avewright/proofwiki-math huggingface dataset:
Theorem: Difference between Two Squares equal to Repunit

The sequence of differences of two squares that each make a repunit begins:
{{begin-eqn}}
{{eqn | l = 1^2 - 0^2
      | r = 1
      | c = 
}}
{{eqn | l = 6^2 - 5^2
      | r = 11
      | c = 
}}
{{eqn | l = 20^2 - 17^2
      | r = 111
      | c = 
}}
{{eqn | l = 56^2 - 45^2
      | r = 1111
      | c = 
}}
{{eqn | l = 56^2 - 55^2
      | r = 111
      | c = 
}}
{{eqn | l = 156^2 - 115^2
      | r = 11 \, 111
      | c = 
}}
{{eqn | l = 340^2 - 67^2
      | r = 111 \, 111
      | c = 
}}
{{eqn | l = 344^2 - 65^2
      | r = 111 \, 111
      | c = 
}}
{{eqn | l = 356^2 - 125^2
      | r = 111 \, 111
      | c = 
}}
{{end-eqn}}

Proof:
Let $x^2 - y^2 = R_n$ for some $n$, where $R_n$ denotes the $n$-digit repunit.
From Integer as Difference between Two Squares:
:$R_n$ has at least two distinct divisors of the same parity that multiply to $R_n$.
Then from Difference of Two Squares:
:$x = \dfrac {a + b} 2$
:$y = \dfrac {a - b} 2$
where:
:$R_n = a b$
for all $a, b$ where:
:$a b = R_n$
:$a$ and $b$ are of the same parity.
Here we have that $R_n$ is odd.
So both $a$ and $b$ are always odd and therefore always of the same parity.
It remains to perform the calculations and evaluate the examples.
Theorem: Smallest 3-Digit Permutable Prime

The smallest $3$-digit permutable prime is $113$.

Proof:
:$113$ is prime.
:$131$ is prime.
:$311$ is prime.
Consider the $3$-digit primes smaller than $113$:
:$101, 103, 107, 109$
They all contain a zero.
Thus, for each of these, at least one permutation ends in a zero.
Hence it is divisible by $10$ and so is not prime.
{{qed}}
Theorem: Infinite Group has Infinite Number of Subgroups

Let $\struct {G, \circ}$ be an infinite group.
Then $\struct {G, \circ}$ has an infinite number of distinct subgroups.

Proof:
There are two cases to consider: either $\struct {G, \circ}$ has an infinite cyclic subgroup, or it does not.
Theorem: Group is Finite iff Finite Number of Subgroups

Let $\struct {G, \circ}$ be a group.
Then $G$ is finite {{iff}} $\struct {G, \circ}$ has a  finite number of subgroups.

Proof:
=== Necessary Condition ===
Suppose that $\struct {G, \circ}$ is a finite group.
Let $\struct {H, \circ}$ be a subgroup of $\struct {G, \circ}$.
$H \subseteq G$ by definition.
Therefore:
:$H \in \powerset G$
where $\powerset G$ denotes the power set of $G$.
By Power Set of Finite Set is Finite, $\powerset G$ is finite.
It is seen that the set of all subgroups form a subset of $\powerset G$.
The result then follows from Subset of Finite Set is Finite.
{{qed|lemma}}
Theorem: 3-Digit Permutable Primes

The $3$-digit permutable primes are:
:$311, 199, 337$
and their anagrams, and no other.

Proof:
It is confirmed that:
:$113, 131, 311$ are all prime
:$199, 919, 991$ are all prime
:$337, 373, 733$ are all prime.
From Digits of Permutable Prime, all permutable primes contain digits in the set:
:$\left\{ {1, 3, 7, 9}\right\}$
The sum of a $3$-digit repdigit number is divisible by $3$.
By Divisibility by 3 it follows that all $3$-digit repdigit numbers are divisible by $3$ and therefore composite.
Hence the possibly permutable primes are:
:$113, 117, 119, 133, 137, 139, 177, 179, 199, 337, 339, 377, 379, 399, 779, 799$
and their anagrams.
We eliminate $113, 199, 337$ from this list, as it has been established that they are permutable primes.
Of those remaining, the following are composite:
{{begin-eqn}}
{{eqn | l = 117
      | r = 3^2 \times 13
}}
{{eqn | l = 119
      | r = 7 \times 17
}}
{{eqn | l = 133
      | r = 7 \times 19
}}
{{eqn | l = 177
      | r = 3 \times 59
}}
{{eqn | l = 339
      | r = 3 \times 113
}}
{{eqn | l = 377
      | r = 13 \times 29
}}
{{eqn | l = 779
      | r = 19 \times 41
}}
{{eqn | l = 799
      | r = 17 \times 47
}}
{{end-eqn}}
It remains to demonstrate that at least one anagram of the remaining numbers:
:$137, 139, 179, 379$
is composite.
We find that:
{{begin-eqn}}
{{eqn | l = 371
      | r = 7 \times 53
}}
{{eqn | l = 319
      | r = 11 \times 29
}}
{{eqn | l = 791
      | r = 7 \times 113
}}
{{eqn | l = 793
      | r = 13 \times 61
}}
{{end-eqn}}
All contenders are eliminated except for the established permutable primes $113, 119, 337$ and their anagrams.
{{qed}}
Theorem: Digits of Permutable Prime

Let $p$ be a permutable prime with more than $1$ digit.
Then $p$ contains only digits from the set:
:$\left\{ {1, 3, 7, 9}\right\}$

Proof:
First note that from 3-Digit Permutable Primes, both $337$ and $199$ are permutable primes.
Hence it follows that all the elements of $\left\{ {1, 3, 7, 9}\right\}$ appear in at least one permutable prime.
Let $p$ contain an even digit $d$.
Then at least one anagram $p'$ of $p$ has $d$ at the end.
From Divisibility by 2, it follows that $p'$ is even.
As $p$ has more than $1$ digit it follows that $p'$ is composite.
Let $p$ contain the digit $5$.
Then at least one anagram $p'$ of $p$ has $5$ at the end.
From Divisibility by 5, it follows that $p'$ is divisible by $5$.
Except for $5$ itself, all integers which are divisible by $5$ are composite.
As $p$ has more than $1$ digit it follows that $p'$ is composite.
The result follows by definition of permutable prime.
{{qed}}
Category:Permutable Primes
fg1nktn31ejo7yn7ruv3oli7m6hih4k
Theorem: Divisibility by 5

An integer $N$ expressed in decimal notation is divisible by $5$ {{iff}} the units digit of $N$ is divisible by $5$.
That is:
:$N = \sqbrk {a_n \ldots a_2 a_1 a_0}_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $5$ 
{{iff}}:
:$a_0$ is divisible by $5$.

Proof:
Let $N$ be divisible by $5$.
Then:
{{begin-eqn}}
{{eqn | l = N
      | o = \equiv
      | r = 0 \pmod 5
}}
{{eqn | ll= \leadstoandfrom
      | l = \sum_{k \mathop = 0}^n a_k 10^k
      | o = \equiv
      | r = 0 \pmod 5
}}
{{eqn | ll= \leadstoandfrom
      | l = a_0 + 10 \sum_{k \mathop = 1}^n a_k 10^{k - 1}
      | o = \equiv
      | r = 0 \pmod 5
}}
{{eqn | ll= \leadstoandfrom
      | l = a_0
      | o = \equiv
      | r = 0 \pmod 5
      | c = as $10 \equiv 0 \pmod 5$
}}
{{end-eqn}}
{{qed}}
Theorem: Prime Repdigit Number is Repunit

Let $b \in \Z_{>0}$ be an integer greater than $1$.
Let $n \in \Z$ expressed in base $b$ be a repdigit number with more than $1$ digit.
Let $n$ be prime.
Then $n$ is a repunit (in base $b$).

Proof:
Let $n$ be a repdigit number with $k$ digits.
Then by the Basis Representation Theorem:
:$\ds n = \sum_{j \mathop = 0}^k m b^j$
for some $m$ such that $1 \le m < b$.
Let $m \ge 2$.
Then:
:$\ds n = m \sum_{j \mathop = 0}^k b^j$
and so has $m$ as a divisor.
Hence $n$ is not prime.
The result follows by the Rule of Transposition.
{{qed}}
Category:Repdigit Numbers
Category:Repunits
Category:Repunit Primes
g7obedralse47mgmfg5fez4upod57c4
Theorem: 2-Digit Permutable Primes

The $2$-digit permutable primes are:
:$11, 13, 17, 37, 79$
and their anagrams, and no other.

Proof:
It is confirmed that:
:$13$ and $31$ are both prime
:$17$ and $71$ are both prime
:$37$ and $73$ are both prime.
:$79$ and $97$ are both prime.
From Digits of Permutable Prime, all permutable primes contain digits in the set:
:$\left\{ {1, 3, 7, 9}\right\}$
By Prime Repdigit Number is Repunit, all $2$-digit repdigit numbers are composite except for $11$ which is prime.
From Repunit Prime is Permutable Prime, it follows that $11$ is a permutable prime.
Apart from $11$ whose status is known, the possibly permutable primes are:
:$13, 17, 19, 37, 39, 79$
and their anagrams.
We eliminate $13, 17, 37, 79$ from this list, as it has been established that they are permutable primes.
Of those remaining, the following is composite:
{{begin-eqn}}
{{eqn | l = 39
      | r = 3 \times 13
}}
{{end-eqn}}
It remains to demonstrate that the anagram of $19$, that is, $91$, composite.
We find that:
{{begin-eqn}}
{{eqn | l = 91
      | r = 7 \times 13
}}
{{end-eqn}}
All contenders are eliminated except for the established permutable primes $11, 13, 17, 37, 79$ and their anagrams.
{{qed}}
Theorem: Prime Gaps of 14

The following pairs of consecutive prime numbers are those whose difference is $14$:
:$\tuple {113, 127}, \tuple {293, 307}, \tuple {317, 331}, \ldots$
{{OEIS|A031932}}

Proof:
Demonstrated by listing the prime gaps.
Theorem: Prime Gaps of 18

After the prime gap of $14$ between the pairs of consecutive prime numbers:
:$\left({113, 127}\right), \left({293, 307}\right), \left({317, 331}\right), \ldots$
the next prime gap which is greater than $14$ is between the pair of consecutive prime numbers:
:$\left({523, 541}\right)$
for a prime gap of $18$.

Proof:
Demonstrated by listing the prime gaps.
Theorem: Smallest Number which is Sum of 4 Triples with Equal Products

The smallest positive integer which is the sum of $4$ 
distinct ordered triples, each of which has the same product, is $118$:
{{begin-eqn}}
{{eqn | l = 118
      | r = 14 + 50 + 54
}}
{{eqn | r = 15 + 40 + 63
      | c = 
}}
{{eqn | r = 18 + 30 + 70
      | c = 
}}
{{eqn | r = 21 + 25 + 72
      | c = 
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = 14 \times 50 \times 54
      | r = \paren {2 \times 7} \times \paren {2 \times 5^2} \times \paren {2 \times 3^3}
}}
{{eqn | r = 2^3 \times 3^3 \times 5^2 \times 7
      | c = 
}}
{{eqn | r = 37 \, 800
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 15 \times 40 \times 63
      | r = \paren {3 \times 5} \times \paren {2^3 \times 5} \times \paren {3^2 \times 7}
}}
{{eqn | r = 2^3 \times 3^3 \times 5^2 \times 7
      | c = 
}}
{{eqn | r = 37 \, 800
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 18 \times 30 \times 70
      | r = \paren {2 \times 3^2} \times \paren {2 \times 3 \times 5} \times \paren {2 \times 5 \times 7}
}}
{{eqn | r = 2^3 \times 3^3 \times 5^2 \times 7
      | c = 
}}
{{eqn | r = 37 \, 800
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 21 \times 25 \times 72
      | r = \paren {3 \times 7} \times \paren {5^2} \times \paren {2^3 \times 3^2}
}}
{{eqn | r = 2^3 \times 3^3 \times 5^2 \times 7
      | c = 
}}
{{eqn | r = 37 \, 800
      | c = 
}}
{{end-eqn}}
Theorem: Interval containing Prime Number of forms 4n - 1, 4n + 1, 6n - 1, 6n + 1

Let $n \in \Z$ be an integer such that $n \ge 118$.
Then between $n$ and $\dfrac {4 n} 3$ there exists at least one prime number of each of the forms:
:$4 m - 1, 4 m + 1, 6 m - 1, 6 m + 1$

Proof:
{{questionable|See below.}}
It is demonstrated that the result is true for $n = 118$:
:$\dfrac {4 \times 118} 3 = 157 \cdotp \dot 3$
The primes between $118$ and $157$ are:
{{begin-eqn}}
{{eqn | l = 127
      | r = 4 \times 32 - 1
}}
{{eqn | l = 
      | r = 6 \times 21 + 1
}}
{{eqn | l = 131
      | r = 4 \times 33 - 1
}}
{{eqn | l = 
      | r = 6 \times 22 - 1
}}
{{eqn | l = 137
      | r = 4 \times 34 + 1
}}
{{eqn | l = 
      | r = 6 \times 23 - 1
}}
{{eqn | l = 139
      | r = 4 \times 35 - 1
}}
{{eqn | l = 
      | r = 6 \times 23 + 1
}}
{{eqn | l = 149
      | r = 4 \times 37 + 1
}}
{{eqn | l = 
      | r = 6 \times 25 - 1
}}
{{eqn | l = 151
      | r = 4 \times 38 - 1
}}
{{eqn | l = 
      | r = 6 \times 25 + 1
}}
{{eqn | l = 157
      | r = 4 \times 39 + 1
}}
{{eqn | l = 
      | r = 6 \times 26 + 1
}}
{{end-eqn}}
So it can be seen that:
:the primes of the form $4 m - 1$ are:
::$127, 131, 139, 151$
:the primes of the form $4 m + 1$ are:
::$137, 139, 157$
:the primes of the form $6 m - 1$ are:
::$131, 137, 149$
:the primes of the form $6 m + 1$ are:
::$127, 139, 151, 157$
and so the conditions of the result are fulfilled for $118$.
But then note we have:
:$\dfrac {4 \times 117} 3 = 156$
and so the primes between $117$ and $157$ are the same ones as between $118$ and $157$, excluding $157$ itself.
Thus the conditions also hold for $117$.
It is puzzling as to why the condition $n \ge 118$ has been applied to the initial statement.
Theorem: Smallest Number to appear 6 Times in Pascal's Triangle

The smallest positive integer greater than $1$ to appear $6$ times in Pascal's Triangle is $120$.

Proof:
We have:
:$\dbinom {120} 1 = \dbinom {16} 2 = \dbinom {10} 3 = \dbinom {10} 7 = \dbinom {16} {14} = \dbinom {120} {119} = 120$
To verify that this is the smallest, we look at binomial coefficients that are no more than $120$.
Observe that for $n > 120$, $1 \le k \le n - 1$:
:$\dbinom n k \ge \dbinom n 1 = n > 120$
For $n > 16$, $2 \le k \le n - 2$:
:$\dbinom n k \ge \dbinom n 2 = \dfrac {n \paren {n - 1}} {2!} > \dfrac {16 \paren {15}} {2!} = 120$
For $n > 10$, $3 \le k \le n - 3$:
:$\dbinom n k \ge \dbinom n 3 = \dfrac {n \paren {n - 1} \paren {n - 2}} {3!} > \dfrac {10 \paren 9 \paren 8} {3!} = 120$
Any number, except $2$, appears twice in the form $\dbinom n 1$ and $\dbinom n {n - 1}$.
Therefore we scout for numbers appearing four times, not as those forms, by listing $\dbinom n k$ for:
:$n \le 10$ and $2 \le k \le n - 2$;
:$n \le 16$ and $k = 2$ or $n - 2$
They are:
:$6$
:$10, 10$
:$15, 20, 15$
:$21, 35, 35, 21$
:$28, 56, 70, 56, 28$
:$36, 84, 126, 126, 84, 36$
:$45, 120, *, *, *, 120, 45$
:$55, 55$
:$66, 66$
:$78, 78$
:$91, 91$
:$105, 105$
:$120, 120$
and we see that the only number appearing four times is $120$.
{{qed}}
Theorem: Smallest Number with 16 Divisors

The smallest positive integer with $16$ divisors is $120$.

Proof:
From {{DCFLink|120}}:
:$\map {\sigma_0} {120} = 16$
The result is a specific instance of Smallest Number with $2^n$ Divisors:
:$120 = 2 \times 3 \times 4 \times 5$
{{qed}}
Theorem: Fermat Set is Diophantine Quadruple

The Fermat set $F = \left\{{1, 3, 8, 120}\right\}$ is a Diophantine quadruple:
:$\forall a, b \in F: a \ne b: a b + 1 = n^2$
for some $n \in \Z$.

Proof:
{{begin-eqn}}
{{eqn | l = 1 \times 3 + 1
      | r = 4
      | c = 
}}
{{eqn | r = 2^2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 1 \times 8 + 1
      | r = 9
      | c = 
}}
{{eqn | r = 3^2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 1 \times 120 + 1
      | r = 121
      | c = 
}}
{{eqn | r = 11^2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 3 \times 8 + 1
      | r = 25
      | c = 
}}
{{eqn | r = 5^2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 3 \times 120 + 1
      | r = 361
      | c = 
}}
{{eqn | r = 19^2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 8 \times 120 + 1
      | r = 961
      | c = 
}}
{{eqn | r = 31^2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Fermat Set cannot be Extended to Diophantine Quintuple

The Fermat set $F = \set {1, 3, 8, 120}$ cannot be extended to a Diophantine quintuple.

Proof:
{{ProofWanted|Can be found in the link below.}}
Theorem: Ratio of 360 to Aliquot Sum

$360$ has the property that its ratio to its aliquot sum is $4 : 9$.

Proof:
The aliquot sum of an integer $n$ is the integer sum  of the aliquot parts of $n$.
That is, the aliquot sum of $360$ is the divisor sum of $360$ minus $360$.
Thus:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {360} - 360
      | r = 1170 - 360
      | c = {{DSFLink|360}}
}}
{{eqn | r = 810
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 4 \times \paren {\map {\sigma_1} {360} - 360}
      | r = 3240
      | c = 
}}
{{eqn | r = 9 \times 360
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Are All Triperfect Numbers Even?/Progress/Minimum Size

It has been established that an odd triperfect number, if one were to exist, would be greater than $10^{70}$.
If it does not have $3$ as a prime factor, then it is greater than $10^{108}$.

Proof:
{{ProofWanted|Details}}
Theorem: 121 is Square Number in All Bases greater than 2

Let $b \in \Z$ be an integer such that $b \ge 3$.
Let $n$ be a positive integer which can be expressed in base $b$ as $\sqbrk {121}_b$.
Then $n$ is a square number.

Proof:
Consider $\sqbrk {11}_b$.
By the Basis Representation Theorem:
:$\sqbrk {11}_b = b + 1$
Thus:
{{begin-eqn}}
{{eqn | l = {\sqbrk {11}_b}^2
      | r = \paren {b + 1}^2
      | c = 
}}
{{eqn | r = b^2 + 2 b + 1
      | c = Square of Sum
}}
{{eqn | r = \sqbrk {121}_b
      | c = 
}}
{{end-eqn}}
Thus:
:$\sqbrk {121}_b = {\sqbrk {11}_b}^2$
and so is a square number.
{{qed}}
Theorem: Cube of 11 is Palindromic

::$11^3 = 1331$

Proof:
From Square of 11 is Palindromic:
:$11^2 = 121$
Thus:
<pre>
 121
x 11
-----
 121
1210
-----
1331
</pre>{{qed}}
Theorem: Fourth Power of 11 is Palindromic

::$11^4 = 14641$

Proof:
From Square of 11 is Palindromic:
:$11^2 = 121$
Then from Square of 121 is Palindromic:
:$121^2 = 14641 = \paren {11^2}^2 = 11^4$
{{qed}}
Theorem: Palindromes in Base 10 and Base 3

The following $n \in \Z$ are palindromic in both decimal and ternary:
:$0, 1, 2, 4, 8, 121, 151, 212, 242, 484, 656, 757, \ldots$
{{OEIS|A007633}}

Proof:
:{| border="1"
|-
! align="center" style = "padding: 2px 10px" | $n_{10}$ 
! align="center" style = "padding: 2px 10px" | $n_3$ 
|-
| align="right" style = "padding: 2px 10px" | $0$
| align="right" style = "padding: 2px 10px" | $0$
|-
| align="right" style = "padding: 2px 10px" | $1$
| align="right" style = "padding: 2px 10px" | $1$
|-
| align="right" style = "padding: 2px 10px" | $2$
| align="right" style = "padding: 2px 10px" | $2$
|-
| align="right" style = "padding: 2px 10px" | $4$
| align="right" style = "padding: 2px 10px" | $11$
|-
| align="right" style = "padding: 2px 10px" | $8$
| align="right" style = "padding: 2px 10px" | $22$
|-
| align="right" style = "padding: 2px 10px" | $121$
| align="right" style = "padding: 2px 10px" | $11 \, 111$
|-
| align="right" style = "padding: 2px 10px" | $151$
| align="right" style = "padding: 2px 10px" | $12 \, 121$
|-
| align="right" style = "padding: 2px 10px" | $212$
| align="right" style = "padding: 2px 10px" | $21 \, 212$
|-
| align="right" style = "padding: 2px 10px" | $242$
| align="right" style = "padding: 2px 10px" | $22 \, 222$
|-
| align="right" style = "padding: 2px 10px" | $484$
| align="right" style = "padding: 2px 10px" | $122 \, 221$
|-
| align="right" style = "padding: 2px 10px" | $656$
| align="right" style = "padding: 2px 10px" | $220 \, 022$
|-
| align="right" style = "padding: 2px 10px" | $757$
| align="right" style = "padding: 2px 10px" | $1 \, 001 \, 001$
|}
{{qed}}
Theorem: Square Numbers which are Sum of Consecutive Powers

The only two square numbers which are the sum of consecutive powers of a positive integer are $121$ and $400$:
:$121 = 3^0 + 3^1 + 3^2 + 3^3 + 3^4 = 11^2$
:$400 = 7^0 + 7^1 + 7^2 + 7^3 = 20^2$

Proof:
:$121 = 1 + 3 + 9 + 27 + 81$
:$400 = 1 + 7 + 49 + 343$
{{ProofWanted|It remains to be shown that these are the only such square numbers.}}
Theorem: Numbers whose Difference equals Difference between Cube and Seventh Power

The following $2$ pairs of integers are the only ones known which exhibit this pattern:
{{begin-eqn}}
{{eqn | l = \size {5^3 - 2^7}
      | r = 5 - 2
}}
{{eqn | l = \size {13^3 - 3^7}
      | r = 13 - 3
}}
{{end-eqn}}

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \size {5^3 - 2^7}
      | r = \size {125 - 128}
      | c = 
}}
{{eqn | r = 3
      | c = 
}}
{{eqn | r = 5 - 2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \size {13^3 - 3^7}
      | r = \size {2197 - 2187}
      | c = 
}}
{{eqn | r = 10
      | c = 
}}
{{eqn | r = 13 - 3
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Reciprocal of 8

The reciprocal of $8$ can be expressed as the summation of the powers of $2$ multiplied by the reciprocal  powers of $10$:
:$\dfrac 1 8 = \ds \sum_{k \mathop \ge 0} \dfrac {2^k} {10^{k + 1} }$
<pre>
.1
  2
   4
    8
    16
     32
      64
      128
       256
        512
        1024
         ....
-------------
.12499999744
-------------
</pre>

Proof:
We have that:
:$\dfrac {2^k} {10^{k + 1} } = \dfrac 1 {10} \paren {\dfrac 1 5}^k$
and so by Sum of Infinite Geometric Sequence:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \dfrac {2^k} {10^{k + 1} }
      | r = \dfrac {1/10} {1 - 1/5}
      | c = 
}}
{{eqn | r = \dfrac 1 {10 \paren {4 / 5} }
      | c = 
}}
{{eqn | r = \dfrac 5 {40}
      | c = 
}}
{{eqn | r = \dfrac 1 8
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Triangles with Integer Area and Integer Sides in Arithmetical Sequence

The triangles with the following sides in arithmetic sequence have integer areas:
:$3, 4, 5$
:$13, 14, 15$
:$15, 28, 41$
:$15, 26, 37$
Their areas are:
:$6, 84, 126, 156$

Proof:
From Heron's Formula, the area $A$ of $\triangle ABC$ is given by:
:$A = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
where $s = \dfrac{a + b + c} 2$ is the semiperimeter of $\triangle ABC$.
For $3, 4, 5$:
{{begin-eqn}}
{{eqn | l = s
      | r = \frac {3 + 4 + 5} 2
      | c = 
}}
{{eqn | r = 6
      | c = 
}}
{{eqn | ll= \leadsto
      | l = A
      | r = \sqrt {6 \paren {6 - 3} \paren {6 - 4} \paren {6 - 5} }
      | c = 
}}
{{eqn | r = \sqrt {6 \times 3 \times 2 \times 1}
      | c = 
}}
{{eqn | r = \sqrt {6 \times 6}
      | c = 
}}
{{eqn | r = 6
      | c = 
}}
{{end-eqn}}
For $13, 14, 15$:
{{begin-eqn}}
{{eqn | l = s
      | r = \frac {13 + 14 + 15} 2
      | c = 
}}
{{eqn | r = 21
      | c = 
}}
{{eqn | ll= \leadsto
      | l = A
      | r = \sqrt {21 \paren {21 - 13} \paren {21 - 14} \paren {21 - 15} }
      | c = 
}}
{{eqn | r = \sqrt {21 \times 8 \times 7 \times 6}
      | c = 
}}
{{eqn | r = \sqrt {\paren {3 \times 7} \times 2^3 \times 7 \times \paren {2 \times 3} }
      | c = 
}}
{{eqn | r = 3 \times 7 \times 2^2
      | c = 
}}
{{eqn | r = 84
      | c = 
}}
{{end-eqn}}
For $15, 28, 41$:
{{begin-eqn}}
{{eqn | l = s
      | r = \frac {15 + 28 + 41} 2
      | c = 
}}
{{eqn | r = 42
      | c = 
}}
{{eqn | ll= \leadsto
      | l = A
      | r = \sqrt {42 \paren {42 - 15} \paren {42 - 28} \paren {42 - 41} }
      | c = 
}}
{{eqn | r = \sqrt {42 \times 27 \times 14 \times 1}
      | c = 
}}
{{eqn | r = \sqrt {\paren {2 \times 3 \times 7} \times 3^3 \times \paren {2 \times 7} }
      | c = 
}}
{{eqn | r = 2 \times 3^2 \times 7
      | c = 
}}
{{eqn | r = 126
      | c = 
}}
{{end-eqn}}
For $15, 26, 37$:
{{begin-eqn}}
{{eqn | l = s
      | r = \frac {15 + 26 + 37} 2
      | c = 
}}
{{eqn | r = 39
      | c = 
}}
{{eqn | ll= \leadsto
      | l = A
      | r = \sqrt {39 \paren {39 - 15} \paren {39 - 26} \paren {39  - 37} }
      | c = 
}}
{{eqn | r = \sqrt {39 \times 24 \times 13 \times 2}
      | c = 
}}
{{eqn | r = \sqrt {\paren {3 \times 13} \times \paren {2^3 \times 3} \times 13 \times 2}
      | c = 
}}
{{eqn | r = 2^2 \times 3 \times 13
      | c = 
}}
{{eqn | r = 156
      | c = 
}}
{{end-eqn}}
{{expand|The citation below generalises this result, and this page could be turned into it.}}
Theorem: Equivalence of Definitions of Weierstrass E-Function

Let $\mathbf y, \mathbf z, \mathbf w$  be $n$-dimensional vectors.
Let $\mathbf y$ be such that $\map{\mathbf y} a=A$ and $\map{\mathbf y} b=B$.
Let $J$ be a functional such that:
:$\ds J \sqbrk {\mathbf y} = \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$
{{TFAE|def = Weierstrass E-Function}}

Proof:
=== Definition 1 implies Definition 2 ===
By Definition 1:
:$\map E {x, \mathbf y, \mathbf z, \mathbf w} = \map F {x, \mathbf y, \mathbf w} - \map F {x, \mathbf y, \mathbf z} + \paren {\mathbf w - \mathbf z} \map {F_{\mathbf y'} } {x, \mathbf y, \mathbf z}$
By Taylor's Theorem, where expansion is done around $\mathbf w = \mathbf z$ and Lagrange form of remainder is used:
:$\ds \map F {x, \mathbf y, \mathbf w} = \map F {x, \mathbf y, \mathbf z} + \frac {\partial \map F {x, \mathbf y, \mathbf z} } {\partial \mathbf y'} \paren {\mathbf w - \mathbf z} + \frac 1 2 \sum_{i, j \mathop = 1}^n \paren {w_i - z_i} \paren {w_j - z_j} \frac {\partial^2 \map F {x, \mathbf y, \mathbf z + \theta \paren {\mathbf z - \mathbf w} } } {\partial y_i'y_j'}$
where $\theta \in \R: 0 < \theta < 1$.
Insertion of this expansion into the definition for Weierstrass E-Function leads to the desired result.
{{qed|lemma}}
Theorem: Factor of Mersenne Number Mp is of form 2kp + 1

:$q = 2 k p + 1$
for some integer $k$.

Proof:
Let $q \divides M_p$.
Then:
:$2^p \equiv 1 \pmod q$
From Integer to Power of Multiple of Order, the multiplicative order of $2 \pmod q$ divides $p$.
By Fermat's Little Theorem, the multiplicative order of $2 \pmod q$ also divides $q - 1$.
Hence:
:$q - 1 = 2 k p$
{{qed}}
Theorem: Factor of Mersenne Number Mp equivalent to 1 mod p

:$q \equiv 1 \pmod p$

Proof:
Let $q \divides M_p$.
From Factor of Mersenne Number $M_p$ is of form $2 k p + 1$: 
:$q = 2 k p + 1$
and so by definition of congruence modulo an integer:
:$q \equiv 1 \pmod p$
immediately.
{{qed}}