avewright/proofwiki-math huggingface dataset:
Theorem: Conditions for Integral Functionals to have same Euler's Equations

Let $\mathbf y$ be a real $n$-dimensional vector-valued function.
Let $\map F {x, \mathbf y, \mathbf y'}$, $\map \Phi {x, \mathbf y}$  be real functions.
Let $\Phi$ be twice differentiable.
Let:
{{begin-eqn}}
{{eqn | l = \Psi
      | r = \frac {\d \Phi} {\d x}
}}
{{eqn | r = \frac {\partial \Phi} {\partial x} + \sum_{i \mathop = 1}^n \frac {\partial \Phi} {\partial y_i} y_i'
}}
{{end-eqn}}
Let $J_1$, $J_2$ be functionals such that:
:$\ds J_1 \sqbrk {\mathbf y} = \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$
:$\ds J_2 \sqbrk {\mathbf y} = \int_a^b \paren {\map F {x, \mathbf y, \mathbf y'} + \map \Psi {x, \mathbf y, \mathbf y'} } \rd x$ 
Then $J_1$ and $J_2$ have same Euler's Equations.

Proof:
According to Necessary Condition for Integral Functional to have Extremum for given function/Dependent on N Functions:
Euler's Equations for functional $J_1$ are: 
:$\ds F_{\mathbf y} - \frac \d {\d x} F_{\mathbf y'} = 0$
Equivalently, for $J_2$ we have
{{begin-eqn}}
{{eqn | l = \paren {F_{\mathbf y} + \Psi_{\mathbf y} } - \map {\frac \d {\d x} } {F_{\mathbf y'} + \Psi_{\mathbf y'} }
      | r = \paren {F_{\mathbf y} - \frac \d {\d x} F_{\mathbf y'} } + \paren {\Psi_{\mathbf y} - \frac \d {\d x} \Psi_{\mathbf y'} }
}}
{{eqn | r = 0
      | c = condition for the existence of extremum
}}
{{end-eqn}}
Furthermore:
{{begin-eqn}}
{{eqn | l = \Psi_{\mathbf y}
      | r = \frac {\partial^2 \Phi} {\partial \mathbf y \partial x} + \frac \partial {\partial \mathbf y} \sum_{j \mathop = 1}^n \frac {\partial \Phi} {\partial y_j} y_j'
}}
{{eqn | r = \frac {\partial^2 \Phi} {\partial \mathbf y \partial x} + \sum_{j \mathop = 1}^n \frac {\partial^2 \Phi} {\partial \mathbf y \partial y_j} y_j'
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \Psi_{\mathbf y'}
      | r = \frac {\partial^2 \Phi} {\partial \mathbf y' \partial x} + \frac \partial {\partial \mathbf y'} \sum_{j \mathop = 1}^n \frac {\partial \Phi} {\partial y_j} y_j'
}}
{{eqn | r = 0 + \sum_{j \mathop = 1}^n \paren {\frac {\partial^2 \Phi} {\partial \mathbf y' \partial y_j} y_j' + \frac {\partial \Phi} {\partial y_j} \frac {\partial y_j'} {\partial \mathbf y'} }
}}
{{eqn | r = \frac {\partial \Phi} {\partial \mathbf y}
      | c = as $\frac {\partial y_i'} {\partial y_j'} = \delta_{i j} $, where $\delta_{i j}$ is the Kronecker Delta
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \frac {\d \Psi_{\mathbf y'} } {\d x}
      | r = \frac {\partial \Psi_{\mathbf y'} } {\partial x} + \sum_{j \mathop = 1}^n \frac {\partial \Psi_{\mathbf y'} } {\partial y_j} \frac {\d y_j} {\d x} + \sum_{j \mathop = 1}^n \frac {\partial \Psi_{\mathbf y'} } {\partial y_j'} \frac {\d y_j'} {\d x}
}}
{{eqn | r = \frac {\partial^2 \Phi} {\partial x \partial \mathbf y} + \sum_{j \mathop = 1}^n \frac {\partial^2 \Phi} {\partial \mathbf y \partial y_j} y_j'
}}
{{end-eqn}}
Since $\Phi$ is twice differentiable, by Schwarz-Clairaut Theorem partial derivatives commute and:
:$\Psi_{\mathbf y} - \dfrac \d {\d x} \Psi_{\mathbf y'} = 0$
Therefore, $J_1$ and $J_2$ have same Euler's Equations.
{{qed}}
Theorem: Square of Odd Number as Difference between Triangular Numbers

Let $n \in \Z_{\ge 0}$ be a positive integer.
Then:
:$\exists a, b \in \Z_{\ge 0}: \paren {2 n + 1}^2 = T_a - T_b$
where:
:$T_a$ and $T_b$ are triangular numbers
:$T_a$ and $T_b$ are coprime.
That is, the square of every odd number is the difference between two coprime triangular numbers.

Proof:
{{begin-eqn}}
{{eqn | l = T_a - T_b
      | r = \dfrac {a^2 + a} 2 - \dfrac {b^2 + b} 2
      | c = Closed Form for Triangular Numbers
}}
{{end-eqn}}
Let $a = 3b + 1$
{{begin-eqn}}
{{eqn | l = T_{3 b + 1} - T_b
      | r = \dfrac {\paren {3 b + 1}^2 + 3 b + 1} 2 - \dfrac {b^2 + b} 2
      | c =
}}
{{eqn | r = \dfrac {\paren {3 b + 1}^2 + 3 b + 1 - b^2 + b} 2
      | c = 
}}
{{eqn | r = \dfrac {9 b^2 + 6 b + 1 + 3 b + 1 - b^2 - b} 2
      | c = 
}}
{{eqn | r = \dfrac {8 b^2 + 8 b + 2} 2
      | c = 
}}
{{eqn | r = 4 b^2 + 4 b + 1
      | c = 
}}
{{eqn | r = \paren {2 b + 1}^2
      | c = 
}}
{{end-eqn}}
The square of every odd number can be made through this method.
{{begin-eqn}}
{{eqn | l = T_{3 b + 1}
      | r = \dfrac {\paren {3 b + 1} ^ 2 + 3 b + 1} 2
      | c =
}}
{{eqn | r = \dfrac {9 b^2 + 9 b + 2} 2
      | c = 
}}
{{eqn | r = 9 \paren {\dfrac {b^2 + b} 2} + 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = T_{3 b + 1} - 9 T_b
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \gcd \set {T_{3 b + 1}, T_b}
      | r = 1
      | c = Bézout's Identity
}}
{{end-eqn}}
Both triangular numbers are coprime.
{{qed}}
Theorem: Meet in Inclusion Ordered Set

Let $P = \struct {X, \subseteq}$ be an inclusion ordered set.
Let $A, B \in X$ such that
:$A \cap B \in X$
Then $A \wedge B = A \cap B$

Proof:
By Intersection is Subset:
:$A \cap B \subseteq A$ and $A \cap B \subseteq B$
By definition:
:$A \cap B$ is a lower bound for $\set {A, B}$
We will prove that
:$\forall C \in X: C$ is a lower bound for $\set {A, B} \implies C \subseteq A \cap B$
Let $C \in X$ such that:
:$C$ is a lower bound for $\set {A, B}$.
By definition of lower bound:
:$C \subseteq A$ and $C \subseteq B$
Thus by Intersection is Largest Subset:
:$C \subseteq A \cap B$
{{qed|lemma}}
By definition of infimum:
:$\inf \set {A, B} = A \cap B$
Thus by definition of meet:
:$A \wedge B = A \cap B$
{{qed}}
Theorem: Element is Meet Irreducible iff Complement of Element is Irreducible

Let $T = \struct {S, \tau}$ be a topological space.
Let $P = \struct {\tau, \preceq}$ be an ordered set
where $\mathord \preceq = \mathord \subseteq \cap \paren {\tau \times \tau}$
Let $A \in \tau$ such that
:$A \ne \top_P$
where $\top_P$ denotes the greatest element in $P$.
Then $A$ is meet irreducible in $P$ {{iff}} $\relcomp S A$ is irreducible
where $\relcomp S A$ denotes the relative complement of $A$ relative to $S$.

Proof:
=== Sufficient Condition ===
Let $A$ be meet irreducible in $P$.
By Top in Ordered Set of Topology:
:$A \ne S$
By Relative Complement of Empty Set and Relative Complement of Relative Complement:
:$\relcomp S A \ne \O$
Thus by definition:
:$\relcomp S A$ is non-empty.
Thus by definition:
:$\relcomp S A$ is closed.
Let $B, C$ be closed subsets of $S$ such that
:$\relcomp S A = B \cup C$
By De Morgan's Laws (Set Theory)/Relative Complement/Complement of Union and Relative Complement of Relative Complement:
:$A = \relcomp S B \cap \relcomp S C$
By definition of closed set:
:$\relcomp S B \in \tau$ and $\relcomp S C \in \tau$
By definition of topological space:
:$\relcomp S B \cap \relcomp S C \in \tau$
By Meet in Inclusion Ordered Set:
:$\relcomp S B \wedge \relcomp S C = A$
By definition of meet irreducible:
:$A = \relcomp S B$ or $A = \relcomp S C$
Thus by Relative Complement of Relative Complement:
:$\relcomp S A = B$ or $\relcomp S A = C$
{{qed|lemma}}
Theorem: Square of Triangular Number equals Sum of Sequence of Cubes

:$\ds \sum_{i \mathop = 1}^n i^3 = {T_n}^2$
where $T_n$ denotes the $n$th triangular number.

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^3
      | r =  \frac {n^2 \paren {n + 1}^2} 4
      | c = Sum of Sequence of Cubes
}}
{{eqn | r = \paren {\frac {n \paren {n + 1} } 2}^2
      | c = 
}}
{{eqn | r = {T_n}^2
      | c = Closed Form for Triangular Numbers
}}
{{end-eqn}}
{{qed}}
Theorem: Square of Triangular Number equals Sum of Sequence of Cubes

:$\ds \sum_{i \mathop = 1}^n i^3 = {T_n}^2$
where $T_n$ denotes the $n$th triangular number.

Proof:
The proof proceeds by induction.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 1}^n i^3 = {T_n}^2$
=== Basis for the Induction ===
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^1 i^3
      | r = 1^3
      | c = 
}}
{{eqn | r = 1
      | c = 
}}
{{eqn | r = \paren {\frac {1 \paren {1 + 1} } 2}^2
      | c = 
}}
{{eqn | r = {T_1}^2
      | c = Closed Form for Triangular Numbers
}}
{{end-eqn}}
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds \sum_{i \mathop = 1}^k i^3 = {T_k}^2$
from which it is to be shown that:
:$\ds \sum_{i \mathop = 1}^{k + 1} i^3 = {T_{k + 1} }^2$
=== Induction Step ===
This is the induction step:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^{k + 1} i^3
      | r = \sum_{i \mathop = 1}^k i^3 + \paren {k + 1}^3
      | c = 
}}
{{eqn | r = {T_k}^2 + \paren {k + 1}^3
      | c = Induction Hypothesis
}}
{{eqn | r = {T_k}^2 + \paren { {T_{k + 1} }^2 - {T_k}^2}
      | c = Cube Number as Difference between Squares of Triangular Numbers
}}
{{eqn | r = {T_{k + 1} }^2
      | c = 
}}
{{end-eqn}}
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{> 0}: \ds \sum_{i \mathop = 1}^n i^3 = {T_n}^2$
{{qed}}
Theorem: Sum of Sequence of Cubes divides 3 times Sum of Sequence of Fifth Powers

:$\ds \sum_{i \mathop = 1}^n i^3 \divides 3 \sum_{i \mathop = 1}^n i^5$
where $\divides$ denotes divisibility.

Proof:
{{begin-eqn}}
{{eqn | l = 3 \sum_{i \mathop = 1}^n i^5
      | r = {T_n}^2 \paren {4 T_n - 1}
      | c = Sum of Sequence of Fifth Powers
}}
{{eqn | r = k {T_n}^2
      | c = where $k = 4 T_n - 1$
}}
{{eqn | ll= \leadsto
      | l = {T_n}^2
      | o = \divides
      | r = 3 \sum_{i \mathop = 1}^n i^5
      | c = {{Defof|Divisor of Integer}}
}}
{{eqn | ll= \leadsto
      | l = \sum_{i \mathop = 1}^n i^3
      | o = \divides
      | r = 3 \sum_{i \mathop = 1}^n i^5
      | c = Square of Triangular Number equals Sum of Sequence of Cubes
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Sequence of Fifth Powers

:$\ds \sum_{i \mathop = 1}^n i^5 = \dfrac { {T_n}^2 \paren {4 T_n - 1} } 3$
where $T_n$ denotes the $n$th triangular number.

Proof:
<onlyinclude>
The proof proceeds by induction.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 1}^n i^5 = \dfrac { {T_n}^2 \paren {4 T_n - 1} } 3$
Theorem: Sum of Adjacent Sequences of Triangular Numbers

{{begin-eqn}}
{{eqn | l = T_1 + T_2 + T_3
      | r = T_4
      | c = 
}}
{{eqn | l = T_5 + T_6 + T_7 + T_8
      | r = T_9 + T_{10}
      | c = 
}}
{{eqn | l = T_{11} + T_{12} + T_{13} + T_{14} + T_{15}
      | r = T_{16} + T_{17} + T_{18}
      | c = 
}}
{{end-eqn}}
and so on.
The $n$th line of the pattern can be written as:
:$\ds \sum_{k \mathop = n^2 + n - 1}^{n^2 + 2 n} T_n = \sum_{k \mathop = n^2 + 2 n + 1}^{n^2 + 3 n} T_n$

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = n^2 + n - 1}^{n^2 + 2 n} T_n
      | r = \sum_{k \mathop = 1}^{n^2 + 2 n} T_n - \sum_{k \mathop = 1}^{n^2 + n - 2} T_n
}}
{{eqn | r = H_{n^2 + 2 n} - H_{n^2 + n - 2}
      | c = {{Defof|Tetrahedral Number}}
}}
{{eqn | r = \frac {\paren {n^2 + 2 n} \paren {n^2 + 2 n + 1} \paren {n^2 + 2 n + 2} } 6 - \frac {\paren {n^2 + n - 2} \paren {n^2 + n - 1} \paren {n^2 + n} } 6
      | c = Closed Form for Tetrahedral Numbers
}}
{{eqn | r = \frac {n \paren {n + 1} \paren {n + 2} } 6 \paren {\paren {n + 1} \paren {n^2 + 2 n + 2} - \paren {n - 1} \paren {n^2 + n - 1} }
}}
{{eqn | r = \frac {n \paren {n + 1} \paren {n + 2} } 6 \paren {3 n^2 + 6 n + 1}
}}
{{eqn | r = \frac {n \paren {n + 1} \paren {n + 2} } 6 \paren {\paren {n + 3} \paren {n^2 + 3 n + 1} - \paren {n + 1} \paren {n^2 + 2 n + 2} }
}}
{{eqn | r = \frac {\paren {n^2 + 3 n} \paren {n^2 + 3 n + 1} \paren {n^2 + 3 n + 2} } 6 - \frac {\paren {n^2 + 2 n} \paren {n^2 + 2 n + 1} \paren {n^2 + 2 n + 2} } 6
}}
{{eqn | r = H_{n^2 + 3 n} - H_{n^2 + 2 n}
      | c = Closed Form for Tetrahedral Numbers
}}
{{eqn | r = \sum_{k \mathop = 1}^{n^2 + 3 n} T_n - \sum_{k \mathop = 1}^{n^2 + 2 n} T_n
      | c = {{Defof|Tetrahedral Number}}
}}
{{eqn | r = \sum_{k \mathop = n^2 + 2 n + 1}^{n^2 + 3 n} T_n
}}
{{end-eqn}}
{{qed}}
Theorem: Fourth Power is Sum of 2 Triangular Numbers

Let $n \in \Z$ be an integer.
Then:
:$\exists a, b \in \Z_{\ge 0}: n^4 = T_a + T_b$
where $T_a$ and $T_b$ are triangular numbers.
That is, the $4$th power of an integer equals the sum of two triangular numbers.

Proof:
{{begin-eqn}}
{{eqn | l = T_{n^2 - n - 1} + T_{n^2 + n - 1}
      | r = \frac {\paren {n^2 - n - 1} \paren {n^2 - n} } 2 + \frac {\paren {n^2 + n - 1} \paren {n^2 + n} } 2
      | c = Closed Form for Triangular Numbers
}}
{{eqn | r = \frac {n \paren {n^2 - n - 1} \paren {n - 1} } 2 + \frac {n \paren {n^2 + n - 1} \paren {n + 1} } 2
      | c = factoring
}}
{{eqn | r = \frac {n \paren {n^3 - n^2 - n - n^2 + n + 1} } 2 + \frac {n \paren {n^3 + n^2 - n + n^2 + n - 1} } 2
      | c = multiplying out the numerators
}}
{{eqn | r = \frac {n \paren {\paren {n^3 - 2 n^2 + 1} + \paren {n^3 + 2 n^2 - 1} } } 2
      | c = simplifying
}}
{{eqn | r = n^4
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Square of Triangular Numbers as Sum of Triangular Numbers

:${T_n}^2 = T_n + T_{n - 1} T_{n + 1}$
where $T_n$ denotes the $n$th triangular number.

Proof:
{{begin-eqn}}
{{eqn | l = T_n + T_{n - 1} T_{n + 1}
      | r = \frac {n \paren {n + 1} } 2 + \paren {\frac {\paren {n - 1} n} 2 \frac {\paren {n + 1} \paren {n + 2} } 2}
      | c = Closed Form for Triangular Numbers
}}
{{eqn | r = \frac {2 n^2 + 2 n + \paren {n^2 - n} \paren {n^2 + 3 n + 2} } 4
      | c = 
}}
{{eqn | r = \frac {2 n^2 + 2 n + n^4 - n^3 + 3 n^3 - 3 n^2 + 2 n^2 - 2 n} 4
      | c = 
}}
{{eqn | r = \frac {n^4 + 2 n^3 + n^2} 4
      | c = 
}}
{{eqn | r = \paren {\frac {n \paren {n + 1} } 2}^2
      | c = 
}}
{{eqn | r = {T_n}^2
      | c = Closed Form for Triangular Numbers
}}
{{end-eqn}}
{{qed}}
Theorem: Product of Consecutive Triangular Numbers

:$2 T_n T_{n - 1} = T_{n^2 - 1}$
where $T_n$ denotes the $n$th triangular number.

Proof:
{{begin-eqn}}
{{eqn | l = 2 T_n T_{n - 1}
      | r = 2 \paren {\frac {n \paren {n + 1} } 2} \paren {\frac {\paren {n - 1} n} 2}
      | c = Closed Form for Triangular Numbers
}}
{{eqn | r = \frac {\paren {n^2 + n} \paren {n^2 - n} } 2
      | c = 
}}
{{eqn | r = \frac {n^4 - n^2} 2
      | c = Difference of Two Squares
}}
{{eqn | r = \frac {n^2 \paren {n^2 - 1} } 2
      | c = 
}}
{{eqn | r = T_{n^2 - 1}
      | c = Closed Form for Triangular Numbers
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Sequence of Reciprocals of Triangular Numbers

:$\ds \sum_{k \mathop \ge 1} \dfrac 1 {T_k} = 2$
where $T_k$ denotes the $k$th triangular number.

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 1} \dfrac 1 {T_k}
      | r = \sum_{k \mathop \ge 1} \dfrac 2 {k \paren {k + 1} }
      | c = Closed Form for Triangular Numbers
}}
{{eqn | r = 2 \sum_{k \mathop \ge 1} \dfrac 1 {k \paren {k + 1} }
      | c = 
}}
{{eqn | r = 2 \times 1
      | c = {{Corollary|Sum from 1 to n of 1 over r(r+1)|disp = Sum from $1$ to $n$ of $\dfrac 1 {r \paren {r + 1} }$}}
}}
{{end-eqn}}
{{qed}}
Theorem: Triangular Number whose Square is Triangular

The only triangular number with less than $660$ digits, whose square is also triangular, is $6$.

Proof:
We have that:
::${T_3}^2 = 6^2 = 36 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8$
To establish that it is the only one yet known can be established by an exhaustive search.
{{qed}}
Theorem: Factors of Integer Congruent to 5 modulo 6

Let $m$ be an positive integer.
Let $m \equiv 5 \pmod 6$.
Then $m$ has two divisors whose sum is divisible by $6$.

Proof:
{{questionable|This is so trivial it's pointless. Both $m$ and $1$ are factors, totalling $6 n$, trivially a multiple of $6$. I expect this should be "prime factors".}}
Theorem: Top in Ordered Set of Topology

Let $T = \left({S, \tau}\right)$ be a topological space.
Let $P = \left({\tau, \subseteq}\right)$ be an inclusion ordered set of $\tau$.
Then $P$ is bounded above and $\top_P = S$
where $\top_P$ denotes the greatest element in $P$.

Proof:
By definition of topological space:
:$S \in \tau$
By definition of subset:
:$\forall A \in \tau: A \subseteq S$
Hence $P$ is bounded above.
Thus by definition of the greatest element:
:$\top_P = S$
{{qed}}
Theorem: Triangular Numbers which are also Square

Let $A_n$ be the $n$th non-negative integer whose square is also a triangular number.
Then:
:$A_n = \begin {cases} 0 & : n = 0 \\ 1 & : n = 1 \\ 6 A_{n - 1} - A_{n - 2} & : n > 1 \end {cases}$

Proof:
Let $n \in \Z_{>0}$ be such that $n^2$ is a triangular number.
Then we have:
{{begin-eqn}}
{{eqn | q = \exists m \in \Z_{>0}
      | l = n^2
      | r = \dfrac {m \paren {m + 1} } 2
      | c = Closed Form for Triangular Numbers
}}
{{eqn | ll= \leadstoandfrom
      | l = 2 n^2
      | r = m^2 + m
      | c = 
}}
{{eqn | r = \dfrac {\paren {2 m + 1}^2 - 1} 4
      | c = Completing the Square
}}
{{eqn | ll= \leadstoandfrom
      | l = 8 n^2 + 1
      | r = \paren {2 m + 1}^2
      | c = simplifying
}}
{{eqn | ll= \leadstoandfrom
      | l = x^2 - 8 y^2
      | r = 1
      | c = setting $x = 2 m + 1$ and $y = n$
}}
{{end-eqn}}
This is a Pellian equation.
From Pell's Equation: $x^2 - 8 y^2 = 1$, the smallest positive integral solution is:
:$\tuple {x, y} = \tuple {3, 1}$
Thus we have that the sequence of $n$ such that $n^2$ is triangular is the sequence of numerators of the Continued Fraction Expansion of $\sqrt 8$ whose indices are even.
Let $\tuple {x, y} = \tuple {p_k, q_k}$ be a  solution to $x^2 - 8 y^2 = 1$.
Then from Continued Fraction Expansion of $\sqrt 8$ we have:
{{begin-eqn}}
{{eqn | l = p_{k + 1}
      | r = a_{k + 1} p_k + p_{k - 1}
      | c = 
}}
{{eqn | r = 4 p_k + p_{k - 1}
      | c = as $a_{k + 1} = 4$
}}
{{eqn | l = p_{k + 2}
      | r = a_{k + 2} p_{k + 1} + p_k
      | c = 
}}
{{eqn | r = p_{k + 1} + p_k
      | c = as $a_{k + 2} = 1$
}}
{{eqn | r = 4 p_k + p_{k - 1} + p_k
      | c = from above
}}
{{eqn | r = 5 p_k + p_{k - 1}
      | c = simplfying
}}
{{eqn | l = p_{k + 3}
      | r = a_{k + 3} p_{k + 2} + p_{k + 1}
      | c = 
}}
{{eqn | r = 4 p_{k + 2} + p_{k + 1}
      | c = as $a_{k + 3} = 4$
}}
{{eqn | l = p_{k + 4}
      | r = a_{k + 4} p_{k + 3} + p_{k + 2} 
      | c = 
}}
{{eqn | r = p_{k + 3} + p_{k + 2} 
      | c = as $a_{k + 4} = 1$
}}
{{eqn | r = 4 p_{k + 2} + p_{k + 1} + p_{k + 2} 
      | c = from above
}}
{{eqn | r = 5 p_{k + 2} + p_{k + 1}
      | c = simplifying
}}
{{eqn | r = 5 p_{k + 2} + \paren {p_{k + 2} - p_k}
      | c = substituting for $p_{k + 1}$
}}
{{eqn | r = 6 p_{k + 2} + p_k
      | c = simplifying
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Index of Square Triangular Number from Preceding

Let $T_n$ be the $n$th triangular number.
Let $T_n$ be square.
Then $T_{4 n \paren {n + 1} }$ is also square.

Proof:
{{begin-eqn}}
{{eqn | l = T_{4 n \paren {n + 1} }
      | r = \frac {\paren {4 n \paren {n + 1} } \paren {4 n \paren {n + 1} + 1} } 2
      | c = Closed Form for Triangular Numbers
}}
{{eqn | ll= \leadsto
      | l = T_{4 n \paren {n + 1} }
      | r = \frac {n \paren {n + 1} } 2 \times 4 \paren {4 n \paren {n + 1} + 1}
      | c = Extract $T_n$
}}
{{eqn | ll= \leadsto
      | l = T_{4 n \paren {n + 1} }
      | r = T_n \times \paren {16 n^2 + 16 n + 4}
      | c = Simplify
}}
{{eqn | ll= \leadsto
      | l = T_{4 n \paren {n + 1} }
      | r = T_n \times \paren {4 n + 2}^2
      | c = Factorise
}}
{{end-eqn}}
The product of two squares is also square.
Let $T_n$ be square.
Therefore $T_{4 n \paren {n + 1} }$ is also square.
{{qed}}
Theorem: Triangular Number cannot be Cube

Let $T_n$ be the $n$th triangular number such that $n > 1$.
Then $T_n$ cannot be a cube.

Proof:
Suppose $T_n = x^3$ for some $x \in \Z$.
Then by Odd Square is Eight Triangles Plus One:
:$\exists y \in \Z: 8 T_n + 1 = \paren {2 x}^3 + 1 = y^2$
By Cube which is One Less than a Square:
:$2 x = 2$, $y = 3$
giving the unique solution:
:$T_n = 1^3 = 1$
{{qed}}
Theorem: Numbers of form 31 x 16^n are sum of 16 4th Powers

Let $m \in \Z$ be an integer of the form $31 \times 16^n$ for $n \in \Z_{\ge 0}$.
Then in order to express $m$ as the sum of fourth powers, you need $16$ of them.

Proof:
We have:
:$31 \times 16^n = \paren {2^{n + 1} }^4 + 15 \times \paren {2^n}^4$
so every integer of the form $31 \times 16^n$ for $n \in \Z_{\ge 0}$ can be expressed as the sum of 16 fourth powers.
Now we show that we cannot use less than $16$ fourth powers.
Observe that for an even number $2 k$:
{{begin-eqn}}
{{eqn | l = \paren {2 k}^4
      | r = 16 k^4
}}
{{eqn | o = \equiv
      | r = 0
      | rr= \pmod {16}
}}
{{end-eqn}}
For an odd number $2 k + 1$:
{{begin-eqn}}
{{eqn | l = \paren {2 k + 1}^4
      | r = 16 k^4 + 32 k^3 + 24 k^2 + 8 k + 1
      | c = Binomial Theorem
}}
{{eqn | o = \equiv
      | r = 8 k \paren {k + 1} + 1
      | rr= \pmod {16}
}}
{{eqn | o = \equiv
      | r = 1
      | rr= \pmod {16}
      | c = since $k \paren {k + 1}$ is even
}}
{{end-eqn}}
It is obvious that for $n = 0$, $31$ requires $16$ fourth powers to express:
:$31 = 2^4 + 15 \times 1^4$
{{AimForCont}} for some $n > 0$, $31 \times 16^n$ requires less than $16$ fourth powers to express.
Let $m$ be the smallest of those $n$.
Suppose $x$ of the summands are odd, where $x < 16$.
By the above, we must have:
:$31 \times 16^m \equiv x \pmod {16}$
Since $31 \times 16^m$ is divisible by $16$, $x = 0$.
Hence each summand is even.
Dividing each summand by $2$ gives a representation of $31 \times 16^{m - 1}$ as a sum of less than $16$ fourth powers.
This contradicts the minimality condition on $m$.
Hence the result by Proof by Contradiction.
{{qed}}
Category:Hilbert-Waring Theorem
Category:16
hh2o62vrvx6vo7d9cpy9xxvp3znff4i
Theorem: Equality of Integers to the Power of Each Other

$2$ and $4$ are the only pair of positive integers $m, n$ such that $m \ne n$ such that:
:$m^n = n^m$
Thus:
$2^4 = 4^2 = 16$

Proof:
{{WLOG}} suppose $m > n$.
Write $m = n + x$, where $x$ is an integer.
Then:
{{begin-eqn}}
{{eqn | l = m^n
      | r = n^m
}}
{{eqn | l = \paren {n + x}^n
      | r = n^{n + x}
}}
{{eqn | l = \paren {1 + \frac x n}^n
      | r = n^x
      | c = dividing both sides by $n^n$
}}
{{end-eqn}}
From Real Sequence $\paren {1 + \dfrac x n}^n$ is Convergent:
:$\paren {1 + \dfrac x n}^n$ is increasing and has limit $e^x$.
Hence:
:$n^x < e^x$
This forces $n = 2$.
We have:
:$m^2 = 2^m$
showing that $m$ is a power of $2$.
Write $m = 2^k$. 
Then:
:$2^{2 k} = 2^{2^k}$
giving:
:$k = 2^{k - 1}$
By Bernoulli's Inequality:
:$2^{k - 1} \ge 1 + k - 1 = k$
where equality holds {{iff}} $k - 1 = 0$ or $k - 1 = 1$.
We can skip Bernoulli's Inequality by induction on $k$ for $k > 2$.
Either way, this gives:
:$k = 1$ or $2$
:$m = 2$ or $4$.
We reject $m = 2$ since $n = 2$.
Hence $2^4 = 4^2$ is the only solution.
{{qed}}
Theorem: Element of Ordered Set of Topology is Dense iff is Everywhere Dense

Let $T = \struct {S, \tau}$ be a topological space.
Let $P = \struct {\tau, \preceq}$ be an ordered set
where $\mathord \preceq = \mathord \subseteq \cap \paren {\tau \times \tau}$
Let $A \in \tau$.
Then $A$ is a dense element in $P$ {{iff}} $A$ is everywhere dense.

Proof:
=== Sufficient Condition ===
Assume that
:$A$ is a dense element in $P$.
By Bottom in Ordered Set of Topology:
:$\bot_P = \O$
We will prove that
:for every open subset $U$ of $S$: $U \ne \O \implies U \cap A \ne \O$
Let $U$ be an open subset of $S$ such that
:$U \ne \O$
By definition of open set:
:$U \in \tau$
By definition of topological space:
:$U \cap A \in \tau$
By Meet in Inclusion Ordered Set:
:$U \wedge A = U \cap A$
Thus by definition of dense element:
:$U \cap A \ne \O$
{{qed|lemma}}
By Characterization of Closure by Open Sets:
:$A^- = S$
where $A^-$ denotes the topological closure of $A$.
Hence $A$ is everywhere dense.
{{qed|lemma}}
Theorem: Square whose Perimeter equals its Area

The $4 \times 4$ square is the only square whose area in square units equals its perimeter in units.
The area and perimeter of this square are $16$.

Proof:
Let $S$ be a square whose area equals its perimeter.
Let $A$ be the area of $S$.
Let $P$ be the perimeter of $S$.
Let $b$ be the length of one side of $S$.
From Area of Square:
:$A = b^2$
From Perimeter of Rectangle:
:$P = 2 b + 2 b = 4 b$
Setting $A = P$
:$b^2 = 4 b$
and so:
:$b = 4$
and so:
:$A = 16 = P$
{{qed}}