2N$. This result is really powerful; for example, observing that $2^1||2N$ yields Euler's well-known form of an odd perfect number after some algebra.

With these tools in our toolbox, the proof is almost immediate. Let $p^\alpha$, $q^\beta$, and $r^\gamma$ be three odd prime powers with $p<q<r$ and assume that $N=p^\alpha q^\beta r^\gamma$ is an odd perfect number. By the same token as above, we see that

$$\begin{align*}\sigma(N)=\sigma(p^\alpha q^\beta r^\gamma) &= \frac{p^{\alpha+1}-1}{p-1} \cdot \frac{q^{\beta+1}-1}{q-1} \cdot \frac{r^{\gamma+1}-1}{r-1} < \frac{p^\alpha p}{p-1} \cdot \frac{q^\beta q}{q-1} \cdot \frac{r^\gamma r}{r-1}\\\implies& \frac{\sigma(N)}{N} = \frac{\sigma(p^\alpha q^\beta r^\gamma)}{p^\alpha q^\beta r^\gamma} < \frac{p}{p-1} \cdot \frac{q}{q-1} \cdot \frac{r}{r-1}.\end{align*}$$

Suppose that $p=5$. Then $\frac{\sigma(N)}{N}<\frac{5}{4}\cdot\frac{7}{6}\cdot\frac{11}{10}=\frac{77}{48}<2$, which is impossible. Hence $p=3$.

Suppose that $q=7$. Then $\frac{\sigma(N)}{N}<\frac{3}{2}\cdot\frac{7}{6}\cdot\frac{11}{10}=\frac{77}{40}<2$, which is impossible. Hence $q=5$.

Suppose that $r=17$. Then $\frac{\sigma(N)}{N}<\frac{3}{2}\cdot\frac{5}{4}\cdot\frac{17}{16}=\frac{255}{128}<2$, which is impossible. Hence $r \le 13$.

In consequence, we have three possible cases to consider: $r=7$, $r=11$, and $r=13$.

In the first case, we have $N=3^\alpha 5^\beta 7^\gamma$. By the factor chain method, $3^1||N$ implies $4|2N$ and $7^1||N$ implies $8|2N$, neither of which is possible. Hence $\alpha\ge2$ and $\gamma\ge2$. It follows that

$$\frac{\sigma(N)}{N} \ge \frac{\sigma(3^2\cdot5\cdot7^2)}{3^2\cdot5\cdot7^2} = \left(1 + \frac{1}{3} + \frac{1}{3^2}\right)\left(1 + \frac{1}{5}\right)\left(1 + \frac{1}{7} + \frac{1}{7^2}\right)=\frac{494}{245}>2,$$

which is impossible, but we already knew that $105=3\cdot5\cdot7$ cannot divide an odd perfect number.

In the second case, we have $N=3^\alpha 5^\beta 11^\gamma$. Suppose that $\beta=1$. It follows that

$$\frac{\sigma(N)}{N} = \frac{\sigma(3^\alpha\cdot5\cdot11^\gamma)}{3^\alpha\cdot5\cdot11^\gamma} < \frac{3}{2} \cdot \frac{6}{5} \cdot \frac{11}{10} = \frac{99}{50} < 2,$$

which is impossible (could we know that beforehand?). Hence $\beta\ge2$. By the factor chain method, $\alpha\ge4$ and $\gamma\ge4$. It follows that

$$\frac{\sigma(N)}{N} \ge \frac{\sigma(3^4\cdot5^2\cdot11^4)}{3^4\cdot5^2\cdot11^4} = \left(1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4}\right)\left(1 + \frac{1}{5} + \frac{1}{5^2}\right)\left(1 + \frac{1}{11} + \frac{1}{11^2} + \frac{1}{11^3} + \frac{1}{11^4}\right)=\frac{99851}{49005}>2,$$

which is impossible, but we already knew that $825=3\cdot5^2\cdot11$ cannot divide an odd perfect number.

In the third case, we have $N=3^\alpha 5^\beta 13^\gamma$. $\beta=1$ implies deficiency, so $\beta\ge2$. Now, $\alpha=2$ also implies deficiency, so $\alpha\ge4$. By the factor chain method, we must have $\gamma\ge2$, which this time implies abundance. Hence the assumption leads to a contradiction; furthermore, $8775=3^3\cdot5^2\cdot13$ cannot divide an odd perfect number. We are finally done.

Unfortunately, proving that $\omega(N)\neq4$ is not this easy by hand. We can similarly show that $p=3$, but then we already have two possible choices for $q$. Of course, one could just write a program and loop through all the possibilities, but Sylvester did it successfully in 1888, unlikely by the aid of computers. Hence I wonder if this algorithm can be significantly improved, or if there are other more efficient techniques that achieve the same goal.

Answer 4

Test answer without links

EDIT now with quotes

let $N$ be an odd "perfect number". It is not difficult to show.

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut sed pharetra erat, in lacinia nunc. Curabitur eget dictum massa. Sed ac nunc eu elit fermentum condimentum nec sed dui. Duis vel semper enim, ut molestie leo. Donec viverra velit enim. Nulla congue fermentum aliquam. Sed a arcu elit. Nulla sollicitudin eget urna a ultricies. Donec varius blandit augue, a mattis ipsum bibendum sed. Nullam auctor luctus risus sed eleifend. Proin pulvinar nisi pulvinar, tristique velit vitae, adipiscing leo. Nunc nisl orci, pulvinar nec tellus a, cursus consequat nunc.

Nullam semper velit quis massa fringilla rutrum. Nam et metus nec lorem mollis bibendum. Fusce sem justo, congue pharetra ultrices sit amet, pulvinar id leo. Donec posuere in lorem at rhoncus. Vestibulum mollis vestibulum posuere. Nullam dapibus euismod nulla, sed mattis enim lobortis commodo. Etiam luctus tellus ut tellus cursus, sed sollicitudin augue vehicula. Nunc condimentum, nibh eget consectetur ornare, elit metus tempus sapien, non fringilla sem nulla ut purus. Nulla id tellus porttitor, tristique sem sit amet, imperdiet orci. Morbi vehicula lobortis velit in facilisis. Integer a sollicitudin eros.

let $N$ be an odd "perfect number". It is not difficult to show.

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut sed pharetra erat, in lacinia nunc. Curabitur eget dictum massa. Sed ac nunc eu elit fermentum condimentum nec sed dui. Duis vel semper enim, ut molestie leo. Donec viverra velit enim. Nulla congue fermentum aliquam. Sed a arcu elit. Nulla sollicitudin eget urna a ultricies. Donec varius blandit augue, a mattis ipsum bibendum sed. Nullam auctor luctus risus sed eleifend. Proin pulvinar nisi pulvinar, tristique velit vitae, adipiscing leo. Nunc nisl orci, pulvinar nec tellus a, cursus consequat nunc.

Nullam semper velit quis massa fringilla rutrum. Nam et metus nec lorem mollis bibendum. Fusce sem justo, congue pharetra ultrices sit amet, pulvinar id leo. Donec posuere in lorem at rhoncus. Vestibulum mollis vestibulum posuere. Nullam dapibus euismod nulla, sed mattis enim lobortis commodo. Etiam luctus tellus ut tellus cursus, sed sollicitudin augue vehicula. Nunc condimentum, nibh eget consectetur ornare, elit metus tempus sapien, non fringilla sem nulla ut purus. Nulla id tellus porttitor, tristique sem sit amet, imperdiet orci. Morbi vehicula lobortis velit in facilisis. Integer a sollicitudin eros.

let $N$ be an odd "perfect number". It is not difficult to show.

Answer 5

Test answer with links

let $N$ be an odd perfect number. It is not difficult to show.

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut sed pharetra erat, in lacinia nunc.let $N$ be an odd perfect number. It is not difficult to show.

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut sed pharetra erat, in lacinia nunc.let $N$ be an odd perfect number. It is not difficult to show.

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut sed pharetra erat, in lacinia nunc.let $N$ be an odd perfect number. It is not difficult to show.

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut sed pharetra erat, in lacinia nunc.let $N$ be an odd perfect number. It is not difficult to show.

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut sed pharetra erat, in lacinia nunc.let $N$ be an odd perfect number. It is not difficult to show.

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut sed pharetra erat, in lacinia nunc.let $N$ be an odd perfect number. It is not difficult to show.

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut sed pharetra erat, in lacinia nunc.let $N$ be an odd perfect number. It is not difficult to show.

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut sed pharetra erat, in lacinia nunc.let $N$ be an odd perfect number. It is not difficult to show.

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut sed pharetra erat, in lacinia nunc.let $N$ be an odd perfect number. It is not difficult to show.

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut sed pharetra erat, in lacinia nunc.let $N$ be an odd perfect number. It is not difficult to show.

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut sed pharetra erat, in lacinia nunc.let $N$ be an odd perfect number. It is not difficult to show.

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut sed pharetra erat, in lacinia nunc.