avewright/proofwiki-math huggingface dataset:
Theorem: Mittag-Leffler Expansion for Cotangent Function

:$\ds \pi \cot \pi z = \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \frac z {z^2 - n^2}$
where:
:$z \in \C$ is not an integer
:$\cot$ is the cotangent function.

Proof:
From Euler's Reflection Formula:
:$\forall x \notin \Z: \map \Gamma x \map \Gamma {1 - x} = \dfrac \pi {\map \sin {\pi x} }$
Taking the logarithm of both sides:
{{begin-eqn}}
{{eqn | l = \map \ln {\map {\Gamma} x } + \map \ln {\map {\Gamma} {1 - x} }
      | r = \map \ln {\pi } - \map \ln {\map \sin {\pi x} }
      | c = Sum of Logarithms/Natural Logarithm and Difference of Logarithms
}}
{{end-eqn}}
Taking the derivative of both sides:
{{begin-eqn}}
{{eqn | l = \frac {\map {\Gamma'} x} {\map \Gamma x} - \frac {\map {\Gamma'} {1 - x} } {\map \Gamma {1 - x} }
      | r = -\frac 1 {\map \sin {\pi x} } \map \cos {\pi x} \pi
      | c = Derivative of Composite Function, Derivative of Natural Logarithm Function and Derivative of Sine Function
}}
{{eqn | r = -\pi \map \cot {\pi x}
      | c = {{defof|cotangent}}
}}
{{end-eqn}}
We now have:
{{begin-eqn}}
{{eqn | l = \pi \map \cot {\pi x}
      | r =  \frac {\map {\Gamma'} {1 - x} } {\map \Gamma {1 - x} } - \frac {\map {\Gamma'} x} {\map \Gamma x}
      | c = multiplying both sides by $-1$
}}
{{eqn | r = \paren {-\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {n - x } } } - \paren {-\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {x + n - 1} } }
      | c = Reciprocal times Derivative of Gamma Function
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \paren {\frac 1 {x + n - 1} - \frac 1 {n - x } }
      | c = Linear Combination of Convergent Series
}}
{{eqn | r = \frac 1 x + \sum_{n \mathop = 1}^\infty \paren {\frac 1 {x + n } - \frac 1 {n - x } }
      | c = reindexing the sum
}}
{{eqn | r = \frac 1 x + \sum_{n \mathop = 1}^\infty \paren {\frac 1 {x + n } + \frac 1 {x - n } }
      | c = 
}}
{{eqn | r = \dfrac 1 x + 2 x \sum_{n \mathop = 1}^\infty \frac 1 {x^2 - n^2}
      | c = Difference of Two Squares
}}
{{end-eqn}}
{{qed}}
Theorem: Laurent Series Expansion for Cotangent Function

{{begin-eqn}}
{{eqn | l = \pi \cot \pi z
      | r = \frac 1 z - 2 \sum_{n \mathop = 1}^\infty \map \zeta {2 n} z^{2 n - 1}
}}
{{eqn | r = \frac 1 z - 2 \paren {\dfrac {\pi^2 } 6 z + \dfrac {\pi^4 } {90 } z^3 + \dfrac {\pi^6 } {945 } z^5 + \cdots}
}}
{{eqn | r = \frac 1 z - \dfrac {\pi^2 } 3 z - \dfrac {\pi^4 } {45 } z^3 - \dfrac {2 \pi^6 } {945 } z^5 - \cdots
}}
{{end-eqn}}
where:
:$z \in \C$ such that $\cmod z < 1$
:$\zeta$ is the Riemann Zeta function.

Proof:
From Mittag-Leffler Expansion for Cotangent Function:
{{begin-eqn}}
{{eqn | l = \pi \cot \pi z
      | r = \frac 1 z + 2 \sum_{k \mathop = 1}^\infty \frac z {z^2 - k^2}
}}
{{end-eqn}}
Factoring $-\dfrac 1 {k^2}$: 
{{begin-eqn}}
{{eqn | l = \pi \cot \pi z
      | r = \frac 1 z + 2 \sum_{k \mathop = 1}^\infty \paren {\frac z {k^2} } \paren {\frac 1 {\frac {z^2} {k^2} - 1} }
}}
{{eqn | r = \frac 1 z - 2 \sum_{k \mathop = 1}^\infty \paren {\frac z {k^2} } \paren {\frac 1 {1 - \frac {z^2} {k^2} } }
}}
{{end-eqn}}
Taking $\cmod z < 1$, and noting that $k \ge 1$, we have, by Sum of Infinite Geometric Sequence:
:$\ds \pi \cot \pi z = \frac 1 z - 2 \sum_{k \mathop = 1}^\infty \frac z {k^2} \cdot \sum_{n \mathop = 1}^\infty \paren {\frac {z^2} {k^2} }^{n - 1}$
from which:
{{begin-eqn}}
{{eqn | l = \pi \cot \pi z
      | r = \frac 1 z - 2 \sum_{k \mathop = 1}^\infty \sum_{n \mathop = 1}^\infty \frac {z^{2 n - 2} \cdot z} {k^{2 n - 2} \cdot k^2}
}}
{{eqn | r = \frac 1 z - 2 \sum_{k \mathop = 1}^\infty \sum_{n \mathop = 1}^\infty \frac 1 {k^{2 n} } \cdot z^{2 n - 1}
}}
{{eqn | r = \frac 1 z - 2 \sum_{n \mathop = 1}^\infty \sum_{k \mathop = 1}^\infty \frac 1 {k^{2 n} } \cdot z^{2 n - 1}
      | c = Product of Absolutely Convergent Series
}}
{{eqn | r = \frac 1 z - 2 \sum_{n \mathop = 1}^\infty \map \zeta {2 n} z^{2 n - 1}
      | c = {{Defof|Riemann Zeta Function}}
}}
{{end-eqn}}
{{qed}}
Theorem: Equivalence of Definitions of Removable Discontinuity of Real Function

Let $A \subseteq \R$ be a subset of the real numbers.
Let $f: A \to \R$ be a real function.
Let $f$ be discontinuous at $a \in A$.
{{TFAE|def = Removable Discontinuity of Real Function|view = removable discontinuity}}
=== Definition 1  ===
{{Definition:Removable Discontinuity of Real Function/Definition 1}}
=== Definition 2  ===
{{Definition:Removable Discontinuity of Real Function/Definition 2}}

Proof:
=== Lemma ===
{{:Equivalence of Definitions of Removable Discontinuity of Real Function/Lemma}}{{qed|lemma}}
Let $A \subseteq \R$ be a subset of the real numbers.
Let $f: A \to \R$ be a real function.
Let $f$ be discontinuous at $a \in A$.
For any $b \in \R$, define the function $f_b$ by:
:$\map {f_b} x = \begin {cases} \map f x &: x \ne a \\ b &: x = a \end {cases}$
Then $\map {f_b} x = \map f x$ for every $x \ne a$.
Theorem: Equivalence of Definitions of Polynomial Function on Subset of Ring

Let $R$ be a commutative ring with unity.
Let $S \subset R$ be a subset.
{{TFAE|def = Polynomial Function/Ring |view = polynomial function}}

Proof:
=== 1 implies 2 ===
Let $\map P X \in R \sqbrk X$ be the polynomial:
:$P = \ds \sum_{k \mathop = 0}^n a_k \cdot X^k$
where $\sum$ denotes indexed summation.
We show that $\map P \iota = f$.
{{finish}}
Theorem: Monomials form Basis of Polynomial Ring/One Variable

Let $R$ be a commutative ring with unity.
Let $R \sqbrk X$ be a polynomial ring over $R$ in the variable $X$.
Then the monomials of $R \sqbrk X$ are a basis of $R \sqbrk X$ as a module over $R$.

Proof:
Follows from:
* Polynomial is Linear Combination of Monomials
* Monomials of Polynomial Ring are Linearly Independent
{{qed}}
Category:Polynomial Rings
Category:Monomials
ibj1ycvcvta6451o5b2hkeptfdnrhkb
Theorem: Equivalence of Definitions of Polynomial in Ring Element

Let $R$ be a commutative ring.
Let $S$ be a subring with unity of $R$.
Let $x\in R$.
{{TFAE|def = Polynomial in Ring Element}}

Proof:
{{proof wanted}}
Category:Polynomial Theory
imnf88w7v83rtn8p2kas888887ipv94
Theorem: Union of Blocks is Set of Points

Let $\struct {X, \BB}$ be a pairwise balanced design.
That is, let $\struct {X, \BB}$ be a design, with $\size X \ge 2$, and the number of occurrences of each pair of distinct points in $\BB$ be $\lambda$ for some $\lambda > 0$ constant.
Then the set union of all the subset elements in $\BB$ is precisely $X$.

Proof:
Let $X = \set {x_1, x_2, \ldots, x_v}$.
Let $\BB = \multiset {y_1, y_2,\ldots, y_b}$, where the notation denotes a multiset.
Let $Y = \ds \bigcup_{i \mathop = 1}^b y_i$.
We shall show that $Y \subseteq X$ and $X \subseteq Y$.
Theorem: Equality of Monomials of Polynomial Ring in One Variable

Let $R$ be a commutative ring with unity.
Let $R \sqbrk X$ be a polynomial ring in one variable $X$ over $R$.
Let $k, l \in \N$ be distinct natural numbers.
Then the mononomials $X^k$ and $X^l$ are distinct, where $X^k$ denotes the $k$th power of $X$.

Proof:
By:
* Uniqueness of Polynomial Ring in One Variable
* Homomorphism Preserves Indexed Products
we may assume $R \sqbrk X$ is the ring of sequences of finite support over $R$, and $X$ is the sequence $\sequence {0, 1, 0, 0 \ldots}$.
One verifies that, for $k \ge 0$, $X^k$ is the sequence with $\map {X^k} l = \delta_{k l}$, where $\delta$ is the Kronecker delta.
{{theoremWanted|verify this somewhere else}}
Thus $X^k \ne X^l$ if $k \ne l$.
{{qed}}
Theorem: Ring Isomorphic to Polynomial Ring is Polynomial Ring/One Variable

Let $R$ be a commutative ring with unity.
Let $R \sqbrk X$ be a polynomial ring in one variable $X$ over $R$.
Let $\iota : R \to R \sqbrk X$ denote the canonical embedding.
Let $S$ be a commutative ring with unity and $f: R \sqbrk X \to S$ be a ring isomorphism.
Then $\struct {S, f \circ \iota, \map f X}$ is a polynomial ring in one variable $\map f X$ over $R$.

Proof:
{{proof wanted}}
Category:Polynomial Rings
4awnnfe99b8800jwaxzarzp3xfiq024
Theorem: Mittag-Leffler Expansion for Cosecant Function

:$\ds \pi \cosec \pi z = \frac 1 z + 2\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac z {z^2 - n^2}$

Proof:
{{begin-eqn}}
{{eqn | l = \pi \cosec \pi z
      | r = \map \Gamma z \map \Gamma {1 - z}
      | c = Euler's Reflection Formula
}}
{{eqn | r = \map \Beta {z, 1 - z}
      | c = {{Defof|Beta Function|index = 3}}
}}
{{eqn | r = \int_{\mathop \to 0}^{\mathop \to 1} t^{z - 1} \paren {1 - t}^{-z} \rd t
      | c = {{Defof|Beta Function|index = 1}}
}}
{{eqn | r = \int_{\mathop \to 0}^{\mathop \to \infty} \paren {\dfrac x {1 + x} }^{z - 1} \paren {1 - \dfrac x {1 + x} }^{-z} \dfrac {\d x} {\paren {1 + x}^2}
      | c = substituting $x = \dfrac t {1 - t}$
}}
{{eqn | r = \int_{\mathop \to 0}^{\mathop \to \infty} \dfrac {x^{z - 1} } {1 + x} \rd x
}}
{{eqn | r = \int_{\mathop \to 0}^{\mathop \to 1} \dfrac {x^{z - 1} } {1 + x} \rd x + \int_{\mathop \to 1}^{\mathop \to \infty} \dfrac {x^{z - 1} } {1 + x} \rd x
      | c = Sum of Integrals on Adjacent Intervals for Integrable Functions
}}
{{eqn | r = \int_{\mathop \to 0}^{\mathop \to 1} \dfrac {u^{z - 1} } {1 + u} \rd u - \int_{\mathop \to 1}^{\mathop \to 0} \dfrac {\paren {1 / u}^{z - 1} } {1 + 1/u} \frac {\rd u} {u^2}
      | c = substituting $u = x$ for the first integral and $u = \dfrac 1 x$ for the second integral
}}
{{eqn | r = \int_{\mathop \to 0}^{\mathop \to 1} \dfrac {u^{z - 1} } {1 + u} \rd u + \int _0^1 \dfrac {u^{-z} } {1 + u} \rd u
      | c = Reversal of Limits of Definite Integral
}}
{{eqn | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^1 u^{n - 1 + z} \rd u + \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^1 u^{n - z} \rd u
      | c = Power Series Expansion for Reciprocal of 1 + x with $x = u$
}}
{{eqn | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac 1 {n + z} + \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac 1 {n - z + 1}
      | c = Integral of Power
}}
{{eqn | r = \dfrac 1 z + \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\dfrac 1 {n + z} - \dfrac 1 {n - z} }
      | c = reindexing and combining convergent series
}}
{{eqn | r = \dfrac 1 z - 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \dfrac z {n^2 - z^2}
      | c = Difference of Two Squares
}}
{{eqn | r = \dfrac 1 z + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \dfrac z {z^2 - n^2}
}}
{{end-eqn}}
{{qed}}
{{Namedfor|Magnus Gustaf Mittag-Leffler|cat = Mittag-Leffler}}
Theorem: Upper Sum Never Smaller than Lower Sum for any Pair of Subdivisions

Let $\closedint a b$ be a closed real interval.
Let $f$ be a bounded real function defined on $\closedint a b$.
Let $P$ and $Q$ be finite subdivisions of $\closedint a b$.
Let $\map L P$ be the lower Darboux sum of $f$ on $\closedint a b$ with respect to $P$.
Let $\map U Q$ be the upper Darboux sum of $f$ on $\closedint a b$ with respect to $Q$.
Then $\map L P \le \map U Q$.

Proof:
Let $P' = P \cup Q$.
We observe:
:$P'$ is either equal to $P$ or finer than $P$
:$P'$ is either equal to $Q$ or finer than $Q$
We find:
:$\map L P \le \map L {P'}$ by Lower Sum of Refinement
:$\map L {P'} \le \map U {P'}$ by Upper Darboux Sum Never Smaller than Lower Darboux Sum
:$\map U {P'} \le \map U Q$ by Upper Sum of Refinement
By combining these inequalities, we conclude:
:$\map L P \le \map U Q$
{{qed}}
Theorem: Set Finite iff Surjection from Initial Segment of Natural Numbers

Let $S$ be a set.
Then $S$ is finite {{iff}} for some $n \in \N$ there exists a surjection $f: \N_{<n} \to S$.
Here, $\N_{<n}$ denotes an initial segment of $\N$.

Proof:
=== Necessary Condition ===
Suppose that $S$ is finite.
By definition, this means there exists a bijection $f: \N_{<n} \to S$.
Then $f$ is a fortiori also the sought surjection.
{{qed|lemma}}
Theorem: Set Finite iff Injection to Initial Segment of Natural Numbers

Let $S$ be a set.
Then $S$ is finite {{iff}} for some $n \in \N$ there exists an injection $f: S \to \N_{< n}$.
Here, $\N_{< n}$ denotes an initial segment of $\N$.

Proof:
=== Necessary Condition ===
Suppose that $S$ is finite.
By definition, this means there exists a bijection $f: S \to \N_{< n}$.
Then $f$ is a fortiori also the sought injection.
{{qed|lemma}}
Theorem: Largest Integer whose Digits taken in Pairs all form Distinct Primes

The largest integer which has the property that every pair of its digits taken together is a distinct prime number is $619 \, 737 \, 131 \, 179$.

Proof:
Let $p$ be such an integer.
Apart from the first digit, each of its digits forms the second digit of a $2$-digit prime number.
Thus, apart from the first digit, $p$ cannot contain $2$, $4$, $5$, $6$, $8$ or $0$.
$0$ cannot of course be the first digit either.
Thus, apart from the first pair of its digits, the $2$-digit prime numbers contained in $p$ can only ever be in the set:
:$S = \left\{ {11, 13, 17, 19, 31, 37, 71, 73, 79, 97}\right\}$
Let it be assumed that the largest such $p$ contains all of the above.
There are $10$ elements of $S$.
Each of the elements of $S$ apart from one shares both digits with one other element of $S$, apart from the final one in $p$.
Thus the elements of $S$ contribute $11$ digits of $p$ in total.
The first digit of the first pair contributes an extra $1$ digit to $p$.
Thus $p$ contains $12$ digits.
Only $1$ element of $S$ begins with $9$, but $2$ of them end with $9$.
Therefore $9$ must be the last digit of $p$.
Only $3$ elements of $S$ end in $1$, but $4$ of them begin with $1$.
Thus the first pair of digits of $p$ must be a prime number which does not appear in $S$ whose $2$nd digit is $1$.
Thus the first $2$ digits of $p$ can only be $41$ or $61$.
Let us construct $p$ by starting from the left and use the greedy algorithm consisting of select the largest digit available.
:$61$ is larger than $41$, so $p$ can be assumed to start with $61$.
Using this algorithm, we select the elements of $S$ in the following order:
:$61$, $19$, $97$ (which is forced), $73$, $37$, $71$
The next element of $S$ cannot be $17$, because that would force $79$ to follow.
But as $79$ is the only remaining element of $S$ ending in $9$, $79$ must be at the end of $p$.
So $71$ must be followed by $11$ or $13$.
Continuing to use the greedy algorithm:
:$61$, $19$, $97$ $73$, $37$, $71$, $13$, $31$ (forced)
For the same reason as above, we cannot then select $17$, as this will be followed by $79$ which cannot be followed.
So we continue with the greedy algorithm:
:$61$, $19$, $97$ $73$, $37$, $71$, $13$, $31$, $11$, $17$, $79$
and $p$ is complete:
:$619 \, 737 \, 131 \, 179$
As a bonus, $p$ is itself a prime number.
{{qed}}
Theorem: Finite Product of Finite Sets is Finite

Let $\sequence {S_n}$ be a sequence of finite sets.
Let $\ds \prod_{k \mathop = 1}^n S_k$ be their Cartesian product.
Then $\ds \prod_{k \mathop = 1}^n S_k$ is also a finite set.

Proof:
{{ProofWanted|Boring induction from Product of Finite Sets is Finite}}
Theorem: Infimum of Upper Sums Never Smaller than Lower Sum

:$\inf_P \map U P \ge \map L S$

Proof:
From Upper Sum Never Smaller than Lower Sum for any Pair of Subdivisions, $\map L S$ is a lower bound for the real set:
:$T = \leftset {\map U P: P}$ is a finite subdivision of $\rightset {\closedint a b}$
Since $\inf_P \map U P$ is the infumum of $T$:
:$\inf_P \map U P \ge \map L S$
Hence the result.
{{qed}}
Category:Real Analysis
on8f8ubn8gqikqzy89y5ahhm0qzot8n
Theorem: Supremum of Lower Sums Never Greater than Upper Sum

:$\sup_P \map L P \le \map U S$

Proof:
From Upper Sum Never Smaller than Lower Sum for any Pair of Subdivisions, $\map U S$ is an upper bound for the real set:
:$T = \leftset {\map L P: P}$ is a finite subdivision of $\rightset {\closedint a b}$
Since $\sup_P \map L P$ is the supremum of $T$:
:$\sup_P \map L P \le \map U S$
Hence the result.
{{qed}}
Category:Real Analysis
ih4lq2hcxvku8a24x1v53oanxowe5tf
Theorem: Properties of 5,559,060,566,555,523

$3^{33} = 5 \, 559 \, 060 \, 566 \, 555 \, 523$ has the following properties:
: It has a remarkably large number of $5$s (half of its digits).
: Multiply it by $2$, $4$ or $6$, and the result has $10$ of a particular digit.
: Multiply it by $8$, and $9$ of the digits of the result are $4$.

Proof:
{{begin-eqn}}
{{eqn | l = 5 \, 559 \, 060 \, 566 \, 555 \, 523 \times 2
      | r = 11 \, 118 \, 121 \, 133 \, 111 \, 046
      | c = $10$ $1$s
}}
{{eqn | l = 5 \, 559 \, 060 \, 566 \, 555 \, 523 \times 4
      | r = 22 \, 236 \, 242 \, 266 \, 222 \, 092
      | c = $10$ $2$s
}}
{{eqn | l = 5 \, 559 \, 060 \, 566 \, 555 \, 523 \times 6
      | r = 33 \, 354 \, 363 \, 399 \, 333 \, 138
      | c = $10$ $3$s
}}
{{eqn | l = 5 \, 559 \, 060 \, 566 \, 555 \, 523 \times 8
      | r = 44 \, 472 \, 484 \, 532 \, 444 \, 184
      | c = $9$ $4$s
}}
{{end-eqn}}
{{qed}}
Theorem: First Occurrence of Prime Gap of 864

The first occurrence of a prime gap of $864$ is between $6 \, 505 \, 941 \, 701 \, 960 \, 039$ and $6 \, 505 \, 941 \, 701 \, 960 \, 903$.

Proof:
$6 \, 505 \, 941 \, 701 \, 960 \, 039$ is a prime number.
$6 \, 505 \, 941 \, 701 \, 960 \, 903$ is a prime number.
It can be checked that all numbers between these two are composite.
It can also be checked by computer that there is no prime gap of $864$ between smaller prime numbers.
{{qed}}
Theorem: Reciprocal of 19 from Sum of Powers of 2 Backwards

The decimal expansion of the reciprocal of $19$ can be constructed by summing the powers of $2$, offset progressively backwards by $1$ digit:
<pre>
                         1
                        2
                       4
                      8
                    16
                   32
                  64
                128
               256
              512
            1024
           2048
          4096
         8192
       16384
      32768
     65536
   131072
+ 262144
 ......
--------------------------
.......1052631578947368421
</pre>
while:
:$\dfrac 1 {19} = 0 \cdotp \dot 05263 \, 15789 \, 47368 \, 42 \dot 1$

Proof:
We confirm that from Reciprocal of $19$:
:$\dfrac 1 {19} = 0 \cdotp \dot 05263 \, 15789 \, 47368 \, 42 \dot 1$
The construction above can be expressed as the sum:
:$\ds \sum_{k \mathop \ge 0} \paren {2^k \times 10^k} = \sum_{k \mathop \ge 0} 20^k$
Obviously this sum does not converge.
However we can still analyse the last $N$ digits from this construction for every positive integer $N$.
From $k \ge N$ onwards, the last $N$ digits will not be affected by further additions of $20^k$ because:
:$\forall k \ge N: 20^k \equiv 20^{k - N} \times 2^N \times 10^N \equiv 0 \pmod {10^N}$
By Number times Recurring Part of Reciprocal gives 9-Repdigit/Generalization, the last $N$ digits of this number coincides with the decimal expansion of the reciprocal of $19$ {{iff}}:
:$\ds \sum_{k \mathop = 0}^{N - 1} 20^k \times 19 \equiv -1 \pmod {10^N}$
We have:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{N - 1} 20^k \times 19
      | r = \frac {20^N - 1} {20 - 1} \times 19
      | c = Sum of Geometric Sequence
}}
{{eqn | r = 20^N - 1
}}
{{eqn | o = \equiv
      | r = -1 \pmod {10^N}
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: General Fibonacci Sequence whose Terms are all Composite

The general Fibonacci sequence $\left\langle{a_n}\right\rangle$ defined as:
:<nowiki>$a_n = \begin{cases}
r & : n = 0 \\
s & : n = 1 \\
a_{n - 2} + a_{n - 1} & : n > 1
\end{cases}$</nowiki>
where:
:$r = 62 \, 638 \, 280 \, 004 \, 239 \, 857$
:$s = 49 \, 463 \, 435 \, 743 \, 205 \, 655$
is such that:
: $r$ and $s$ are coprime
: all its terms are composite.

Proof:
{{ProofWanted|Knuth proves it in the citation given. Needs to be transcribed, but it's boring and I can't be bothered.}}
Theorem: Superset of Infinite Set is Infinite

Let $S$ be an infinite set.
Let $T \supseteq S$ be a superset of $S$.
Then $T$ is also infinite.

Proof:
Suppose $T$ were finite.
Then by Set Finite iff Injection to Initial Segment of Natural Numbers, there is an injection $f: T \to \N_{<n}$ for some $n \in \N$.
But then by Restriction of Injection is Injection, also the restriction of $f$ to $S$:
:$f {\restriction_S}: S \to \N_{<n}$
is an injection.
Again by Set Finite iff Injection to Initial Segment of Natural Numbers, this contradicts the assumption that $S$ is infinite.
Hence $T$ is infinite.
{{qed}}
Theorem: Prime Decomposition of 2^58+1

The number $2^{58} + 1$ has the prime decomposition:
:$2^{58} + 1 = 5 \times 107 \, 367 \, 629 \times 536 \, 903 \, 681$

Proof:
From Aurifeuillian Factorization of 2 Mod 4th Power of Two plus 1, we have:
:$2^{4 n + 2} + 1 = \paren {2^{2 n + 1} - 2^{n + 1} + 1} \paren {2^{2 n + 1} + 2^{n + 1} + 1}$
Setting $n = 14$:
{{begin-eqn}}
{{eqn | l = 2^{58} + 1
      | r = \paren {2^{29} - 2^{15} + 1} \paren {2^{29} + 2^{15} + 1}
      | c = 
}}
{{eqn | r = \paren {536 \, 870 \, 912 - 32 \, 768 + 1} \paren {536 \, 870 \, 912 + 32 \, 768 + 1}
      | c = 
}}
{{eqn | r = 536 \, 838 \, 145 \times 536 \, 903 \, 681
      | c = 
}}
{{eqn | r = 5 \times 107 \, 367 \, 629 \times 536 \, 903 \, 681
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Squares as Product of Factors with Square Roots

:$x^2 + y^2 = \paren {x + \sqrt {2 x y} + y} \paren {x - \sqrt {2 x y} + y}$

Proof:
{{begin-eqn}}
{{eqn | l = \paren {x + \sqrt {2 x y} + y} \paren {x - \sqrt {2 x y} + y}
      | r = x^2 - x \sqrt {2 x y} + x y + x \sqrt {2 x y} - \sqrt {2 x y} \sqrt {2 x y} + y \sqrt {2 x y} + x y - y \sqrt {2 x y} + y^2
      | c = 
}}
{{eqn | r = x^2 + \paren {x - x} \sqrt {2 x y} + 2 x y - 2 x y + \paren {y - y} \sqrt {2 x y} + y^2
      | c = 
}}
{{eqn | r = x^2 + y^2
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Algebra
Category:Sums of Squares
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Theorem: Repunit Expressed using Power of 10

The repunit number $R_n$ can be expressed as:
:$R_n = \dfrac {10^n - 1} 9$

Proof:
{{begin-eqn}}
{{eqn | l = \dfrac {10^n - 1} 9
      | r = \dfrac {10^n - 1} {10 - 1}
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^{n - 1} 10^k
      | c = Sum of Geometric Sequence
}}
{{eqn | r = 1 + 10 + 100 + \ldots + 10^{n - 2} + 10^{n - 1}
      | c = 
}}
{{end-eqn}}
The result follows from the Basis Representation Theorem.
{{qed}}
Theorem: Repunit cannot be Square

A repunit (apart from the trivial $1$) cannot be a square.

Proof:
Let $m$ be a repunit with $r$ digits such that $r > 1$.
By definition, $m$ is odd.
Thus from Square Modulo 4, if $m$ were square it would be of the form:
:$m \equiv 1 \pmod 4$.
$m$ is of the form $\ds \sum_{k \mathop = 0}^{r - 1} 10^k$ where $r$ is the number of digits.
Thus for $r \ge 2$:
{{begin-eqn}}
{{eqn | l = m
      | r = 11 + 100 s
      | c = for some $s \in \Z$
}}
{{eqn | r = \paren {2 \times 4} + 3 + 4 \times \paren {25 s}
      | c = 
}}
{{eqn | r = 3 + 4 t
      | c = for some $t \in \Z$
}}
{{end-eqn}}
Hence:
:$m \equiv 3 \pmod 4$
and so cannot be square.
{{qed}}