avewright/proofwiki-math huggingface dataset:
Theorem: Tableau Extension Lemma/General Statement/Proof 1

Let $T$ be a finite propositional tableau.
Let its hypothesis set $\mathbf H$ be finite.
{{:Tableau Extension Lemma/General Statement}}

Proof:
Let $T_{\mathbf H'}$ be the finite propositional tableau obtained by replacing the hypothesis set $\mathbf H$ of $T$ with $\mathbf H \cup \mathbf H'$.
By the Tableau Extension Lemma, $T_{\mathbf H'}$ has a finished extension $T'$.
By definition of extension, $T_{\mathbf H'}$ is a rooted subtree of $T'$.
But $T_{\mathbf H'}$ and $T$ are equal when considered as rooted trees.
The result follows.
{{qed}}
Theorem: Factors of Sums of Powers of 100,000

All integers $n$ of the form:
:$n = \ds \sum_{k \mathop = 0}^m 10^{5 k}$ for $m \in \Z_{> 0}$
are composite.

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^1 10^{5 k}
      | r = 100 \, 001
      | c = 
}}
{{eqn | r = 11 \times 9091
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^2 10^{5 k}
      | r = 10 \, 000 \, 100 \, 001
      | c = 
}}
{{eqn | r = 3 \times 31 \times 37 \times 2 \, 906 \, 161
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^3 10^{5 k}
      | r = 1 \, 000 \, 010 \, 000 \, 100 \, 001
      | c = 
}}
{{eqn | r = 11 \times 101 \times 3541 \times 9091 \times 27961
      | c = 
}}
{{end-eqn}}
Notice that:
:$\ds \sum_{k \mathop = 0}^m 10^{5 k} \times R_5 = R_{5 \paren {m + 1} }$
where $R_i$ is the $i$th repunit.
Suppose $p \divides m + 1$, where $p$ is a prime that is not $5$.
By Divisors of Repunit with Composite Index:
:$R_p \divides R_{5 \paren {m + 1} }$
By Prime not Divisor implies Coprime, $p$ and $5$ are coprime.
By Condition for Repunits to be Coprime, $R_p$ and $R_5$ are coprime.
By Euclid's Lemma:
:$R_p \divides \dfrac {R_{5 \paren {m + 1} } } {R_5} = n$
Suppose $25 \divides m + 1$.
By Divisors of Repunit with Composite Index:
:$R_5 \divides R_{25}$
and:
:$R_{25} \divides R_{5 \paren {m + 1} }$
So we have:
:$\dfrac {R_{25}} {R_5} \divides \dfrac {R_{5 \paren {m + 1} } } {R_5} = n$
The final case is $m + 1 = 5$.
This is the case $n = 100 \, 001 \, 000 \, 010 \, 000 \, 100 \, 001$.
We have:
:$n = 21 \, 401 \times 25 \, 601 \times 182 \, 521 \, 213 \, 001$
Thus all cases are covered.
{{qed}}
Theorem: Continuous Function on Compact Subspace of Euclidean Space is Bounded

Let $\R^n$ be the $n$-dimensional Euclidean space.
Let $S \subseteq \R^n$ be a compact subspace of $\R^n$.
Let $f: S \to \R$ be a continuous function.
Then $f$ is bounded in $\R$.

Proof:
An application of Continuous Function on Compact Space is Bounded.
Theorem: Points Defined by Adjacent Pairs of Digits of Reciprocal of 13 lie on Hyperbola

Consider the digits that form the recurring part of the reciprocal of $13$:
:$\dfrac 1 {13} = 0 \cdotp \dot 07692 \dot 3$
Take the digits in ordered pairs, and treat them as coordinates of a Cartesian plane.
It will be found that they all lie on a hyperbola:
:600px

Proof:
:600px
Let the points be labelled to simplify:
:$A := \left({0, 7}\right)$
:$B := \left({7, 6}\right)$
:$C := \left({6, 9}\right)$
:$D := \left({9, 2}\right)$
:$E := \left({2, 3}\right)$
:$F := \left({3, 0}\right)$
{{finish|Just too tedious to contemplate.}}
Theorem: Number times Recurring Part of Reciprocal gives 9-Repdigit

Let a (strictly) positive integer $n$ be such that the decimal expansion of its reciprocal has a recurring part of period $d$ and no non-recurring part.
Let $m$ be the integer formed from the $d$ digits of the recurring part.
Then $m \times n$ is a $d$-digit repdigit number consisting of $9$s.

Proof:
Let $x = \dfrac 1 n = \sqbrk {0. mmmm \dots}$.
Then:
:$10^d x = \sqbrk {m.mmmm \dots}$
Therefore:
{{begin-eqn}}
{{eqn | l = 10^d x - x
      | r = \sqbrk {m.mmmm \dots} - \sqbrk {0. mmmm \dots}
}}
{{eqn | ll= \leadsto
      | l = \frac 1 n \paren {10^d - 1}
      | r = m
}}
{{eqn | ll= \leadsto
      | l = m n
      | r = 10^d - 1
}}
{{end-eqn}}
which is the $d$-digit repdigit number consisting of $9$s.
{{qed}}
Theorem: Reciprocal of 142,857

:$\dfrac 1 {142 \, 857} = 0 \cdotp \dot 00000 \, \dot 7$

Proof:
Performing the calculation using long division:
<pre>
       0.000007000007...
      ------------------
142857)1.000000000000...
       1 000000
         999999
       --------
              1000000
               999999
              -------
</pre>
Theorem: Quotient of Group by Itself

Let $G$ be a group.
Let $G / G$ be the quotient group of $G$ by itself.
Then:
:$G / G \cong \set e$
That is, the quotient of a group by itself is isomorphic to the trivial group.

Proof:
Let the homomorphism $\phi: G \to \set e$ be defined as:
:$\forall g \in G: \map \phi g = e$
Then:
:$\map \ker \phi = G$
and:
:$\Img \phi = \set e$
By the First Isomorphism Theorem:
:$G / \map \ker \phi \cong \Img \phi$
Hence the result:
:$G / G \cong \set e$
{{qed}}
Theorem: Integer whose Digits when Grouped in 3s add to Multiple of 999 is Divisible by 999

Let $n$ be an integer which has at least $3$ digits when expressed in decimal notation.
Let the digits of $n$ be divided into groups of $3$, counting from the right, and those groups added.
Then the result is equal to a multiple of $999$ {{iff}} $n$ is divisible by $999$.

Proof:
{{refactor|The below sentence should be on the mistake page. Need to go back to the source work to clarify what we have.|level = medium}}
The mistake is either ''and conversely'' or ''equal to $999$'', since $999 \, 999$ is an easy counterexample.
Here we will show that the result is equal to '''a multiple of''' $999$ {{iff}} $n$ is divisible by $999$.
Write $n = \ds \sum_{i \mathop = 0}^k a_i 10^{3 i}$, where $0 \le a_i < 1000$.
This divides the digits of $n$ into groups of $3$.
Then the statement is equivalent to:
:$999 \divides n \iff 999 \divides \ds \sum_{i \mathop = 0}^k a_i$
This statement is true since:
{{begin-eqn}}
{{eqn | l = n
      | r = \sum_{i \mathop = 0}^k a_i 10^{3 i}
}}
{{eqn | r = \sum_{i \mathop = 0}^k a_i 1000^i
}}
{{eqn | o = \equiv
      | r = \sum_{i \mathop = 0}^k a_i 1^i
      | rr= \pmod {999}
      | c = Congruence of Powers
}}
{{eqn | o = \equiv
      | r = \sum_{i \mathop = 0}^k a_i
      | rr= \pmod {999}
}}
{{end-eqn}}
{{qed}}
Theorem: Number which is Sum of Subfactorials of Digits

The only integer which is the sum of the subfactorials of its digits is $148 \, 349$:
:$148 \, 349 = \mathop !1 + \mathop !4 + \mathop !8 + \mathop !3 \mathop + \mathop !4 \mathop + \mathop !9$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = 148 \, 349
      | r = 0 + 9 + 14 \, 833 + 2 + 9 + 133 \, 496
      | c = 
}}
{{eqn | r = \mathop !1 + \mathop !4 + \mathop !8 + \mathop !3 \mathop + \mathop !4 \mathop + \mathop !9
      | c = 
}}
{{end-eqn}}
A computer search can verify solutions under $10^6$ (that is, with no more than $6$ digits) in seconds.
Let $n$ be a $k$-digit number, for $k \ge 7$.
Then the sum of the subfactorials of its digits is not more than $\mathop !9 \times k$.
But we have:
{{begin-eqn}}
{{eqn | l = n
      | o = \ge
      | r = 10^{k - 1}
}}
{{eqn | r = 10^6 \times 10^{k - 7}
}}
{{eqn | o = \ge
      | r = 10^6 \times \paren {1 + 9 \paren {k - 7} }
      | c = Bernoulli's Inequality
}}
{{eqn | o = >
      | r = 7 \times \mathop !9 \times \paren {9 k - 62}
      | c = $7 \times \mathop !9 = 934472$
}}
{{eqn | r = \mathop !9 \paren {63 k - 62 \times 7}
}}
{{eqn | o = \ge
      | r = \mathop !9 \times k
      | c = $k \ge 7$
}}
{{eqn | o = \ge
      | r = \text{sum of the subfactorials of digits of } n
}}
{{end-eqn}}
So no more numbers have this property.
{{qed}}
Theorem: Integers Representable as Product of both 3 and 4 Consecutive Integers

There are $3$ integers which can be expressed as both $x \paren {x + 1} \paren {x + 2} \paren {x + 3}$ for some $x$, and $y \paren {y + 1} \paren {y + 2}$ for some $y$:
:$24, 120, 175 \, 560$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = 24
      | r = 1 \times 2 \times 3 \times 4
      | c = 
}}
{{eqn | r = 2 \times 3 \times 4
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 120
      | r = 2 \times 3 \times 4 \times 5
      | c = 
}}
{{eqn | r = 4 \times 5 \times 6
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 175 \, 560
      | r = 55 \times 56 \times 57
      | c = 
}}
{{eqn | r = 19 \times 20 \times 21 \times 22
      | c = 
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown there are no more.}}
Theorem: Squares whose Digits form Consecutive Increasing Integers

The sequence of integers whose squares have a decimal representation consisting of the concatenation of $2$ consecutive increasing integers begins:
:$428, 573, 727, 846, 7810, 36 \, 365, 63 \, 636, 326 \, 734, \ldots$
{{OEIS|A030467}}

Proof:
We have:
{{begin-eqn}}
{{eqn | l = 428^2
      | r = 183 \, 184
      | c = 
}}
{{eqn | l = 573^2
      | r = 328 \, 329
      | c = 
}}
{{eqn | l = 727^2
      | r = 528 \, 529
      | c = 
}}
{{eqn | l = 846^2
      | r = 715 \, 716
      | c = 
}}
{{eqn | l = 7810^2
      | r = 6099 \, 6100
      | c = 
}}
{{eqn | l = 36 \, 365^2
      | r = 13224 \, 13225
      | c = 
}}
{{eqn | l = 63 \, 636^2
      | r = 40495 \, 40496
      | c = 
}}
{{eqn | l = 326 \, 734^2
      | r = 106755 \, 106756
      | c = 
}}
{{end-eqn}}
They can be determined by inspection.
{{qed}}
Theorem: Smallest Fifth Power which is Sum of 6 Fifth Powers

The smallest fifth power which is the sum of $6$ fifth powers is $12^5 = 248 \, 832$:
:$12^5 = 4^5 + 5^5 + 6^5 + 7^5 + 9^5 + 11^5$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = 12^5
      | r = 248 \, 832
      | c = 
}}
{{eqn | r = 1024 + 3125 + 7776 + 16 \, 807 + 59 \, 049 + 161 \, 051
      | c = 
}}
{{eqn | r = 4^5 + 5^5 + 6^5 + 7^5 + 9^5 + 11^5
      | c = 
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown that this is the smallest.}}
Theorem: Prime Numbers Embedded in Digits of Pi

The sequence of prime numbers that can be found starting from the beginning of the decimal expansion of $\pi$ (pi) begins:
:$3, 31, 314 \, 159, 31 \, 415 \, 926 \, 535 \, 897 \, 932 \, 384 \, 626 \, 433 \, 832 \, 795 \, 028 \, 841, \ldots$
{{OEIS|A005042}}

Proof:
By inspection.
Theorem: 333,667 is Unique Period Prime with Period 9

$333 \, 667$ is a unique period prime whose reciprocal has a period of $9$:
:$\dfrac 1 {333 \, 667} = 0 \cdotp \dot 00000 \, 299 \dot 7$

Proof:
By long division:
<pre>
       0.000002997000002...
      ---------------------
333667)1.000000000000000000
         667334
       --------
         3326660
         3003003
         -------
          3236570
          3003003
          -------
           2335670
           2335669
           -------
                 1000000
                  667334
                 -------
                  ......
</pre>
It remains to be shown that $333 \, 667$ is the ''only'' prime number whose reciprocal has a period of $9$.
We have that:
:$333 \, 667 \nmid 10$
From Period of Reciprocal of Prime, the period of such a prime is the order of $10$ modulo $p$.
That is, the smallest integer $d$ such that:
:$10^d \equiv 1 \pmod p$
From the above long division we know that the period of $\dfrac 1 {333 \, 667}$ is $9$, so $10^9 \equiv 1 \pmod {333 \, 667}$.
The only other possible primes $p$ whose reciprocals might have a period of $9$ must also satisfy:
:$10^9 \equiv 1 \pmod p$
that is:
:$p \divides 10^9 - 1$
Consider:
{{begin-eqn}}
{{eqn | l = 10^9 - 1
      | r = 999 \, 999 \, 999
      | c = 
}}
{{eqn | r = 3^4 \times 37 \times 333 \, 667
      | c = prime factorization
}}
{{end-eqn}}
Therefore the only other possible primes whose reciprocals might have a period of $9$ are $3$ and $37$.
From Reciprocal of $37$:
:$\dfrac 1 {37} = 0 \cdotp \dot 02 \dot 7$
and trivially:
:$\dfrac 1 3 = 0 \cdotp \dot 3$
which has a period of $1$.
As required, the only prime number whose reciprocal has a period of $9$ is $333 \, 667$.
{{qed}}
Theorem: Cube which can be Represented as Sum of 3, 4, 5, 6, 7 or 8 Cubes

:$351 \, 120^3$ can be represented as the sum of $3$, $4$, $5$, $6$, $7$ or $8$ cubes.

Proof:
{{begin-eqn}}
{{eqn | l = 351120^3
      | r = 175560^3 + 234080^3 + 292600^3
}}
{{eqn | r = 2 \times 87780^3 + 204820^3 + 321860^3
}}
{{eqn | r = 2 \times 87780^3 + 175560^3 + 2 \times 263340^3
}}
{{eqn | r = 3 \times 117040^3 + 3 \times 234080^3
}}
{{eqn | r = 2 \times 58520^3 + 117040^3 + 3 \times 175560^3 + 292600^3
}}
{{eqn | r = 8 \times 175560^3
}}
{{end-eqn}}
These representations are not necessarily unique.
{{qed}}
Theorem: Prime Gaps of 100

The following pairs of consecutive prime numbers are those whose difference is $100$:
:$\tuple {396 \, 733, 396 \, 833}, \ldots$
{{expand|Only know the first pair so far. Research needed to find the next one(s).}}

Proof:
Demonstrated by listing the prime gaps.
{{qed}}
Theorem: Property of 490,689

The number $490 \, 689$ can be expressed as the sum of $3$ cubes in $2$ different ways:
:$490 \, 689 = 4^3 + 60^3 + 65^3 = 8^3 + 25^3 \times 78^3$
while at the same time the products of the contributory cube roots of each of those $2$ ways are equal:
:$4 \times 60 \times 65 = 8 \times 25 \times 78$

Proof:
{{begin-eqn}}
{{eqn | l = 490 \, 689
      | r = 64 + 216 \, 000 + 274 \, 625
      | c = 
}}
{{eqn | r = 4^3 + 60^3 + 65^3
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 490 \, 689
      | r = 512 + 15 \, 625 + 474 \, 552
      | c = 
}}
{{eqn | r = 8^3 + 25^3 + 78^3
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = 15 \, 600
      | r = 2^4 \times 3 \times 5^2 \times 13
      | c = 
}}
{{eqn | r = 2^2 \times \left({2^2 \times 3 \times 5}\right) \times \left({5 \times 13}\right)
      | c = 
}}
{{eqn | r = 4 \times 60 \times 65
      | c = 
}}
{{eqn | r = 2^3 \times 5^2 \times \left({2 \times 3 \times 13}\right)
      | c = 
}}
{{eqn | r = 8 \times 25 \times 78
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: 510,510 is Product of 4 Consecutive Fibonacci Numbers

$510 \, 510$ can be expressed as the product of $4$ distinct consecutive Fibonacci numbers:
:$510 \, 510 = 13 \times 21 \times 34 \times 55$
and is also the $7$th primorial:
:$510 \, 510 = 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17$

Proof:
By observation:
{{begin-eqn}}
{{eqn | l = 510 \, 510
      | r = 13 \times 21 \times 34 \times 55
      | c = 
}}
{{eqn | r = 13 \times \paren {3 \times 7} \times \paren {2 \times 17} \times \paren {5 \times 11}
      | c = 
}}
{{eqn | r = 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Tableau Confutation contains Finite Tableau Confutation

Let $\mathbf H$ be a countable set of WFFs of propositional logic.
Let $T$ be a tableau confutation of $\mathbf H$.
Then there exists a finite rooted subtree of $T'$ that is also a tableau confutation of $\mathbf H'$.

Proof:
For each node $v \in T$, let $\map p v$ be the path from $v$ to $r_T$, the root of $T$.
This path is unique by Path in Tree is Unique.
Let $\VV$ be the subtree of $T$ consisting those nodes $v$ of $T$ such that $\map p v$ is not contradictory.
{{AimForCont}} that $\VV$ were infinite.
Then by König's Tree Lemma, $\VV$ has an infinite branch $\Gamma$.
Since $\VV \subseteq T$, it follows that $\Gamma$ is also a branch of $T$.
However, by construction, it is impossible that $\Gamma$ is contradictory.
This contradicts that $T$ is a tableau confutation.
Hence $\VV$ is finite.
Next, define a finite propositional tableau $T'$ by:
:$v \in T' \iff \map \pi v \in \VV$
that is, the rooted tree formed by $\VV$ and all its children.
Then by construction, for each leaf node $v$ of $T'$, we have that $v \notin \VV$.
That is, that $\map p v$ is a contradictory branch of $T'$.
By Leaf of Rooted Tree is on One Branch, every branch of $T'$ is contradictory.
Hence $T'$ is a tableau confutation of $\mathbf H'$, as desired.
{{qed}}
Theorem: Two-Sided Prime/Sequence

The complete sequence of two-sided primes is:
:$2, 3, 5, 7, 23, 37, 53, 73, 313, 317, 373, 797, 3137, 3797, 739 \, 397$

Proof:
{{ProofWanted}}
Category:Two-Sided Primes
ct1xp4yykscy4hdf7tndy9q4udrwcmo
Theorem: Set of 7 Anagrams which are Square

The following integers are all anagrams, and all square:
:$1 \, 048 \, 576, 1 \, 056 \, 784, 1 \, 085 \, 764, 5 \, 740 \, 816, 5 \, 764 \, 801, 6 \, 754 \, 801, 7 \, 845 \, 601$

Proof:
{{begin-eqn}}
{{eqn | l = 1 \, 048 \, 576
      | r = 1024^2
}}
{{eqn | l = 1 \, 056 \, 784
      | r = 1028^2
}}
{{eqn | l = 1 \, 085 \, 764
      | r = 1042^2
}}
{{eqn | l = 5 \, 740 \, 816
      | r = 2396^2
}}
{{eqn | l = 5 \, 764 \, 801
      | r = 2401^2
}}
{{eqn | l = 6 \, 754 \, 801
      | r = 2599^2
}}
{{eqn | l = 7 \, 845 \, 601
      | r = 2801^2
}}
{{end-eqn}}
{{qed}}
Theorem: Burnside's Lemma

Let $G$ be a finite group acting on a set $X$.
Let $X / G$ be the set of orbits under this action.
For $x \in X$, let $\Stab x$ be the stabilizer of $x$ by $G$.
For $g \in G$, let $X^g$ denotes the set of all elements of $X$ which are fixed by $g$:
:$X^g := \set {x \in X: g x = x}$
Then:
:$\size {X / G} = \dfrac 1 {\order G} \ds \sum_{g \mathop \in G} \size {X^g}$
In words, the number of orbits equals the average number of fixed elements.

Proof:
{{begin-eqn}}
{{eqn | l = \frac 1 {\order G} \sum_{g \mathop \in G} \size {X^g}
      | r = \frac 1 {\order G} \sum_{g \mathop \in G} \size {\set {x \in X: g x = x} }
      | c = by definition
}}
{{eqn | r = \frac 1 {\order G} \sum_{x \mathop \in X} \size {\set {g \in G: g x = x} }
      | c = Same summation, different indexing
}}
{{eqn | r = \frac 1 {\order G} \sum_{x \mathop \in X} \order {\Stab x}
      | c = {{Defof|Stabilizer}}
}}
{{eqn | r = \frac 1 {\order G} \sum_{x \mathop \in X} \frac {\order G} {\order {\Orb x} }
      | c = Orbit-Stabilizer Theorem
}}
{{eqn | r = \sum_{x \mathop \in X} \frac 1 {\order {\Orb x} }
      | c = 
}}
{{eqn | r = \sum_{\Orb x \mathop \in X / G} \paren {\sum_{x \mathop \in \Orb x} \frac 1 {\order {\Orb x} } }
      | c = 
}}
{{eqn | r = \sum_{\Orb x \mathop \in X / G} 1
      | c = 
}}
{{eqn | r = \order {X / G}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Smallest Cunningham Chain of the First Kind of Length 7

The smallest Cunningham chain of the first kind of length $7$ is:
:$\left({1 \, 122 \, 659, 2 \, 245 \, 319, 4 \, 490 \, 639, 8 \, 981 \, 279, 17 \, 962 \, 559, 35 \, 925 \, 119, 71 \, 850 \, 239}\right)$

Proof:
Let $C$ denote the sequence in question.
We have that:
:$\dfrac {1 \, 122 \, 659 - 1} 2 = 561 \, 329 = 83 \times 6763$
and so is not prime.
Thus $1 \, 122 \, 659$ is not a safe prime, as is required for $C$ to be a Cunningham chain of the first kind.
Then:
{{begin-eqn}}
{{eqn | l = 2 \times 561 \, 329 + 1
      | r = 1 \, 122 \, 659
      | c = which is the $87 \, 359$th prime
}}
{{eqn | l = 2 \times 1 \, 122 \, 659 + 1
      | r = 2 \, 245 \, 319
      | c = which is the $165 \, 760$th prime
}}
{{eqn | l = 2 \times 2 \, 245 \, 319 + 1
      | r = 4 \, 490 \, 639
      | c = which is the $315 \, 347$th prime
}}
{{eqn | l = 2 \times 4 \, 490 \, 639 + 1
      | r = 8 \, 981 \, 279
      | c = which is the $601 \, 286$th prime
}}
{{eqn | l = 2 \times 8 \, 981 \, 279 + 1
      | r = 17 \, 962 \, 559
      | c = which is the $1 \, 149 \, 096$th prime
}}
{{eqn | l = 2 \times 17 \, 962 \, 559 + 1
      | r = 35 \, 925 \, 119
      | c = which is the $2 \, 199 \, 933$rd prime
}}
{{eqn | l = 2 \times 35 \, 925 \, 119 + 1
      | r = 71 \, 850 \, 239
      | c = which is the $4 \, 220 \, 407$th prime
}}
{{eqn | l = 2 \times 71 \, 850 \, 239 + 1
      | r = 143 \, 700 \, 479
      | c = 
}}
{{eqn | r = 13 \times 47 \times 479 \times 491
      | c = and so is not prime
}}
{{end-eqn}}
Establishing that this is indeed the smallest such Cunningham chain of the first kind of length $7$ can be done by a computer search.
{{qed}}
Theorem: Square of Repunit times Sum of Digits

The following pattern emerges:
{{begin-eqn}}
{{eqn | l = 121 \times \paren {1 + 2 + 1}
      | r = 22^2
}}
{{eqn | l = 12 \, 321 \times \paren {1 + 2 + 3 + 2 + 1}
      | r = 333^2
}}
{{eqn | l = 1 \, 234 \, 321 \times \paren {1 + 2 + 3 + 4 + 3 + 2 + 1}
      | r = 4444^2
}}
{{end-eqn}}
and so on, up until $999 \, 999 \, 999^2$ after which the pattern breaks down.

Proof:
From Square of Repunit:
{{begin-eqn}}
{{eqn | l = 121
      | r = 11^2
}}
{{eqn | l = 12 \, 321
      | r = 111^2
}}
{{eqn | l = 1 \, 234 \, 321
      | r = 1111^2
}}
{{end-eqn}}
and so on.
Then from 1+2+...+n+(n-1)+...+1 = n^2:
{{begin-eqn}}
{{eqn | l = 1 + 2 + 1
      | r = 2^2
}}
{{eqn | l = 1 + 2 + 3 + 2 + 1
      | r = 3^2
}}
{{eqn | l = 1 + 2 + 3 + 4 + 3 + 2 + 1
      | r = 4^2
}}
{{end-eqn}}
and so on.
Then:
{{begin-eqn}}
{{eqn | l = 11^2 \times 2^2
      | r = 22^2
}}
{{eqn | l = 111^2 \times 3^2
      | r = 333^2
}}
{{eqn | l = 1111^2 \times 4^2
      | r = 4444^2
}}
{{end-eqn}}
The pattern breaks down after $9$:
:$1 \, 111 \, 111 \, 111^2 = 1 \, 234 \, 567 \, 900 \, 987 \, 654 \, 321$
{{qed}}
Theorem: Factorisation of Quintic x^5 - x + n into Irreducible Quadratic and Irreducible Cubic

The quintic $x^5 - x + n$ can be factorized into the product of an irreducible quadratic and an an irreducible cubic {{iff}} $n$ is in the set:
:$\set {\pm 15, \pm 22 \, 440, \pm 2 \, 759 \, 640}$

Proof:
We have that:
{{begin-eqn}}
{{eqn | l = x^5 - x \pm 15
      | r = \paren {x^2 \pm x + 3} \paren {x^3 \mp x^2 \mp 2 x \pm 5}
}}
{{eqn | l = x^5 - x \pm 22440
      | r = \paren {x^2 \mp 12 x + 55} \paren {x^3 \pm 12 x^2 + 89 x \pm 408}
}}
{{eqn | l = x^5 - x \pm 2 \, 759 \, 640
      | r = \paren {x^2 \pm 12 x + 377} \paren {x^3 \mp 12 x^2 - 233 x \pm 7320}
}}
{{end-eqn}}
{{ProofWanted}}
Theorem: Factorial as Product of Consecutive Factorials

The only factorials which are the product of consecutive factorials are:
{{begin-eqn}}
{{eqn | l = 0!
      | r = 0! \times 1!
      | c = 
}}
{{eqn | l = 1!
      | r = 0! \times 1!
      | c = 
}}
{{eqn | l = 2!
      | r = 1! \times 2!
      | c = 
}}
{{eqn | r = 0! \times 1! \times 2!
      | c = 
}}
{{eqn | l = 10!
      | r = 6! \times 7!
      | c = 
}}
{{end-eqn}}

Proof:
Suppose $m, n \in \N$ and $m > n$.
Write $\map F {n, m} = n! \paren {n + 1}! \cdots m!$.
Suppose we have $\map F {n, m} > r!$ for some $r \in \N$.
Suppose further that there is a prime $p$ where $m < p \le r$.
We claim that $\map F {n, m}$ cannot be a factorial of any number.
{{AimForCont}} $\map F {n, m} = s!$ for some $s \in \N$.
Since $s! > r!$, we must have $r! \divides s!$.
Since $p \le r$, we must have $p \divides r!$.
Thus we have $p \divides s!$.
However, since $n, n + 1, \dots, m < p$, we must have $p \nmid k!$ for each $n \le k \le m$.
Thus $p \nmid \map F {n, m} = s!$, which is a contradiction.
Hence $\map F {n, m}$ cannot be a factorial of some number.
We have the following lemmata: