avewright/proofwiki-math huggingface dataset:
Theorem: Fourier Series/Identity Function over Minus Pi to Pi

For $x \in \openint {-\pi} \pi$:
:$\ds x = 2 \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin n x$

Proof:
By Fourier Series for Identity Function over Symmetric Range, the function $f: \openint {-\lambda} \lambda \to \R$ defined as:
:$\forall x \in \openint {-\lambda} \lambda: \map f x = x$
has a Fourier series:
:$\map f x \sim \dfrac {2 \lambda} \pi \ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin \frac {n \pi x} \lambda$
Substituting for $\lambda = \pi$ gives:
:$\ds x = 2 \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin n x$
as required. 
{{qed}}
Theorem: Fourier Series/Fourth Power of x over Minus Pi to Pi

:$\ds x^4 = \frac {\pi^4} 5 + \sum_{n \mathop = 1}^\infty \frac {8 n^2 \pi^2 - 48} {n^4} \cos n \pi \cos n x$

Proof:
Since $x^4 = \paren {-x}^4$, $x^4$ is an even function.
By Fourier Series for Even Function over Symmetric Range, the Fourier series of $\map f x$ can be expressed as:
:$x^4 \sim \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty a_n \cos n x$
where for all $n \in \Z_{> 0}$:
{{begin-eqn}}
{{eqn | l = a_n 
      | r = \dfrac 2 \pi \ds \int_0^\pi x^4 \cos n x \rd x
}}
{{eqn | r = \dfrac 2 \pi \paren {\intlimits {\frac {\sin n x} n x^4 + \frac {4 \cos n x} {n^2} x^3 - \frac {12 \sin n x} {n^3} x^2 - \frac {24 \cos n x} {n^4} x + \frac {24 \sin n x} {n^5} } {x \mathop = 0} {x \mathop = \pi} }
      | c = Primitive of $x^4 \cos a x$
}}
{{eqn | r = \dfrac 2 \pi \paren {\frac {4 \pi^3 \cos n \pi} {n^2} - \frac {24 \pi \cos n \pi} {n^4} }
      | c = 
}}
{{eqn | r = \frac {8 n^2 \pi^2 - 48} {n^4} \cos n \pi
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = a_0
      | r = \dfrac 2 \pi \int_0^\pi x^4 \rd x
}}
{{eqn | r = \dfrac 2 \pi \intlimits {\frac {x^5} 5} {x \mathop = 0} {x \mathop = \pi}
      | c = Primitive of Power
}}
{{eqn | r = \frac {2 \pi^4} 5
}}
{{end-eqn}}
This gives:
:$x^4 \sim \dfrac {\pi^4} 5 + \ds \sum_{n \mathop = 1}^\infty \frac {8 n^2 \pi^2 - 48} {n^4} \cos n \pi \cos n x$
{{qed}}
Category:Examples of Fourier Series
2ut11g6bnt76257lccj1ezo4g5b323t
Theorem: Fourier Series/Absolute Value of x over Minus Pi to Pi

For $x \in \openint {-\pi} \pi$:
:$\ds \size x = \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {2 n - 1} x} {\paren {2 n - 1}^2}$

Proof:
By definition, the absolute value function is an even function:
:$\size {-x} = x = \size x$
Thus by Fourier Series for Even Function over Symmetric Range, $\size x$ can be expressed as:
:$\ds \size x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos n x$
where for all $n \in \Z_{\ge 0}$:
:$a_n = \ds \frac 2 \pi \int_0^\pi \size x \cos n x \rd x$
On the real interval $\openint 0 \pi$:
:$\size x = x$
and so for all $n \in \Z_{\ge 0}$:
:$a_n = \ds \frac 2 \pi \int_0^\pi x \cos n x \rd x$
Thus Half-Range Fourier Cosine Series for Identity Function over $\openint 0 \pi$ can be applied directly.
So for $x \in \openint {-\pi} \pi$:
:$\ds \size x = \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {2 n - 1}  x} {\paren {2 n - 1}^2}$
{{qed}}
Theorem: Fourier Series/Absolute Value of x over Minus Pi to Pi

For $x \in \openint {-\pi} \pi$:
:$\ds \size x = \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {2 n - 1} x} {\paren {2 n - 1}^2}$

Proof:
By Fourier Series for Absolute Value Function over Symmetric Range, the function $f: \openint {-\lambda} \lambda \to \R$ defined as:
:$\forall x \in \openint {-\lambda} \lambda: \map f x = \size x$
has a Fourier series:
:$\map f x \sim \dfrac \lambda 2 - \ds \dfrac {4 \lambda} {\pi^2} \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} \cos \dfrac {\paren {2 n + 1} \pi x} \lambda$
Substituting for $\lambda = \pi$ gives:
:$\size x = \dfrac \pi 2 - \ds \dfrac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {2 n - 1}  x} {\paren {2 n - 1}^2}$
as required. 
{{qed}}
Theorem: Fourier Series/Pi minus x over 0 to 2 Pi

:$\ds \pi - x = 2 \sum_{n \mathop = 1}^\infty \frac {\sin n x} n$

Proof:
By definition of Fourier series:
:$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$
where:
{{begin-eqn}}
{{eqn | l = a_n
      | r = \dfrac 1 \pi \int_0^{2 \pi} \map f x \cos n x \rd x
}}
{{eqn | l = b_n
      | r = \dfrac 1 \pi \int_0^{2 \pi} \map f x \sin n x \rd x
}}
{{end-eqn}}
for all $n \in \Z_{>0}$.
Thus:
{{begin-eqn}}
{{eqn | l = a_0
      | r = \frac 1 \pi \int_0^{2 \pi} \map f x \rd x
      | c = Cosine of Zero is One
}}
{{eqn | r = \frac 1 \pi \int_0^{2 \pi} \pi - x \rd x
      | c = Definition of $f$
}}
{{eqn | r = \frac 1 \pi \intlimits {\pi x - \frac 1 2 x^2} 0 {2 \pi}
      | c = Primitive of Power
}}
{{eqn | r = 2 \pi^2 - \frac 4 2 \pi^2
      | c = 
}}
{{eqn | r = 0
}}
{{end-eqn}}
{{qed|lemma}}
For $n > 0$:
{{begin-eqn}}
{{eqn | l = a_n
      | r = \frac 1 \pi \int_0^{2 \pi} \map f x \cos n x \rd x
      | c = 
}}
{{eqn | r = \frac 1 \pi \int_0^{2 \pi} \paren {\pi - x} \cos n x \rd x
      | c = Definition of $f$
}}
{{eqn | r = \pi \int_0^{2 \pi} \cos n x \rd x - \frac 1 \pi \int_0^{2 \pi} x \cos n x \rd x
      | c = 
}}
{{eqn | r = - \frac 1 \pi \int_0^{2 \pi} x \cos n x \rd x
      | c = Integral over $2 \pi$ of $\cos n x$
}}
{{eqn | r = - \frac 1 \pi \intlimits {\frac {\cos n x} {n^2} + \frac {x \sin n x} n} 0 {2 \pi}
      | c = Primitive of $x \cos n x$
}}
{{eqn | r = - \frac 1 \pi \paren {\cos 0 - \cos 2 \pi}
      | c = Sine of Integer Multiple of Pi
}}
{{eqn | r = 0
      | c = Sine and Cosine are Periodic on Reals
}}
{{end-eqn}}
{{qed|lemma}}
Now for the $\sin n x$ terms:
{{begin-eqn}}
{{eqn | l = b_n
      | r = \frac 1 \pi \int_0^{2 \pi} \map f x \sin n x \rd x
      | c = 
}}
{{eqn | r = \frac 1 \pi \int_0^{2 \pi} \paren {\pi - x} \sin n x \rd x
      | c = Definition of $f$
}}
{{eqn | r = \int_0^{2 \pi} \sin n x \rd x - \frac 1 \pi \int_0^{2 \pi} x \sin n x \rd x
      | c = 
}}
{{eqn | r = - \frac 1 \pi \int_0^{2 \pi} x \sin n x \rd x
      | c = Integral over $2 \pi$ of $\sin n x$
}}
{{eqn | r = - \frac 1 \pi \intlimits {\frac {\sin n x} {n^2} - \frac {x \cos n x} n} 0 {2 \pi}
      | c = Primitive of $x \sin n x$
}}
{{eqn | r = \frac 1 \pi \frac {2 \pi \cos 2 n \pi} n
      | c = Sine of Multiple of Pi
}}
{{eqn | r = \frac 2 n
      | c = Cosine of Multiple of Pi
}}
{{end-eqn}}
Finally:
{{begin-eqn}}
{{eqn | l = \map f x
      | o = \sim
      | r = \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}
      | c = 
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \frac 2 n \sin n x
      | c = substituting for $a_0$, $a_n$ and $b_n$ from above
}}
{{eqn | r = 2 \sum_{n \mathop = 1}^\infty \frac {\sin n x} n 
      | c = rearranging
}}
{{end-eqn}}
{{qed}}
Theorem: Cube of 20 is Sum of Sequence of 4 Consecutive Cubes

:$20^3 = \ds \sum_{k \mathop = 11}^{14} k^3$
That is:
:$20^3 = 11^3 + 12^3 + 13^3 + 14^3$

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^{14} k^3
      | r = \paren {\dfrac {14 \paren {14 + 1} } 2}^2
      | c = Sum of Sequence of Cubes
}}
{{eqn | r = 11 \, 025
      | c = 
}}
{{eqn | l = \sum_{k \mathop = 1}^{10} k^3
      | r = \paren {\dfrac {10 \paren {10 + 1} } 2}^2
      | c = Sum of Sequence of Cubes
}}
{{eqn | r = 3025
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sum_{k \mathop = 11}^{14} k^3
      | r = \sum_{k \mathop = 1}^{14} k^3 - \sum_{k \mathop = 1}^{10} k^3
      | c = 
}}
{{eqn | r = 11 \, 025 - 3025
      | c = 
}}
{{eqn | r = 8000
      | c = 
}}
{{eqn | r = 20^3
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Mapping is Continuous implies Mapping Preserves Filtered Infima in Lower Topological Lattice

Let $T = \struct {S, \preceq, \tau}$ and $Q = \struct {X, \preceq', \tau'}$ be complete topological lattices with lower topologies.
Let $f: S \to X$ be a mapping such that:
:$f$ is a continuous mapping.
Then $f$ preserves filtered infima.

Proof:
Define $B := \set {\relcomp S {x^\succeq}: x \in S}$
By definition of lower topology:
:$B$ is an analytic sub-basis.
Let $F$ be a filtered subset of $S$ such that:
:$F$ admits an infimum in $T$.
Thus by definition of complete lattice:
:$f \sqbrk F$ admits an infimum in $Q$.
We will prove that:
:$\forall A \in B: \inf F \in A \implies F \cap A \ne \O$
Let $A \in B$ be such that:
:$\inf F \in A$
By definition of $B$:
:$\exists x \in S: A = \relcomp S {x^\succeq}$
By definition of relative complement:
:$\inf F \notin x^\succeq$
By definition of upper closure of element:
:$x \npreceq \inf F$
By definition of infimum:
:$x$ is not a lower bound for $F$.
By definition of lower bound:
:$\exists y \in F: x \npreceq y$
By definition of upper closure of element:
:$y \notin x^\succeq$
By definition of relative complement:
:$y \in A$
By definitions of intersection and non-empty set:
:$F \cap A \ne \O$
{{qed|lemma}}
Then by If Infimum of Filtered Subset belongs to Element of Sub-Basis then Subset and Element Intersect implies Infimum of Subset belongs to Closure of Subset:
:$\inf F \in F^-$
We will prove that:
:$f$ is an increasing mapping.
Let $x, y \in S$ be such that:
:$x \preceq y$
By definition of reflexivity:
:$\map f x \preceq' \map f x$
By definition of upper closure of element:
:$\map f x \in \paren {\map f x}^{\succeq'}$
By definition of preimage of set:
:$x \in f^{-1} \sqbrk {\paren {\map f x}^{\succeq'} }$
By Complement of Upper Closure of Element is Open in Lower Topology:
:$\paren {\map f x}^{\succeq'}$ is closed.
By Continuity by Closed Sets:
:$f^{-1} \sqbrk {\paren {\map f x}^{\succeq'} }$ is closed.
By Closed Subset is Upper Section in Lower Topology:
:$f^{-1} \sqbrk {\paren {\map f x}^{\succeq'} }$ isan  upper section.
By definition of upper section:
:$y \in f^{-1} \sqbrk {\paren {\map f x}^{\succeq'} }$
By definition of preimage of set:
:$\map f y \in \paren {\map f x}^{\succeq'}$
Thus by definition of upper closure of element:
:$\map f x \preceq' \map f y$
{{qed|lemma}}
We will prove that:
:$\map f {\inf F}$ is lower bound for $f \sqbrk F$
By definition of infimum:
:$\map f {\inf F} \preceq' \map \inf {f \sqbrk F}$
We will prove that:
:$F \subseteq f^{-1} \sqbrk {\paren {\map \inf {f \sqbrk F} }^{\succeq'} }$
{{qed}}
Theorem: 3 Numbers in A.P. whose 4th Powers are Sum of Four 4th Powers

The following triplets of integers in arithmetic sequence with common difference of $60$ can all be expressed as the sum of four $4$th powers:
:$\tuple {8373, 8433, 8493}, \tuple {8517, 8577, 8637}, \ldots$

Proof:
{{begin-eqn}}
{{eqn | l = 8373^4
      | r = 4450^4 + 5500^4 + 5670^4 + 7123^4
}}
{{eqn | l = 8433^4
      | r = 4730^4 + 4806^4 + 5230^4 + 7565^4
}}
{{eqn | l = 8493^4
      | r = 524^4 + 4910^4 + 5925^4 + 7630^4
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 8517^4
      | r = 1642^4 + 3440^4 + 6100^4 + 7815^4
}}
{{eqn | l = 8577^4
      | r = 1050^4 + 2905^4 + 5236^4 + 8230^4
}}
{{eqn | l = 8637^4
      | r = 3450^4 + 3695^4 + 5780^4 + 8012^4
}}
{{end-eqn}}
{{qed}}
The internal structure of these numbers reveals an interesting pattern:
{{begin-eqn}}
{{eqn | l = 8373
      | r = 3 \times 2791
}}
{{eqn | l = 8433
      | r = 3^2 \times 937
}}
{{eqn | l = 8493
      | r = 3 \times 19 \times 149
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 8517
      | r = 3 \times 17 \times 169
}}
{{eqn | l = 8577
      | r = 3^3 \times 953
}}
{{eqn | l = 8637
      | r = 3 \times 2879
}}
{{end-eqn}}
Theorem: Intersection of Chain of Prime Ideals of Commutative Ring is Prime Ideal

Let $R$ be a commutative ring.
Let $\Spec R$ be the spectrum of $R$, ordered by the subset relation.
Let $\set {P_\alpha}_{\alpha \mathop \in A}$ be a non-empty chain of prime ideals of $\Spec R$.
Let $\ds P = \bigcap_{\alpha \mathop \in A} P_\alpha$ be their intersection.
Then $P$ is a prime ideal of $R$.

Proof:
By Intersection of Ring Ideals is Ideal, $P$ is an ideal of $R$.
We show that $P$ is prime.
Let $a, b\in R$ with $a, b \notin P$.
We show that $a b \notin P$.
Because $a \notin P$, there exists $\alpha \in A$ with $a \notin P_\alpha$.
Because $b \notin P$, there exists $\beta \in A$ with $b \notin P_\beta$.
Because $\set {P_\gamma}_{\gamma \mathop \in A}$ is totally ordered, $P_\alpha \subseteq P_\beta$ or $P_\beta \subseteq P_\alpha$.
{{WLOG}}, we can assume $P_\alpha \subseteq P_\beta$, which gives us $a, b \notin P_\alpha$.
Since $P_\alpha$ is an prime ideal, $a b \notin P_\alpha$.
Thus $a b \notin P$.
{{qed}}
Theorem: Number of Different Ways to play First n Moves in Chess

The sequence formed from the number of ways to play the first $n$ moves in chess begins:
:$20, 400, 8902, 197 \, 742, \ldots$
{{OEIS|A007545}}
The count for the fourth move is already ambiguous, as it depends on whether only legal moves count, or whether all moves, legal or illegal, are included.
The count as given here does include illegal moves in addition to legal ones.

Proof:
There are $20$ ways to make the $1$st move by White:
:Each of the $8$ pawns may be moved either $1$ or $2$ squares forward, making $16$ moves
:Each of the $2$ knights may be moved to either of $2$ squares before it, making $4$ moves.
For each of those $20$ first moves by White, Black has the same $20$ options.
Thus there are $20 \times 20$ possible different games after the $2$nd move.
To count the $3$rd moves, one needs to consider cases.
First note that after the $1$st move, whatever it was, there are $7$ pawns on the $2$nd rank, each of which can be moved $1$ or $2$ squares forward, making $14$ moves for each of those $400$ possibilities.
Note that if a knight was one of the pieces to have moved first, the pawn in the square behind where it ends up cannot move -- hence the $7$ pawns on the $2$nd rank that can move.
Thus there are $400 \times 14 = 5600$ possible moves involving a so-far unmoved pawn.
For each of the $400$ positions, there are exactly $8$ which consist of two pawns in opposition on the $4$th and $5$th rank.
There are also another $4 \times 20 = 80$ positions in which white moved a knight.
For all other $400 - 88 = 312$ positions, the already-moved pawn has the option of moving another square forward.
This gives another $312$ options for the $3$rd move.
We now need to take into account the possibility that White may be able to capture a Black pawn.
{{finish|Each case needs to be investigated.}}
Theorem: Largest Product of Pandigital Factors

The largest integer that can be obtained by multiplying $2$ integers which between them use all the digits from $1$ to $9$ is:
:$843 \, 973 \, 902 = 9642 \times 87531$

Proof:
{{ProofWanted|probably quite simple}}
Theorem: Sequence of 9 Primes of form 4n+1

The following sequence of $9$ consecutive prime numbers are all of the form $4 n + 1$:
:$11 \, 593, 11 \, 597, 11 \, 617, 11 \, 621, 11 \, 633, 11 \, 657, 11 \, 677, 11 \, 681, 11 \, 689$

Proof:
{{begin-eqn}}
{{eqn | l = 11 \, 593
      | r = 4 \times 2898 + 1
}}
{{eqn | l = 11 \, 597
      | r = 4 \times 2899 + 1
}}
{{eqn | l = 11 \, 617
      | r = 4 \times 2904 + 1
}}
{{eqn | l = 11 \, 621
      | r = 4 \times 2905 + 1
}}
{{eqn | l = 11 \, 633
      | r = 4 \times 2908 + 1
}}
{{eqn | l = 11 \, 657
      | r = 4 \times 2914 + 1
}}
{{eqn | l = 11 \, 677
      | r = 4 \times 2919 + 1
}}
{{eqn | l = 11 \, 681
      | r = 4 \times 2920 + 1
}}
{{eqn | l = 11 \, 689
      | r = 4 \times 2922 + 1
}}
{{end-eqn}}
It remains to be noted that:
:the prime number before $11 \, 593$ is $11 \, 587$ which is $4 \times 2897 - 1$
:the prime number after $11 \, 689$ is $11 \, 699$ which is $4 \times 2925 - 1$
confirming that they are not of the form $4 n + 1$.
{{qed}}
Theorem: Smallest Penholodigital Square

The smallest penholodigital square number is:
:$11 \, 826^2 = 139 \, 854 \, 276$

Proof:
Let $n$ be the smallest positive integer whose square is penholodigital.
First it is noted that the smallest penholodigital number is $123 \, 456 \, 789$.
Hence any square penholodigital number must be at least as large as that.
Thus we can can say that:
:$n \ge \ceiling {\sqrt {123 \, 456 \, 789} } = 11 \, 112$
where $\ceiling {\, \cdot \,}$ denotes the ceiling function.
It remains to be demonstrated that no positive integer between $11 \, 112$ and $11 \, 826$ has a penholodigital square.
{{ProofWanted|This is over $700$ numbers. The task can be filtered by, for example, disregarding all $n$ ending in $1$ and $9$ because their squares will both begin and end in $1$ at this low range of the $10000$s, and of course all $n$ ending in $0$ because their squares will end in $0$.}}
Theorem: Locally Compact Space is Weakly Locally Compact

Let $T = \struct {S, \tau}$ be a locally compact topological space.
Then $T$ is weakly locally compact.

Proof:
We are given $T$ is locally compact, so every point of $S$ has a neighborhood basis consisting of compact sets.
Let $x \in S$ be arbitrary. 
From Topological Space is Neighborhood of all its Points, $S$ is a neighborhood of $x$ in $S$.
So, every point contained in $S$ has a compact neighborhood.
Hence, $T$ is weakly locally compact.
{{qed}}
Theorem: Topological Space is Neighborhood of all its Points

Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$.
Then $S$ is a neighborhood of $x$.

Proof:
By the definition of the topology $\tau$, $S$ is an open set.
From Set is Open iff Neighborhood of all its Points, $S$ is a neighborhood of $x$.
{{qed}}
Theorem: Equivalence of Definitions of Noetherian Module

Let $A$ be a commutative ring with unity.
Let $M$ be an $A$-module.
{{TFAE|def = Noetherian Module}}

Proof:
{{tidy}}
Theorem: Particular Point Space is Locally Compact

Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is locally compact.

Proof:
{{Recall|Particular Point Topology|particular point space}}
{{:Definition:Particular Point Topology}}
{{Recall|Locally Compact Space|locally compact space}}
{{:Definition:Locally Compact Space}}
Let $x \in S$.
Consider the set $\set {p, x}$.
From the definition of particular point topology, $\set {p, x}$ is open in $T$.
By Finite Topological Space is Compact, $\set {p, x}$ is compact.
Let $N$ be a neighborhood of $x$.
Then:
:$\exists U \in \tau_p: x \in U \subseteq N$.
From the definition of particular point topology, since $U \ne \O$, we must have $p \in U$.
Therefore $\set {p, x} \subseteq U \subseteq N$.
Since $N$ is arbitrary, $\set {\set {p, x} }$ is a neighborhood basis for $x$.
The result follows from definition of a locally compact space.
{{qed}}
Theorem: Largest Integer not Sum of Two Abundant Numbers

The largest integer which is not the sum of $2$ abundant numbers is $20 \, 161$.

Proof:
First we show that for $1 < k < 90$, $315 k$ is abundant.
If $k$ is divisible by $3, 5, 7$, note that:
:$945, 1575, 2205$
are all abundant, and $315 k$ is a multiple of at least one of them.
Hence $315 k$ is abundant by Multiple of Abundant Number is Abundant.
If $k$ is not divisible by $3, 5, 7$:
Let $p$ be a prime such that $p \divides k$.
Then:
{{begin-eqn}}
{{eqn | l = \frac {\map \sigma {315 p} } {315 p}
      | r = \frac 1 {315 p}  \paren {1 + 3 + 3^2} \paren {1 + 5} \paren {1 + 7}  \paren {1 + p}
}}
{{eqn | r = \frac {208} {105} \paren {1 + \frac 1 p}
}}
{{eqn | o = >
      | r = \frac {208} {105} \paren {1 + \frac 1 {90} }
      | c = $p < 90$
}}
{{eqn | o = >
      | r = 2
}}
{{end-eqn}}
hence $315 p$ and $315 k$ are abundant.
Since $88$ and $315$ are coprime:
:$88 = 2^3 \times 11$
:$315 = 3^2 \times 5 \times 7$
By Largest Number not Expressible as Sum of Multiples of Coprime Integers, all numbers greater than or equal to:
:$\paren {88 - 1} \paren {315 - 1} = 27 \, 318$
can be expressed as a sum of multiples of $88$ and $315$.
Hence for $n \ge 27 \, 318 + 315 \times 2 = 27 \, 948$:
:$\exists s, t \in \N: 90 > t \ge 2: n = 88 s + 315 t$
and both $88 s$ and $315 t$ are abundant for $s > 0$.
For $s = 0$, $t \ge \dfrac {27 \, 948} {315} > 7 = \paren {2 - 1} \paren {3 - 1} + 5$.
By Largest Number not Expressible as Sum of Multiples of Coprime Integers, $t - 5$ can be expressed as a sum of multiples of $2$ and $3$.
Hence:
:$\exists a, b \in \Z_{> 0}: 2 a + 3 b = t$
This gives:
:$n = 630 a + 945 b$
and both $630 a$ and $945 b$ are abundant.
We still need to find representations for $20 \, 162 < n < 27 \, 948$.
We can check this via brute force.
Using Largest Number not Expressible as Sum of Multiples of Coprime Integers/Generalization, we can narrow down our search to numbers that are not divisible by small primes:
Since $\gcd \set {18, 20} = 2$, the largest multiple of $2$ not expressible as a sum of multiples of $18$ and $20$ is:
:$\dfrac {18 \times 20} 2 - 18 - 20 = 142 < 20161$
Since $\gcd \set {12, 945} = 3$, the largest multiple of $3$ not expressible as a sum of multiples of $12$ and $945$ is:
:$\dfrac {12 \times 945} 3 - 12 - 945 = 2823 < 20161$
Since $\gcd \set {20, 945} = 5$, the largest multiple of $5$ not expressible as a sum of multiples of $20$ and $945$ is:
:$\dfrac {20 \times 945} 5 - 20 - 945 = 2815 < 20161$
Since $\gcd \set {56, 945} = 7$, the largest multiple of $7$ not expressible as a sum of multiples of $56$ and $945$ is:
:$\dfrac {56 \times 945} 7 - 56 - 945 = 6559 < 20161$
All numbers involved above are abundant.
Hence we only need to consider $n$ not divisible by $2, 3, 5, 7$.
{{finish|Brute force by computer}}
Theorem: Smallest Integer using Three Words in English Description

The smallest integer which uses exactly $3$ words in its standard (British) English description is:
:$21 \, 000$: '''twenty-one thousand'''
counting hyphenations as separate words.

Proof:
All integers up to $100$ ('''one hundred''') use either $1$ or $2$ words:
:'''one'''
:'''sixty'''
:'''seventeen'''
:'''ninety-eight'''
All integers of the form $100 n$ for $n = 1, 2, \ldots 9$ use exactly $2$ words:
:'''one hundred'''
:'''seven hundred'''
:'''nine hundred'''
In British English, the technique for describing integers from $101$ to $199$, and $201$ to $299$ and so on, is to use '''and''' between the number of hundreds and the rest:
:'''one hundred and one'''
:'''three hundred and thirteen'''
:'''four hundred and twenty-six'''
:'''seven hundred and seventy'''
thus using either $4$ or $5$ words.
All integers of the form $1000 n$ for $n = 1, 2, \ldots 10$ use exactly $2$ words:
:'''two thousand'''
:'''five thousand'''
:'''eight thousand'''
:'''twelve thousand'''
:'''nineteen thousand'''
:'''twenty thousand'''
Similarly with hundreds, the technique for describing integers of the form $1000 m + n$ for $1 \le n \le 99$ is to use '''and''' between the number of thousands and the rest:
:'''five thousand and eighteen'''
:'''sixteen thousand and forty-eight'''
:'''thirty-seven thousand and sixty'''
thus using either $4$ or $5$ words.
All other numbers between $1100$ and $20 \, 999$ trivially use more than $3$ words:
:'''four thousand, eight hundred'''
:'''sixteen thousand, one hundred and seventy-seven'''
:'''twenty thousand, nine hundred and ninety-nine'''
and so on.
The smallest integer to use exactly $2$ words is $21$:
:'''twenty-one'''
Hence:
:'''twenty-one thousand'''
{{qed}}
Theorem: Smallest Pythagorean Quadrilateral with Integer Sides

The smallest Pythagorean quadrilateral in which the sides of the $4$ right triangles formed by its sides and perpendicular diagonals are all integers has an area of $21 \, 576$.
The sides of the right triangles in question are:
:$25, 60, 65$
:$91, 60, 109$
:$91, 312, 325$
:$25, 312, 313$

Proof:
:800px
The $4$ right triangles are inspected:
{{begin-eqn}}
{{eqn | l = 25^2 + 60^2
      | r = 625 + 3600
}}
{{eqn | r = 4225
      | c = 
}}
{{eqn | r = 65^2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 91^2 + 60^2
      | r = 8281 + 3600
}}
{{eqn | r = 11 \, 881
      | c = 
}}
{{eqn | r = 109^2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 91^2 + 312^2
      | r = 8281 + 97 \, 344
}}
{{eqn | r = 105 \, 625
      | c = 
}}
{{eqn | r = 325^2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 25^2 + 312^2
      | r = 625 + 97 \, 344
}}
{{eqn | r = 97 \, 969
      | c = 
}}
{{eqn | r = 313^2
      | c = 
}}
{{end-eqn}}
The area of each right triangle is calculated:
{{begin-eqn}}
{{eqn | l = \dfrac {25 \times 60} 2
      | r = 750
}}
{{eqn | l = \dfrac {91 \times 60} 2
      | r = 2730
}}
{{eqn | l = \dfrac {91 \times 312} 2
      | r = 14 \, 196
}}
{{eqn | l = \dfrac {25 \times 312} 2
      | r = 3900
}}
{{end-eqn}}
Thus the total area is:
:$750 + 2730 + 14 \, 196 + 3900 = 21 \, 576$
{{ProofWanted|It needs to be proved that this is the smallest.}}