avewright/proofwiki-math huggingface dataset:
Theorem: Finite Infima Set and Upper Closure is Smallest Filter

Let $L = \struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $X$ be a non-empty subset of $S$.
Then
:$X \subseteq \map {\operatorname {fininfs} } X^\succeq$ and
:for every a filter $F$ in $L$: $\paren {X \subseteq F \implies \map {\operatorname {fininfs} } X^\succeq \subseteq F}$
where
:$\map {\operatorname {fininfs} } X$ denotes the finite infima set of $X$,
:$X^\succeq$ denotes the upper closure of $X$.

Proof:
By Set is Subset of Finite Infima Set:
:$X \subseteq \map {\operatorname {fininfs} } X$
By Upper Closure of Subset is Subset of Upper Closure:
:$X^\succeq \subseteq \map {\operatorname {fininfs} } X^\succeq$
By Set is Subset of Upper Closure:
:$X \subseteq X^\succeq$
Thus by Subset Relation is Transitive:
:$X \subseteq \map {\operatorname {fininfs} } X^\succeq$
Let $F$ be a filter in $L$ such that
:$X \subseteq F$
Let $x \in \map {\operatorname {fininfs} } X^\succeq$
By definition of upper closure of subset:
:$\exists y \in \map {\operatorname{fininfs} } X: y \preceq x$
By definition of finite infima set:
:$\exists A \in \map {\operatorname {Fin} } X: y = \inf A \land A$ admits an infimum
where $\map {\operatorname {Fin} } X$ denotes the set of all finite subsets of $S$.
By Subset Relation is Transitive:
:$A \subseteq F$
By Filtered in Meet Semilattice with Finite Infima:
:$A \ne \O \implies \inf A \in F$
By Infimum of Empty Set is Greatest Element and Top in Filter:
:$A = \O \implies \inf A = \top \in F$
where $\top$ denotes the greatest element of $S$.
So
:$y \in F$
Thus by definition of upper section:
:$x \in F$
{{qed}}
Theorem: Smaller Number of Wythoff Pair is Smallest Number not yet in Sequence

Consider the sequence of Wythoff pairs arranged in sequential order:
:$\tuple {0, 0}, \tuple {1, 2}, \tuple {3, 5}, \tuple {4, 7}, \tuple {6, 10}, \tuple {8, 13}, \ldots$
The first coordinate of each Wythoff pair is the smallest positive integer which has not yet appeared in the sequence.

Proof:
From Sequence of Wythoff Pairs contains all Positive Integers exactly Once Each, every positive integer can be found in the sequence of Wythoff pairs.
From Difference between Terms of Wythoff Pair, the first coordinate is the smaller of the coordinates of the Wythoff pair.
So consider a given Wythoff pair.
Let $p$ be the smallest positive integer which has not yet appeared in the sequence.
It has to appear somewhere.
The terms of both the lower and upper Wythoff sequences are in ascending order.
$n$ must appear in the next Wythoff pair, otherwise a larger positive integer will appear before it when it eventually does appear.
$n$ cannot appear as the second coordinate, or it will be less than whatever number appears as the first coordinate.
Hence $n$ appears as the first coordinate of the next Wythoff pair.
{{qed}}
Theorem: Sequence of Wythoff Pairs contains all Positive Integers exactly Once Each

Consider the sequence of Wythoff pairs arranged in sequential order:
:$\tuple {0, 0}, \tuple {1, 2}, \tuple {3, 5}, \tuple {4, 7}, \tuple {6, 10}, \tuple {8, 13}, \ldots$
Apart from the first Wythoff pair $\tuple {0, 0}$, every positive integer appears in this sequence exactly once.

Proof:
By definition, the $n$th Wythoff pair is $\tuple {\floor {n \phi}, \floor {n \phi^2} }$.
Thus the coordinates of the sequence of Wythoff pairs are the terms of the lower and upper Wythoff sequences.
By definition:
:the lower Wythoff sequence is the Beatty sequence $\BB_x$ on the golden section $\phi$.
:the upper Wythoff sequence is the complementary Beatty sequence on the golden section $\phi$.
Also by definition, the complementary Beatty sequence on $x$ is the integer sequence formed by the integers which are missing from $\BB_x$.
Hence the result.
{{qed}}
Theorem: Zeta of 2 as Product of Fractions with Prime Numerators

{{begin-eqn}}
{{eqn | l = \map \zeta 2
      | r = \prod_p \paren {\frac p {p - 1} } \paren {\frac p {p + 1} }
      | c = 
}}
{{eqn | r = \dfrac 2 1 \times \dfrac 2 3 \times \dfrac 3 2 \times \dfrac 3 4 \times \dfrac 5 4 \times \dfrac 5 6 \times \dfrac 7 6 \times \dfrac 7 8 \times \dfrac {11} {10} \times \dfrac {11} {12} \times \dfrac {13} {12} \times \dfrac {13} {14} \times \cdots
      | c = 
}}
{{end-eqn}}
where:
:$\zeta$ denotes the Riemann zeta function
:$\ds \prod_p$ denotes the continued product over all prime numbers.

Proof:
From Sum of Reciprocals of Powers as Euler Product:
:$\ds \map \zeta z = \prod_p \frac 1 {1 - p^{-z} }$
where $p$ ranges over the prime numbers.
Thus:
{{begin-eqn}}
{{eqn | l = \map \zeta 2
      | r = \prod_p \frac 1 {1 - p^{-2} }
      | c = 
}}
{{eqn | r = \prod_p \frac {p^2} {p^2 - 1}
      | c = multiplying top and bottom by $p^2$
}}
{{eqn | r = \prod_p \frac {p^2} {\paren {p - 1} \paren {p + 1} }
      | c = Difference of Two Squares
}}
{{eqn | r = \prod_p \paren {\frac p {p - 1} } \paren {\frac p {p + 1} }
      | c = 
}}
{{end-eqn}}
which is the result required.
{{qed}}
Theorem: Top in Filter

Let $\struct {S, \preceq}$ be a bounded above ordered set.
Let $F$ be a filter on $S$.
Then $\top \in F$
where $\top$ denotes the greatest element of $S$.

Proof:
By definition of filter in ordered set:
:$F$ is non-empty and upper.
By definition of non-empty set:
:$\exists x: x \in F$
By definition of greatest element:
:$x \preceq \top$
Thus by definition of upper section:
:$\top \in F$
{{qed}}
Theorem: Finite Infima Set and Upper Closure is Filter

Let $P = \struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $X$ be a non-empty subset of $S$.
Then
:$\map {\operatorname {fininfs} } X^\succeq$ is filter in $P$.
where
:$\map {\operatorname {fininfs} } X$ denotes the finite infima set of $X$,
:$X^\succeq$ denotes the upper closure of $X$.

Proof:
By Finite Infima Set and Upper Closure is Smallest Filter:
:$X \subseteq \map {\operatorname {fininfs} } X^\succeq$
By definition of non-empty set:
:$\map {\operatorname {fininfs} } X^\succeq$ is a non-empty set.
We will prove that
:$\map {\operatorname {fininfs} } X$ is filtered.
Let $x, y \in \map {\operatorname {fininfs} } X$
By definition of finite infima set:
:$\exists A \in \map {\mathit {Fin} } X: x = \inf A \land A$ admits an infimum
and
:$\exists B \in \map {\mathit {Fin} } X: y = \inf B \land B$ admits an infimum
where $\map {\mathit {Fin} } X$ denotes the set of all finite subsets of $X$.
Define $C = A \cup B$.
By Union of Subsets is Subset:
:$C \subseteq X$
By Union of Finite Sets is Finite:
:$C$ is finite.
Then
:$C \in \map {\mathit {Fin} } X$
By Existence of Non-Empty Finite Infima in Meet Semilattice:
:$C \ne \O \implies C$ admits an infimum.
By Union is Empty iff Sets are Empty:
:$C = \O \implies A = \O$
So
:$C$ admits an infimum.
By definition of finite infima set:
:$\inf C \in \map {\operatorname {fininfs} } X$
By Set is Subset of Union:
:$A \subseteq C$ and $B \subseteq C$
Thus by Infimum of Subset:
:$\inf C \preceq x$ and $\inf C \preceq y$
Hence $\map {\operatorname {fininfs} } X$ is filtered.
{{qed|lemma}}
By Filtered iff Upper Closure Filtered:
:$\map {\operatorname {fininfs} } X^\succeq$ is filtered.
By Upper Closure is Upper Section:
:$\map {\operatorname {fininfs} } X^\succeq$ is upper.
Hence $\map {\operatorname {fininfs} } X^\succeq$ is filter in $P$.
{{qed}}
Theorem: Divisibility by 2

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

Proof:
Let $N$ be divisible by $2$.
Then:
{{begin-eqn}}
{{eqn | l = N
      | o = \equiv
      | r = 0 \pmod 2
}}
{{eqn | ll= \leadstoandfrom
      | l = \sum_{k \mathop = 0}^n a_k 10^k
      | o = \equiv
      | r = 0 \pmod 2
}}
{{eqn | ll= \leadstoandfrom
      | l = a_0 + 10 \sum_{k \mathop = 1}^n a_k 10^{k - 1}
      | o = \equiv
      | r = 0 \pmod 2
}}
{{eqn | ll= \leadstoandfrom
      | l = a_0
      | o = \equiv
      | r = 0 \pmod 2
      | c = as $10 \equiv 0 \pmod 2$
}}
{{end-eqn}}
{{qed}}
Theorem: Divisibility by Power of 2

Let $r \in \Z_{\ge 1}$ be a strictly positive integer.
An integer $N$ expressed in decimal notation is divisible by $2^r$ {{iff}} the last $r$ digits of $N$ form an integer divisible by $2^r$.
That is:
:$N = [a_n \ldots a_2 a_1 a_0]_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $2^r$ 
{{iff}}:
:$a_0 + a_1 10 + a_2 10^2 + \cdots + a_r 10^r$ is divisible by $2^r$.

Proof:
First note that:
:$10^r = 2^r 5^r$
and so:
:$2^r \divides 10^r$
where $\divides$ denotes divisibility.
Thus:
:$\forall s \in \Z: s \ge r: 2^r \divides 10^s$
but:
:$\forall s \in \Z: s < r: 2^r \nmid 10^s$
Thus let $N$ be divisible by $2^r$.
Then:
{{begin-eqn}}
{{eqn | l = N
      | o = \equiv
      | r = 0 \pmod {2^r}
}}
{{eqn | ll= \leadstoandfrom
      | l = \sum_{k \mathop = 0}^n a_k 10^k
      | o = \equiv
      | r = 0 \pmod {2^r}
}}
{{eqn | ll= \leadstoandfrom
      | l = \sum_{k \mathop = 0}^r a_k 10^r + \sum_{k \mathop = r + 1}^n a_k 10^k
      | o = \equiv
      | r = 0 \pmod {2^r}
}}
{{eqn | ll= \leadstoandfrom
      | l = \sum_{k \mathop = 0}^r a_k 10^r + 10^r \sum_{k \mathop = r + 1}^n a_k 10^{k - r}
      | o = \equiv
      | r = 0 \pmod {2^r}
}}
{{eqn | ll= \leadstoandfrom
      | l = \sum_{k \mathop = 0}^r a_k 10^r
      | o = \equiv
      | r = 0 \pmod {2^r}
      | c = as $10^r \equiv 0 \pmod {2^r}$
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Positive Integer is Sum of Consecutive Positive Integers iff not Power of 2

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then $n$ can be expressed as the sum of $2$ or more consecutive (strictly) positive integers {{iff}} $n$ is not a power of $2$.

Proof:
=== Necessary Condition ===
Let $a$ be the smallest of $m$ consecutive (strictly) positive integers, where $m \ge 2$.
From Sum of Arithmetic Sequence, their sum is $\dfrac {m \paren {2 a + m - 1} } 2$. 
{{AimForCont}} $\dfrac {m \paren {2 a + m - 1} } 2$ is a power of $2$.
Then $m$ and $2 a + m - 1$ must also be powers of $2$.
Since $m \ge 2$, $m$ must be even.
Then $2 a + m - 1$ is odd.
Together with $2 a + m - 1$ being a power of $2$, we have $2 a + m - 1 = 1$.
However, $2 a + m - 1 > 2 \times 0 + 2 - 1 = 1$.
This is a contradiction.
Therefore any sum of $2$ or more consecutive integers cannot be a power of $2$.
{{qed|lemma}}
Theorem: Equivalence of Definitions of Deficient Number

The following definitions of a deficient number are equivalent:

Proof:
By definition of abundance:
:$\map A n = \map {\sigma_1} n - 2 n$
By definition of divisor sum function:
:$\map {\sigma_1} n$ is the sum of all the divisors of $n$.
Thus $\map {\sigma_1} n - n$ is the aliquot sum of $n$.
Hence the result.
{{qed}}
Category:Deficient Numbers
6yqynr752q7u4eoehiw1j4hp7sgqe6f
Theorem: Equivalence of Definitions of Abundant Number

The following definitions of a abundant number are equivalent:

Proof:
By definition of abundance:
:$\map A n = \map {\sigma_1} n - 2 n$
By definition of divisor sum function:
:$\map {\sigma_1} n$ is the sum of all the divisors of $n$.
Thus $\map {\sigma_1} n - n$ is the aliquot sum of $n$.
The result follows.
{{qed}}
Category:Abundant Numbers
c39jf7s0hwcrkg4tdw3re6eyrwnctgd
Theorem: Power of Prime is Deficient

Let $n \in \Z_{>0}$ be a power of a prime number $p$:
:$n = p^k$
for some $k \in \Z_{>0}$.
Then $n$ is deficient.

Proof:
From Divisor Sum of Power of Prime:
:$\map {\sigma_1} n = \dfrac {p^{k + 1} - 1} {p - 1}$
Thus:
{{begin-eqn}}
{{eqn | l = \map A n
      | r = \map {\sigma_1} n - 2 n
      | c = {{Defof|Abundance}}
}}
{{eqn | r = \dfrac {p^{k + 1} - 1} {p - 1} - 2 p^k
      | c = 
}}
{{eqn | r = \dfrac {p^{k + 1} - 1 - \paren {2 p^{k + 1} - 2 p^k} } {p - 1}
      | c = 
}}
{{eqn | r = \dfrac {2 p^k - p^{k + 1} - 1} {p - 1}
      | c = 
}}
{{eqn | r = \dfrac {p^k \paren {2 - p} - 1} {p - 1}
      | c = 
}}
{{eqn | r = \dfrac {p^k - 1} {p - 1} - p^k
      | c = 
}}
{{eqn | r = \dfrac {p^k} {p - 1} - p^k - \dfrac 1 {p - 1}
      | c = 
}}
{{end-eqn}}
If $p = 2$ then $\dfrac {p^k} {p - 1} - p^k = 0$ and so:
:$\dfrac {p^k} {p - 1} - p^k - \dfrac 1 {p - 1} = - \dfrac 1 {p - 1} < 0$
If $p > 2$ then $\dfrac {p^k} {p - 1} < p^k$ and so:
:$\dfrac {p^k} {p - 1} - p^k - \dfrac 1 {p - 1} < 0$
Thus in all cases:
:$\map A n < 0$
and so $n = p^k$ is deficient by definition.
{{qed}}
Theorem: Power of 2 is Almost Perfect

Let $n \in \Z_{>0}$ be a power of $2$:
:$n = 2^k$
for some $k \in \Z_{>0}$.
Then $n$ is almost perfect.

Proof:
{{begin-eqn}}
{{eqn | l = \map A n
      | r = \dfrac {2^k} {2 - 1} - 2^k - \dfrac 1 {2 - 1}
      | c = Power of Prime is Deficient
}}
{{eqn | r = 2^k - 2^k - 1
      | c = 
}}
{{eqn | r = - 1
      | c = 
}}
{{end-eqn}}
and so $n = p^k$ is almost perfect by definition.
{{qed}}
Theorem: Set is Subset of Finite Infima Set

Let $\struct {S, \preceq}$ be an ordered set.
Let $X$ be a subset of $S$.
Then $X \subseteq \map {\operatorname{fininfs} } X$
where $\map {\operatorname{fininfs} } X$ denotes the finite infima set of $X$.

Proof:
Let $x \in X$.
By Infimum of Singleton:
:$\set x$ admits an infimum and $\inf \set x = x$
By definitions of subset and singleton:
:$\set x \subseteq X$
By Singleton is Finite:
:$\set x$ is a finite set.
Thus by definition of finite infima set:
:$x \in \map {\operatorname{fininfs} } X$
{{qed}}
Theorem: Set is Subset of Upper Closure

Let $\struct {S, \preceq}$ be an ordered set.
Let $X$ be a subset of $S$.
Then $X \subseteq X^\succeq$
where $X^\succeq$ denotes the upper closure of $X$.

Proof:
Let $x \in X$.
By definition of reflexivity:
:$x \preceq x$
Thus by definition of upper closure:
:$x \in X^\succeq$
{{qed}}
Theorem: Tamref's Last Theorem

The Diophantine equation:
:$n^x + n^y = n^z$
has exactly one form of solutions in integers:
:$2^x + 2^x = 2^{x + 1}$
for all $x \in \Z$.

Proof:
Since $n^z = n^x + n^y > n^x$ and $n^y$, $z > x,y$.
{{WLOG}} assume that $x \le y < z$.
{{begin-eqn}}
{{eqn | l = n^x + n^y 
      | r = n^z
}}
{{eqn | ll = \leadsto
      | l = 1 + n^{y - x}
      | r = n^{z - x}
      | c = Divide both sides by $n^x$
}}
{{eqn | ll = \leadsto
      | l = 1
      | r = n^{y - x} \paren {n^{z - y} - 1}
}}
{{end-eqn}}
Since both $n^{y - x}$ and $n^{z - y} - 1$ are positive integers, both are equal to $1$.
This gives:
:$y = x$ and $n^{z - y} = 2$
which gives the integer solution:
:$n = 2$, $z - y = 1$
Thus the solutions are:
:$\tuple {n, x, y, z} = \tuple {2, x, x, x + 1}, x \in \Z$
{{qed}}
Theorem: 3^x + 4^y equals 5^z has Unique Solution

The Diophantine equation:
:$3^x + 4^y = 5^z$
has exactly one solution in (strictly) positive integers:
:$3^2 + 4^2 = 5^2$

Proof:
Rewriting our Diophantine equation modulo $4$ we have:
:$\paren {-1}^x + 0^y \equiv 1^z \pmod 4$
Therefore $x$ is even.
Rewriting our Diophantine equation modulo 3 we have:
:$0^x + 1^y \equiv \paren {-1}^z \pmod 3$
Therefore $z$ is even.
Since $x$ and $z$ must both be even, we will rewrite $x$ as $2 r$, $z$ as $2 s$ and notice that $4$ is $2^2$.
We now have:
{{begin-eqn}}
{{eqn	| l = 3^{2 r} + 2^{2 y}
	    | r = 5^{2 s}
        | c = 
}}
{{eqn	| l = 5^{2 s} - 3^{2 r}
	    | r = 2^{2 y}
        | c = rewriting the equation
}}
{{eqn	| l = \paren {5^s + 3^r} \paren {5^s - 3^r}
	    | r = 2^{2 y}
        | c = factoring the {{LHS}}: Difference of Two Squares
}}
{{end-eqn}}
We can see that the {{RHS}} is a product consisting entirely of instances of $2$.
Therefore:
:$\paren {5^s + 3^r} = 2^u$
and:
:$\paren {5^s - 3^r} = 2^v$
where $u > v$ and $u + v = 2 y$.
Solving for $5^s$, we add the two equations and get:
{{begin-eqn}}
{{eqn	| l = \paren {5^s + 3^r} + \paren {5^s - 3^r}
	    | r = 2^u + 2^v
        | c = 
}}
{{eqn	| l = 5^s
	    | r = 2^{u - 1} + 2^{v - 1}
        | c = Dividing both sides by $2$
}}
{{eqn	| r = 2^{v - 1} \paren {2^{u - v} + 1}
        | c = Factoring out $2^{v - 1}$
}}
{{end-eqn}}
Solving for $3^r$, we subtract the two equations and get:
{{begin-eqn}}
{{eqn	| l = \paren {5^s + 3^r} - \paren {5^s - 3^r}
	    | r = 2^u - 2^v
        | c = 
}}
{{eqn	| l = 3^r
	    | r = 2^{u - 1} - 2^{v - 1}
        | c = dividing both sides by $2$
}}
{{eqn	| r = 2^{v - 1} \paren {2^{u - v} - 1}
        | c = factoring out $2^{v - 1}$
}}
{{end-eqn}}
Since both $5^s$ and $3^r$ are odd, $v$ must be equal to $1$.
Let $t = u - v$.
We now have:
:$5^s = \paren {2^t + 1}$
:$3^r = \paren {2^t - 1}$
Let us now consider the second equation Modulo 3:
:$0 \equiv \paren{-1}^t - 1 \pmod 3$
Therefore t must be even. 
We will rewrite the second equation above with $t$ as $2 w$:
{{begin-eqn}}
{{eqn	| l = 3^r
	    | r = \paren {2^{2w} - 1}
        | c = 
}}
{{eqn	| r = \paren {2^w + 1} \paren {2^w - 1}
        | c = Difference of Two Squares
}}
{{end-eqn}}
We can see that the {{LHS}} is a product consisting entirely of instances of $3$.
Therefore:
:$ 3^m = \paren {2^w + 1}$
and:
:$ 3^n = \paren {2^w - 1}$
where $m > n$ and $m + n = r$.
Subtracting the $2$ equations above we get:
{{begin-eqn}}
{{eqn	| l = 3^m - 3^n
	    | r = \paren{2^w + 1} - \paren{2^w - 1}
        | c = 
}}
{{eqn	| r = 2
        | c = 
}}
{{end-eqn}}
Therefore $m = 1$ and $n = 0$.
From above:
:$3^m = \paren {2^w + 1}$
therefore:
:$w = 1$
Recall that $t = 2w$.
Therefore:
:$t = 2$
and:
:$5^s = \paren {2^t + 1}$
:$3^r = \paren {2^t - 1}$
Therefore $r = 1$ and $s = 1$.
Finally recall that $x = 2 r$ and $z = 2 s$.
Therefore we arrive at the only possible solution in (strictly) positive integers:
:$x = 2$, $y = 2$ and $z = 2$
{{qed}}
Theorem: 5^x + 12^y equals 13^z has Unique Solution

The Diophantine equation:
:$5^x + 12^y = 13^z$
has exactly one solution in (strictly) positive integers:
:$5^2 + 12^2 = 13^2$

Proof:
Let $\nu_p: \Q \to \Z \cup \set {+\infty}$ be the $p$-adic valuation on $\Q$.
Suppose $x,y,z$ are positive integers with $5^x+12^y=13^z$.
Reducing modulo $3$ gives us $(-1)^x+0^y\equiv1^z$.
Thus $x$ is even.
Say $x=2a$ for postive integer $a$, so we have $(1):\quad 25^a+12^y=13^z$.
Now reducing modulo $5$ gives us $0^x+2^y\equiv3^z$.
Multiply both sides by $2^z$ to get $2^{y+z}\equiv1^z\equiv1$.
The multiplicative order of $2$ modulo $5$ is $4$, so:
$(2):\quad y+z\equiv0\mod4$.
Next, reduce modulo $13$ to get $(-1)^a+(-1)^y\equiv0$.
Thus:
$(3):\quad a\not\equiv y\mod2$.
We now prove $y\neq1$.
If $y=1$, then $(2)$ gives us $z\equiv3\mod4$, say $z=4c+3$.
Also, $(3)$ gives us $a\equiv0\mod2$, say $a=2r$.
Our equation thus becomes $(25^2)^a+12\equiv13^3\cdot(13^4)^z$.
Taking this modulo $16$, we have $1^a+12\equiv5\cdot1^z$, a contradiction.
Since $y\neq1$ is a nonnegative integer, $y\geq2$.
Thus, taking $(1)$ modulo $8$:
$1^a+0\equiv5^z$,
so $z\equiv0\mod2$.
Using this in $(2)$, we have $y\equiv0\mod2$.
Say $y=2b$ and $z=2c$.
Going back to the original equation, we now have $5^{2a}+12^{2b}=13^{2c}$.
Rearranging:
$2^{4b}3^{2b}=(13^c-5^a)(13^c+5^a)$.
Let $L=13^c-5^a$ and $R=13^c+5^a$.
Note that the only prime factors to divide $R$ are $2$ and $3$ and that $R\geq18$ since $a,c\geq1$..
Modulo $4$, we find $R\equiv2$, so $\nu_2(R)=1$.
If $\nu_3(R)=0$, then $R=2$, so we must have $\nu_3(R)\geq1$.
If $\nu_3(L)\geq1$, then $1\leq\nu_3(R+L)=\nu_3(2\cdot13^c)=0$, a contradiction.
Thus $\nu_3(L)=0$.
Since $\nu_3(2^{4b}3^{2b})=2b$, we have $\nu_3(R)=2b$.
Thus $R=2^1\cdot3^{2b}$, so $L=2^{4b-1}$.
We have $0<2\cdot5^a=R-L=2\cdot3^{2b}-2^{4b-1}$, solving for $b$ gives:
$b<\log_{16/9}(4)<3$,
so $b=1$ or $b=2$.
If $b=1$, then $R=13^c+5^a=18$, so $a=c=1$.
This corresponds with the solution $x=y=z=2$.
If $b=2$, then $R=13^c+5^a=162$.
This has no integer solutions.
Thus the only solution is $x=y=z=2$.
{{qed}}
Theorem: Upper Closure of Subset is Subset of Upper Closure

Let $\left({S, \preceq}\right)$ be an ordered set.
Let $X, Y$ be subsets of $S$.
Then
:$X \subseteq Y \implies X^\succeq \subseteq Y^\succeq$
where $X^\succeq$ denotes the upper closure of $X$.

Proof:
Let $X \subseteq Y$.
Let $x \in X^\succeq$.
By definition of upper closure of subset:
:$\exists y \in X: y \preceq x$
By definition of subset:
:$y \in Y$
Thus by definition of upper closure of subset:
:$x \in Y^\succeq$
{{qed}}
Theorem: Euler's Number as Limit of n over nth Root of n Factorial

:$\ds e = \lim_{n \mathop \to \infty} \dfrac n {\sqrt [n] {n!} }$
where:
:$e$ denotes Euler's number
:$n!$ denotes $n$ factorial.

Proof:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} {n!} {n^n \sqrt n e^{-n} }
      | r = \sqrt {2 \pi}
      | c = Lemma for Stirling's Formula
}}
{{eqn | ll= \leadsto
      | l = e
      | r = \lim_{n \mathop \to \infty} \dfrac {n \paren {2 \pi n}^{1 / 2 n} } {\sqrt [n] {n!} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = e
      | r = \lim_{n \mathop \to \infty} \dfrac {n \sqrt {\paren {2 \pi n}^{1 / n} } } {\sqrt [n] {n!} }
      | c = 
}}
{{end-eqn}}
From Limit of Root of Positive Real Number:
:$\ds \lim_{n \mathop \to \infty} \paren {2 \pi}^{1 / 2 n} = 1$
and from Limit of Integer to Reciprocal Power:
:$n^{1 / n} = 1$
Hence the result.
{{qed}}
Theorem: Continued Fraction Expansion of Euler's Number/Convergents

The convergents of the continued fraction expansion to Euler's number $e$ are:
:$2, 3, \dfrac 8 3, \dfrac {11} 4, \dfrac {19} 7, \dfrac {87} {32}, \dfrac {106} {39}, \dfrac {193} {71}, \dfrac {1264} {465}, \dfrac {1457} {536}, \dfrac {2721} {1001}, \ldots$
{{OEIS-Numerators|A007676}}
{{OEIS-Denominators|A007677}}
These best rational approximations are accurate to $0, 0, 1, 1, 2, 3, 3, 4, 5, 5, \ldots$ decimals.
{{OEIS|A114539}}
The fraction $\dfrac {878} {323}$ is exceptionally easy to remember:
:$\dfrac {878} {323} = 2 \cdotp 71826 \, 625 \ldots$
although this does not occur in the above continued fraction expansion.

Proof:
{{ProofWanted|Calculation needed}}
Theorem: Sine of X over X as Infinite Product

Let $z \in \C$ be a non-zero complex number.
Then:
:$\ds \frac {\sin z} z = \cos \frac z 2 \cos \frac z 4 \cos \frac z 8 \cdots = \prod_{i \mathop = 1}^{\infty} \cos \frac z {2^i}$
where $\sin$ denotes the sine function and $\cos$ denotes the cosine function.

Proof:
First we prove that:
:$\ds \frac {\sin z} z = \paren {\frac {2^n} z} \sin \frac z {2^n} \prod_{i \mathop = 1}^n \cos \frac z {2^i}$
for $n \in \N$.
The proof proceeds by induction.
For all $n \in \N$, let $\map P n$ be the proposition:
:$\ds \frac {\sin z} z = \paren {\frac {2^n} z} \sin \frac z {2^n} \prod_{i \mathop = 1}^n \cos \frac z {2^i}$
Theorem: Maximum of Three Mutually Perpendicular Lines in Ordinary Space

In ordinary space, there can be no more than $3$ straight lines which are pairwise perpendicular.
Thus, in a configuration of $4$ straight lines in space, at least one pair will not be perpendicular to each other.

Proof:
By assumption and popular belief, ordinary space (on the measurable local level) is an instance of a $3$-dimensional Euclidean space.
Hence the results of {{ElementsLink}} can be applied.
Let there be $4$ straight lines $a$, $b$, $c$ and $d$ in space.
Let $a$ be perpendicular to each of $b$, $c$ and $d$.
From Three Intersecting Lines Perpendicular to Another Line are in One Plane, $b$, $c$ and $d$ are all in the same plane.
Let $b$ be perpendicular to $c$ and $d$.
Either $c$ and $d$ are the same line or not.
If $c = d$ then $c$ is not perpendicular to $d$.
Otherwise, from Equal Corresponding Angles implies Parallel Lines, $c$ is parallel to $d$.
In either case $c$ is not perpendicular to $d$.
Hence the result.
{{qed}}