nyuuzyou/wikis huggingface dataset:
Pell's Equation/Examples/8

== Theorem ==
Pell's equation:
:$x^2 - 8 y^2 = 1$
has the positive integral solutions:
and so on.
== Proof ==
From Continued Fraction Expansion of $\sqrt 8$:
:$\sqrt 8 = \sqbrk {2, \sequence {1, 4} }$
The cycle is of length $2$.
By Solution of Pell's Equation, the only solutions of $x^2 - 8 y^2 = 1$ are:
:${p_{2 r} }^2 - 8 {q_{2 r} }^2 = \paren {-1}^{2 r}$
for $r = 1, 2, 3, \ldots$
When $r = 1$ this gives:
:${p_2}^2 - 8 {q_2}^2 = 1$
which is the solution required.
From Convergents of Continued Fraction Expansion of $\sqrt 8$:
:$p_2 = 3$
:$q_2 = 1$
although on that page the numbering goes from $p_0$ to $p_9$, and $q_0$ to $q_9$.
Category:Pell's Equation
Category:8
Pell's Equation/Sequence

== Sequence of Minimum Solutions to Pell's Equation ==
The minimum solutions of $x$ and $y$ for the Diophantine equation $x^2 - n y^2 = 1$ for various values of $n$ are as follows:
:
$n$ 	$x$ 	$y$
$2$ 	$3$ 	$2$
$3$ 	$2$ 	$1$
$5$ 	$9$ 	$4$
$6$ 	$5$ 	$2$
$7$ 	$8$ 	$3$
$8$ 	$3$ 	$1$
$10$	$19$ 	$6$
$11$	$10$ 	$3$
$12$	$7$ 	$2$
$13$	$649$	$180$
$14$	$15$ 	$4$
$15$	$4$ 	$1$
Note the unusual peak at $13$.
Thus, the sequence for $x$ for non-square $n$ begins:
:$3, 2, 9, 5, 8, 3, 19, 10, 7, 649, 15, 4, 33, 17, 170, \ldots$
Similarly, the sequence for $y$ for non-square $n$ begins:
:$2, 1, 4, 2, 3, 1, 6, 3, 2, 180, 4, 1, 8, 4, 39, 2, 12, \ldots$
== Sources ==
* 
Category:Pell's Equation
Pell Number as Sum of Squares

== Theorem ==
Let $P_n$ be a Pell Number:
:$P_n = \begin{cases} 0 & : n = 0 \\ 1 & : n = 1 \\ 2 P_{n - 1} + P_{n - 2} & : \text {otherwise}\end{cases}$
Then:
:$P_{2 n + 1} = P_{n + 1}^2 + P_n^2$
== Proof ==
This proof proceeds by induction.
=== Basis for the Induction ===
$\map P 0$ is the case:
So $\map P 0$ is seen to hold.
$\map P 1$ is the case:
So $\map P 1$ is seen to hold.
=== Induction Hypothesis ===
Now we need to show that, if $\map P {n - 1}$ and $\map P n$ are true, where $n \ge 1$, then it logically follows that $\map P {n + 1}$ is true.
So this is our induction hypotheses:
:$P_{2 n - 1} = P_n^2 + P_{n - 1}^2$
:$P_{2 n + 1} = P_{n + 1}^2 + P_n^2$
from which we are to show:
:$P_{2 n + 3} = P_{n + 2}^2 + P_{n + 1}^2$
=== Induction Step ===
This is our induction step:
The result follows by induction.
Category:Pell Numbers
Pencil of Straight Lines through Intersection of Two Straight Lines

== Theorem ==
Let $u = l_1 x + m_1 y + n_1$.
Let $v = l_2 x + m_2 y + n_2$.
Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$, expressed using the general equations:
The pencil of lines through the point of intersection of $\LL_1$ and $\LL_2$ is given by:
:$\set {u + k v = 0: k \in \R} \cup \set {\LL_2}$
== Proof ==
Let $\LL$ denote an arbitrary straight line through the point of intersection of $\LL_1$ and $\LL_2$.
From Equation of Straight Line through Intersection of Two Straight Lines, $\LL$ can be given by an equation of the form:
:$u + k v = 0$
It remains to be seen that the complete pencil of lines through the point of intersection of $\LL_1$ and $\LL_2$ can be obtained by varying $k$ over the complete set of real numbers $\R$.
We have that $\LL$ can also be given by: 
:$\paren {l_1 x + m_1 y + n_1} - k \paren {l_2 x + m_2 y + n_2} = 0$
That is:
:$\paren {l_1 - k l_2} x + \paren {m_1 - k m_2} y + \paren {n_1 - k n_2} = 0$
Let the slope of $\LL$ be $\tan \psi$ where $\psi$ is the angle $\LL$ makes with the $x$-axis.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $10$. Equation of a straight line through the intersection of two given lines
Category:Pencils
Category:Straight Lines
Pendulum Contained by Cycloid moves along Cycloidal Path

== Theorem ==
Let a pendulum with a flexible rod be suspended from a point $P$.
Let the rod be contained by a pair of bodies shaped as the arcs of a cycloid such that $P$ is the cusp between those two arcs.
:
Then the bob is constrained to move such that its path traces the arc of a cycloid.
== Proof ==
From Evolute of Cycloid is Cycloid, the evolute of a cycloid is another cycloid.
From Curve is Involute of Evolute, the involute of a cycloid is another cycloid as well.
But by the definition of involute, the path defined by the pendulum as described is the involute of the cycloid.
Hence the result.
== Historical Note ==
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid
* : Chapter $\text {B}.23$: Evolutes and Involutes. The Evolute of a Cycloid
Category:Cycloids
Category:Pendulums
Pendulum Contained by Cycloid moves along Cycloidal Path/Historical Note

== Historical Note on Pendulum Contained by Cycloid moves along Cycloidal Path ==
The discovery that a Pendulum Contained by Cycloid moves along Cycloidal Path was made by during his work on developing a reliable and accurate pendulum clock.
Unfortunately, the technique proved impractical, as the energy losses caused by the mechanics of the system compromise the pendulum's ability to swing reliably for a practical length of time.
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid
Category:Historical Notes
Penholodigital Product of Penholodigital Set of Factors

== Observation ==
:$164 \, 597 \, 832 = 6 \times 7 \times 9 \times 524 \times 831$
Both sides are penholodigital.
== Historical Note ==
== Sources ==
* : $164,597,832$
Category:Penholodigital Sets
Category:164,597,832
Penholodigital Product of Penholodigital Set of Factors/Historical Note

== Historical Note on Penholodigital Product of Penholodigital Set of Factors ==
According to in his $1997$ work , this factoid appeared in in the issue dated $25$ January $1996$, reported by a writer named Havard, but this has not been corroborated.
== Sources ==
* : $164,597,832$
Category:Historical Notes
Penholodigital Properties of 123,456,789

== Theorem ==
$123 \, 456 \, 789$ has the following properties:
It is penholodigital, and remains so when multiplied by $2$, $4$, $5$, $7$ and $8$:
== Examples ==
== Also see ==
* Properties of 12,345,679
== Sources ==
* : $123,456,789$
* : $123,456,789$
Category:123,456,789
Category:Penholodigital Integers
Penholodigital Properties of 123,456,789/Examples/8

== Puzzle ==
:Place in a row $9$ digits each different from the others.
:Multiply them by $8$, and the product shall still consist of $9$ different digits.
== Solution ==
:$123 \, 456 , 789 \times 8 = 987 \, 654 \, 312$
== Sources ==
* 
* : Rational Amusements for Winter Evenings: $150$
Category:Penholodigital Integers
Penholodigital Properties of 987,654,321

== Theorem ==
$987 \, 654 \, 321$ has the following properties:
It is penholodigital, and remains so when multiplied by $1$, $2$, $4$, $5$, $7$ and $8$:
Also:
:$987 \, 654 \, 321 - 123 \, 456 \, 789 = 864 \, 197 \, 532$
which is also penholodigital.
== Sources ==
* : $987,654,321$
* : $987,654,321$
Category:987,654,321
Category:Penholodigital Integers
Penholodigital Square Equation

== Theorem ==
The following equations, which include each digit from $1$ to $9$ inclusive, are the only ones of their kind:
== Proof ==
The square of a $2$-digit integer cannot have more than $4$ digits:
:$99^2 = 9801$
The square of a $4$-digit integer has at least $7$ digits:
:$1000^2 = 1 \, 000 \, 000$
Hence we only need to inspect $3$-digit integers, with a corresponding $6$-digit square.
A lower bound is given by $\ceiling {\sqrt {123 \, 456}} = 352$, where $123 \, 456$ is the smallest $6$-digit integer without repeating digits or $0$.
An upper bound is given by $\floor {\sqrt {876 \, 543}} = 936$, where $876 \, 543$ is the largest $6$-digit integer without repeating digits or $9$.
It is seen that it is not necessary to investigate $6$-digit squares beginning with $9$, because its square root would also begin with $9$.
It is noted that integers ending in $1$, $5$ or $6$ have squares ending in those same digits.
Such numbers can be eliminated from our search, as they will duplicate the appearance of those digits.
We also cannot have $0$ as a digit.
Moreover, observe that an integer $\bmod 9$ and its square $\bmod 9$ have the following pattern:
:$\begin{array}{|c|c|c|c|c|c|c|c|c|c|}
\hline n \bmod 9 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline n^2 \bmod 9 & 0 & 1 & 4 & 0 & 7 & 7 & 0 & 4 & 1 \\
\hline
\end{array}$
By Congruence of Sum of Digits to Base Less 1, the sum of digits of $n$ and $n^2 \pmod 9$ are congruent to $n$ and $n^2 \pmod 9$ respectively.
We must require their sum to be congruent to $0 \bmod 9$ since the sum of all their digits is:
:$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 \equiv 0 \pmod 9$
Hence $n \equiv 0$ or $8 \pmod 9$.
There are $52$ numbers left to check:
For $n \equiv 0 \pmod 9$:
For $n \equiv 8 \pmod 9$:
Hence the result by inspection.
== Sources ==
* : $567$
* : $567$
Category:Square Numbers
Category:Recreational Mathematics
Category:Penholodigital Sets
Category:Penholodigital Integers
Pentagonal Number/Sequence

== Definition ==
The sequence of pentagonal numbers, for $n \in \Z_{\ge 0}$, begins:
:$0, 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, 176, \ldots$
== Sources ==
* : $22$
* : Tables: $2$ The First $20$ Pentagonal, Hexagonal, Heptagonal and Octagonal Numbers
* : $22$
* : Tables: $2$ The First $20$ Pentagonal, Hexagonal, Heptagonal and Octagonal Numbers
* 
* 
Category:Pentagonal Numbers
Pentagonal Number as Sum of Triangular Numbers

== Theorem ==
Let $P_n$ be the $n$th pentagonal number.
Then:
:$P_n = T_n + 2 T_{n - 1}$
where $T_n$ is the $n$th triangular number.
== Proof ==
== Illustration ==
:
== Sources ==
* : $15$
* : $15$
Category:Triangular Numbers
Category:Pentagonal Numbers
Pentagonal and Hexagonal Numbers

== Theorem ==
The sequence of positive integers which are simultaneously pentagonal and hexagonal begins:
:$1, 40 \, 755, 1 \, 533 \, 776 \, 805, 57 \, 722 \, 156 \, 241 \, 751, \ldots$
== Proof ==
== Sources ==
* : $40,755$
* : $40,755$
Category:Pentagonal Numbers
Category:Hexagonal Numbers
Pentagonal and Hexagonal Numbers/Mistake

== Source Work ==
:
:The Dictionary
::$40,755$
== Mistake ==
:The first number, after $1$ and $15$, to be simultaneously pentagonal and hexagonal and therefore, also, triangular.
== Correction ==
$15$ is not a pentagonal number as such.
It is, however, a second pentagonal number, and so is classified as a generalized pentagonal number.
If this is what meant, then he should have stated it.
== Sources ==
* : $40,755$
Category:Mistakes/Curious and Interesting Numbers
Percentage/Examples

== Examples of Percentages ==
=== Example: $5 \%$ ===
=== Example: $20 \%$ ===
=== Example: $25 \%$ ===
=== Example: $50 \%$ ===
=== Example: $120 \%$ ===
=== Example: $b - a$ as a Percentage ===
Category:Examples of Percentages
Percentage Decrease/Examples/Arbitrary Example 1

== Examples of Percentage Decrease ==
Let the price of an item $Q$ be reduced from $\$ 500$ to $\$ 400$.
Then the percentage decrease of $Q$ is:
:$\dfrac {500 - 400} {500} \times 100 = 20 \%$
== Sources ==
* 
Category:Examples of Percentage Decrease
Percentage Error/Examples/Arbitrary Example 1

== Examples of Percentage Error ==
Let the true value of a population mean be $45$.
Let the estimate of that population mean, based on a sample, be $47 \cdotp 2$.
Then the percentage error of this estimate is:
:$\dfrac {47 \cdotp 2 - 45} {45} \times 100 = 4 \cdotp 9 \%$
to $2$ significant figures.
== Sources ==
* 
* 
Category:Examples of Percentage Error
Percentage Increase/Examples/Arbitrary Example 1

== Examples of Percentage Decrease ==
Let the price of an item $Q$ be increased from $\$ 400$ to $\$ 500$.
Then the percentage increase of $Q$ is:
:$\dfrac {500 - 400} {400} \times 100 = 25 \%$
== Sources ==
* 
Category:Examples of Percentage Increase
Percentage Point/Examples/Standard Normal Distribution

== Examples of Percentage Points ==
Let $T$ have a standard normal distribution
The lower $5 \%$ percentage point relevant to a one-tail test is $-1 \cdotp 64$.
The upper $5 \%$ percentage point relevant to a one-tail test is $1 \cdotp 64$.
Hence values outside the interval $\openint {-1 \cdotp 64} {1 \cdotp 64}$ indicate significance level of $5 \%$ in a one-tail hypothesis test.
== Sources ==
* 
Category:Examples of Percentage Points
Perfect Digit-to-Digit Invariant/Examples/3435

== Example of Perfect Digit-to-Digit Invariant ==
$3435$ is a perfect digit-to-digit invariant:
:$3435 = 3^3 + 4^4 + 3^3 + 5^5$
== Proof ==
== Sources ==
* 
* : $3435$
* : $3435$
Category:Perfect Digit-to-Digit Invariants
Category:3435
Perfect Digit-to-Digit Invariant/Examples/3435/Mistake

== Source Work ==
:
:The Dictionary
::$3435$
== Mistake ==
:$3435 = 3^3 + 4^4 + 3^3 + 5^5$. This is the only known number with this property, apart from $1^1 = 1$.
== Correction ==
Later on in the same book, in Section $438,579,088$, we see:
:$438,579,088$
::Equal to $4^4 + 3^3 + 8^8 + 5^5 + 7^7 + 9^9 + 0^0 + 8^8 + 8^8$.
::The only other number with this property is $3435$.
== Sources ==
* : $3435$
Category:Mistakes/Curious and Interesting Numbers
Perfect Digit-to-Digit Invariant/Examples/438,579,088

== Example of Perfect Digit-to-Digit Invariant ==
$438 \, 579 \, 088$ is a perfect digit-to-digit invariant:
:$438 \, 579 \, 088 = 4^4 + 3^3 + 8^8 + 5^5 + 7^7 + 9^9 + 0^0 + 8^8 + 8^8$
== Proof ==
== Sources ==
* : $438,579,088$
* : $438,579,088$
Category:Perfect Digit-to-Digit Invariants
Category:438,579,088
Perfect Magic Cube/Examples/Order 7

== Example of Perfect Magic Cube ==
Order $7$ perfect magic cube:
:$\begin{array}{|c|c|c|c|c|c|c|}
\hline 327 & 41 & 98 & 99 & 156 & 213 & 270 \\
\hline 52 & 109 & 166 & 223 & 280 & 330 & 44 \\
\hline 169 & 226 & 283 & 340 & 5 & 62 & 119 \\
\hline 293 & 301 & 8 & 65 & 122 & 179 & 236 \\
\hline 18 & 75 & 132 & 189 & 239 & 247 & 304 \\
\hline 135 & 192 & 200 & 257 & 314 & 28 & 78 \\
\hline 210 & 260 & 317 & 31 & 88 & 145 & 153 \\
\hline
\end{array} \qquad \begin{array}{|c|c|c|c|c|c|c|}
\hline 113 & 170 & 227 & 284 & 341 & 6 & 63 \\
\hline 237 & 294 & 295 & 9 & 66 & 123 & 180 \\
\hline 305 & 19 & 76 & 133 & 183 & 240 & 248 \\
\hline 79 & 136 & 193 & 201 & 258 & 315 & 22 \\
\hline 154 & 204 & 261 & 318 & 32 & 89 & 146 \\
\hline 271 & 328 & 42 & 92 & 100 & 157 & 214 \\
\hline 45 & 53 & 110 & 167 & 224 & 274 & 331 \\
\hline
\end{array}$
:$\begin{array}{|c|c|c|c|c|c|c|}
\hline 249 & 306 & 20 & 77 & 127 & 184 & 241 \\
\hline 23 & 80 & 137 & 194 & 202 & 259 & 309 \\
\hline 147 & 148 & 205 & 262 & 319 & 33 & 90 \\
\hline 215 & 272 & 329 & 36 & 93 & 101 & 158 \\
\hline 332 & 46 & 54 & 111 & 168 & 218 & 275 \\
\hline 57 & 114 & 171 & 228 & 285 & 342 & 7 \\
\hline 181 & 238 & 288 & 296 & 10 & 67 & 124 \\
\hline
\end{array} \qquad \begin{array}{|c|c|c|c|c|c|c|}
\hline 91 & 141 & 149 & 206 & 263 & 320 & 34 \\
\hline 159 & 216 & 273 & 323 & 37 & 94 & 102 \\
\hline 276 & 333 & 47 & 55 & 112 & 162 & 219 \\
\hline 1 & 58 & 115 & 172 & 229 & 286 & 343 \\
\hline 125 & 182 & 232 & 289 & 297 & 11 & 68 \\
\hline 242 & 250 & 307 & 21 & 71 & 128 & 185 \\
\hline 310 & 24 & 81 & 138 & 195 & 203 & 253 \\
\hline
\end{array}$
:$\begin{array}{|c|c|c|c|c|c|c|}
\hline 220 & 277 & 334 & 48 & 56 & 106 & 163 \\
\hline 337 & 2 & 59 & 116 & 173 & 230 & 287 \\
\hline 69 & 126 & 176 & 233 & 290 & 298 & 12 \\
\hline 186 & 243 & 251 & 308 & 15 & 72 & 129 \\
\hline 254 & 311 & 25 & 82 & 139 & 196 & 197 \\
\hline 35 & 85 & 142 & 150 & 207 & 264 & 321 \\
\hline 103 & 160 & 217 & 267 & 324 & 38 & 95 \\
\hline
\end{array} \qquad \begin{array}{|c|c|c|c|c|c|c|}
\hline 13 & 70 & 120 & 177 & 234 & 291 & 299 \\
\hline 130 & 187 & 244 & 252 & 302 & 16 & 73 \\
\hline 198 & 255 & 312 & 26 & 83 & 140 & 190 \\
\hline 322 & 29 & 86 & 143 & 151 & 208 & 265 \\
\hline 96 & 104 & 161 & 211 & 268 & 325 & 39 \\
\hline 164 & 221 & 278 & 335 & 49 & 50 & 107 \\
\hline 281 & 338 & 3 & 60 & 117 & 174 & 231 \\
\hline
\end{array}$
:$\begin{array}{|c|c|c|c|c|c|c|}
\hline 191 & 199 & 256 & 313 & 27 & 84 & 134 \\
\hline 266 & 316 & 30 & 87 & 144 & 152 & 209 \\
\hline 40 & 97 & 105 & 155 & 212 & 269 & 326 \\
\hline 108 & 165 & 222 & 279 & 336 & 43 & 51 \\
\hline 225 & 282 & 339 & 4 & 61 & 118 & 175 \\
\hline 300 & 14 & 64 & 121 & 178 & 235 & 292 \\
\hline 74 & 131 & 188 & 245 & 246 & 303 & 17 \\
\hline
\end{array}$
== Historical Note ==
== Sources ==
* 
* : $8$
* : $8$
Category:Examples of Perfect Magic Cubes
Perfect Number/Examples/28

== Example of Perfect Number ==
$28$ is a perfect number:
:$1 + 2 + 4 + 7 + 14 = 28$
== Proof ==
Thus $28$ is in the form $2^{p - 1} \paren {2^p - 1}$.
$2^3 - 1 = 7$ is prime.
So $28$ is perfect by the Theorem of Even Perfect Numbers.
The aliquot parts of $28$ are enumerated at .
:$1, 2, 4, 7, 14$
== Sources ==
* 
* 
Category:Perfect Numbers/Examples
Category:28
Perfect Number/Examples/496

== Example of Perfect Number ==
$496$ is a perfect number:
:$1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 = 496$
== Proof ==
Thus $496$ is in the form $2^{p - 1} \paren {2^p - 1}$.
$2^5 - 1 = 31$ is prime.
So $496$ is perfect by the Theorem of Even Perfect Numbers.
The aliquot parts of $496$ are enumerated at :
:$1, 2, 4, 8, 16, 31, 62, 124, 248$
== Sources ==
* 
* 
Category:Perfect Numbers/Examples
Category:496
Perfect Number/Examples/6

== Example of Perfect Number ==
$6$ is a perfect number:
:$1 + 2 + 3 = 6$
== Proof ==
Thus $6$ is in the form $2^{n - 1} \paren {2^n - 1}$.
$2^2 - 1 = 3$ is prime.
So $6$ is perfect by the Theorem of Even Perfect Numbers.
The aliquot parts of $6$ are enumerated at :
:$1, 2, 3$
== Sources ==
* 
* 
* 
* 
Category:Perfect Numbers/Examples
Category:6
Perfect Number/Examples/8128

== Example of Perfect Number ==
$8128$ is a perfect number:
:$1 + 2 + 4 + 8 + 16 + 32 + 64 + 127 + 254 + 508 + 1016 + 2032 + 4064 = 8128$
== Proof ==
Thus $8128$ is in the form $2^{p - 1} \paren {2^p - 1}$.
$2^7 - 1 = 127$ is prime.
So $8128$ is perfect by the Theorem of Even Perfect Numbers.
The aliquot parts of $8128$ are enumerated at :
:$1, 2, 4, 8, 16, 32, 64, 127, 254, 508, 1016, 2032, 4064$
Category:Perfect Numbers/Examples
Category:8128
Perfect Number ends in 6 or 28 preceded by Odd Digit

== Theorem ==
Let $n$ be an even perfect number.
Then $n$ ends either in $6$ or $28$ preceded by an odd digit.
== Proof ==
By the Theorem of Even Perfect Numbers:
:$n = 2^{p - 1} \paren {2^p - 1}$
where $p$ is prime.
With the exception of $6 = 2^1 \paren {2^2 - 1}$ and $28 = 2^2 \paren {2^3 - 1}$:
:$p$ is odd and $p > 4$.
We claim that:
:$n$ ends in $\phantom 0 6$ preceded by an odd digit if $p \equiv 1 \pmod 4$
:$n$ ends in $28$ preceded by an odd digit if $p \equiv 3 \pmod 4$
These statements are equivalent to:
:$n \equiv \phantom 0 16 \pmod {\phantom 0 20}$ if $p \equiv 1 \pmod 4$
:$n \equiv 128 \pmod {200}$ if $p \equiv 3 \pmod 4$
By Powers of 16 Modulo 20, we have:
:$2^{4 n} = 16^n \equiv 16 \pmod {20}$ for $n \ge 1$
=== Case $1$: $p \equiv 1 \pmod 4$ ===
Write $p = 4 n + 1$.
Then:
showing our first claim.
=== Case $2$: $p \equiv 3 \pmod 4$ ===
Write $p = 4 n + 3$.
By Powers of 16 Modulo 20, we can write $2^{4 n} = 20 K + 16$ for some $K \in \Z$.
Then:
showing our second claim.
== Sources ==
* : No. $207$
* : $28$
* : $28$
Category:Perfect Numbers
Perfect Number has at least Two Distinct Prime Factors/Proof 1

== Theorem ==
== Proof ==
 the contrary: that $n$ is a perfect number with exactly $1$ prime factor.
Hence let $n = p^k$ where $p$ is prime.
By definition of perfect number:
:$\map {\sigma_1} n = 2 n$
where $\map {\sigma_1} n$ denotes the divisor sum of $n$.
Hence:
That is, $p$ is a divisor of $1$ which is a contradiction.
Hence the result by Proof by Contradiction.
Category:Perfect Number has at least Two Distinct Prime Factors
Perfect Number has at least Two Distinct Prime Factors/Proof 2

== Theorem ==
== Proof ==
 the contrary: that $n$ is a perfect number with exactly $1$ prime factor.
Hence let $n = p^k$ where $p$ is prime.
By definition of perfect number:
:$\map {\sigma_1} n = 2 n$
where $\map {\sigma_1} n$ denotes the divisor sum of $n$.
Hence:
Hence the result by Proof by Contradiction.
Category:Perfect Number has at least Two Distinct Prime Factors
Perfect Number is Ore Number

== Theorem ==
Let $n \in \Z_{>0}$ be a perfect number.
Then $n$ is an Ore number.
== Proof ==
From Harmonic Mean of Divisors in terms of Divisor Count and Divisor Sum, the harmonic mean of the divisors of $n$ is given by:
:$\map H n = \dfrac {n \map {\sigma_0} n} {\map {\sigma_1} n}$
where:
:$\map {\sigma_0} n$ denotes the divisor count function: the number of divisors of $n$
:$\map {\sigma_1} n$ denotes the divisor sum function: the sum of the divisors of $n$.
Let $n$ be a perfect number.
By definition of perfect number:
:$\dfrac {\map {\sigma_1} n} n = 2$
From Divisor Count Function is Odd Iff Argument is Square:
:$\map {\sigma_0} n = 2 k$
for some $k \in \Z$.
Hence:
:$\map H n = \dfrac {2 k} 2 = k$
Hence the result.
== Sources ==
* 
* : $140$
Category:Ore Numbers
Category:Perfect Numbers
Perfect Number is Primitive Semiperfect

== Theorem ==
Let $n \in \Z_{>0}$ be a perfect number.
Then $n$ is also a primitive semiperfect number.
== Proof ==
Let $n$ be perfect.
From Divisor of Perfect Number is Deficient, all divisors of $n$ are deficient.
But from Semiperfect Number is not Deficient, it follows that the divisors of $n$ cannot be semiperfect.
Hence the result, by definition of primitive semiperfect number.
Category:Perfect Numbers
Category:Primitive Semiperfect Numbers
Perfect Number is Sum of Successive Odd Cubes except 6

== Theorem ==
Let $n$ be an even perfect number such that $n \ne 6$.
Then:
:$\ds n = \sum_{k \mathop = 1}^m \paren {2 k - 1}^3 = 1^3 + 3^3 + \cdots + \paren {2 m - 1}^3$
for some $m \in \Z_{>0}$.
That is, every even perfect number is the sum of the sequence of the first $r$ odd cubes, for some $r$.
== Proof ==
From Sum of Sequence of Odd Cubes:
:$1^3 + 3^3 + 5^3 + \cdots + \paren {2 m − 1}^3 = m^2 \paren {2 m^2 − 1}$
By the Theorem of Even Perfect Numbers:
:$n = 2^{r - 1} \paren {2^r - 1}$
for some $r$.
Setting $m = 2^{r - 2}$:
:$m^2 = 2^{r - 1}$
and so:
:$2 m^2 = 2^r$
and it follows that:
:$\ds n = \sum_{k \mathop = 1}^{2^{r - 2} } \paren {2 k - 1}^3$
and hence the result.
When $n = 6$ we have:
:$6 = 2^1 \paren {2^2 - 1}$
leading to $r = 2$ and thence $m^2 = 2$, at which point the formula fails to work.
== Sources ==
* : $6$
* : $28$
* : $6$
:::Note that this result is incorrectly stated.
* : $28$
Category:Perfect Numbers
Category:Sums of Sequences
Category:Cube Numbers
Perfect Number is Sum of Successive Odd Cubes except 6/Mistake

== Source Work ==
:
:The Dictionary
:: $6$
:
:The Dictionary
:: $6$
== Mistake ==
:[ $6$ ] is the only perfect number that is not the sum of successive cubes.
== Correction ==
The statement should be:
:[ $6$ ] is the only even perfect number that is not the sum of successive odd cubes.
For example:
:$496 = 1 + 27 + 125 + 343 = 1^3 + 3^3 + 5^3 + 7^3$
It is appreciated that it is not known whether there exist any odd perfect numbers.
However, for complete accuracy it is necessary to include a qualification of the parity of the perfect numbers for which the statement does in fact hold.
== Sources ==
* : $6$
* : $6$
Category:Mistakes/Curious and Interesting Numbers