avewright/proofwiki-math huggingface dataset:
Theorem: Sequence of Composite Mersenne Numbers

The sequence of Mersenne numbers which are composite begins:
:$2047, 8 \, 388 \, 607, 536 \, 870 \, 911, 137 \, 438 \, 953 \, 471, 2 \, 199 \, 023 \, 255 \, 551,\ldots$
{{OEIS|A065341}}
The sequence of corresponding indices $p$ such that $2^p - 1$ is composite begins:
:$11, 23, 29, 37, 41, 43, 47, 53, 59, 67, 71, 73, 79, 83, \ldots$
{{OEIS|A054723}}
The sequence of corresponding integers $n$ such that the $n$th prime number $p \left({n}\right)$ is such that $2^{p \left({n}\right)} - 1$ is composite begins:
:$5, 9, 10, 12, 13, 14, 15, 16, 17, 19, 20, 21, 22, 23, 25, 26, 27, \ldots$
{{OEIS|A135980}}

Proof:
Established by inspecting the sequence of Mersenne numbers:
:$3, 7, 31, 127, 2047, 8191, 131 \, 071, 524 \, 287, 8 \, 388 \, 607, 536 \, 870 \, 911, 2 \, 147 \, 483 \, 647, \ldots$
and removing from it the sequence of Mersenne primes:
:$3, 7, 31, 127, 8191, 131 \, 071, 524 \, 287, 2 \, 147 \, 483 \, 647, \ldots$
{{qed}}
Theorem: Numbers Reversed when Multiplying by 4

Numbers of the form $\sqbrk {21 \paren 9 78}_{10}$ are reversed when they are multiplied by $4$:
{{begin-eqn}}
{{eqn | l = 2178 \times 4
      | r = 8712
}}
{{eqn | l = 21 \, 978 \times 4
      | r = 87 \, 912
}}
{{eqn | l = 219 \, 978 \times 4
      | r = 879 \, 912
}}
{{end-eqn}}
and so on.

Proof:
Let k represent the number of $9$s in the middle of the number.
For $k > 0$ We can rewrite the number as follows:
{{begin-eqn}}
{{eqn | l = \sqbrk {21 (9) 78}_{10}
      | r = 21 \times 10^{k + 2 } + 900 \sum_{i \mathop = 0}^{k - 1} 10^i + 78
      | c = {{Defof|Geometric Series}}
}}
{{end-eqn}}
Taking numbers of this form and multiplying by $4$ produces:
{{begin-eqn}}
{{eqn | l = 4 \times \paren {21 \times 10^{k + 2 } + 900 \sum_{i \mathop = 0}^{k - 1} 10^i + 78 }
      | r = 84 \times 10^{k + 2 } + 3600 \sum_{i \mathop = 0}^{k - 1} 10^i + 312
}}
{{end-eqn}}
The first part is composed of $k + 4$ digits. The first two digits will be $84$ followed by $k +2$ digits of $0$
{{begin-eqn}}
{{eqn | l = 84 \times 10^{k + 2 } 
      | r = 84 \times 10^{k + 2 } 
}}
{{end-eqn}}
The sum in the middle is composed of $k + 3$ digits. The first digit will be $3$ followed by $k - 1$ digits of $9$ and then the remaining three digits at the end are $600$
{{begin-eqn}}
{{eqn | l = 3600 \sum_{i \mathop = 0}^{k - 1} 10^i
      | r = 399 \cdots 99600
}}
{{end-eqn}}
Summing the three pieces, the final answer will have $k + 4$ digits.
The first digit is $8$ 
followed by $7$ which is the sum of the $4$ from the first part and the $3$ of the middle part
followed by $k$ digits of $9$ where the last $9$ is the sum of the $6$ from the middle part and the $3$ of the last part
and then ending in $12$:
{{begin-eqn}}
{{eqn | l = \sqbrk {21 (9) 78}_{10} \times 4
      | r = \sqbrk {87 (9) 12}_{10}
      | c =
}}
{{end-eqn}}
{{qed}}
Theorem: 17 Consecutive Integers each with Common Factor with Product of other 16

The $17$ consecutive integers from $2184$ to $2200$ have the property that each one is not coprime with the product of the other $16$.

Proof:
We obtain the prime decomposition of all $17$ of these integers:
{{begin-eqn}}
{{eqn | l = 2184
      | r = 2^3 \times 3 \times 7 \times 13
}}
{{eqn | l = 2185
      | r = 5 \times 19 \times 23
}}
{{eqn | l = 2186
      | r = 2 \times 1093
}}
{{eqn | l = 2187
      | r = 3^7
}}
{{eqn | l = 2188
      | r = 2^2 \times 547
}}
{{eqn | l = 2189
      | r = 11 \times 199
}}
{{eqn | l = 2190
      | r = 2 \times 3 \times 5 \times 73
}}
{{eqn | l = 2191
      | r = 7 \times 313
}}
{{eqn | l = 2192
      | r = 2^4 \times 137
}}
{{eqn | l = 2193
      | r = 3 \times 17 \times 43
}}
{{eqn | l = 2194
      | r = 2 \times 1097
}}
{{eqn | l = 2195
      | r = 5 \times 439
}}
{{eqn | l = 2196
      | r = 2^2 \times 3^2 \times 61
}}
{{eqn | l = 2197
      | r = 13^3
}}
{{eqn | l = 2198
      | r = 2 \times 7 \times 157
}}
{{eqn | l = 2199
      | r = 3 \times 733
}}
{{eqn | l = 2200
      | r = 2^3 \times 5^2 \times 11
}}
{{end-eqn}}
It can be seen by inspection that each of the integers in this sequence shares at least one prime factor with at least one other.
It is then worth noting that:
{{begin-eqn}}
{{eqn | l = 2183
      | r = 37 \times 59
}}
{{eqn | l = 2201
      | r = 31 \times 71
}}
{{end-eqn}}
and it can be seen that the sequence can be extended neither upwards nor downwards.
{{qed}}
Theorem: Relational Structure admits Lower Topology

Let $R = \left({S, \preceq}\right)$ be a relational structure.
Then there exists a relational structure with lower topology $T = \left({S, \preceq, \tau}\right)$ such that $T$ is a topological space.

Proof:
Define $B := \left\{ {\complement_S\left({x^\succeq}\right): x \in S}\right\}$
where $x^\succeq$ denotes the upper closure of $x$.
By definition of generated topology:
:$\tau\left({B}\right)$ is a topology on $S$
where $B$ is a sub-basis of $\tau \left({B}\right)$.
Thus by definition of lower topology:
:$T := \left({S, \preceq, \tau\left({B}\right)}\right)$ has a lower topology.
Thus by definition:
:$T$ is a topological space.
{{qed}}
Theorem: Lower and Upper Bounds for Sequences/Warning

Let $\sequence {x_n}$ be a sequence in $\R$.
Let $x_n \to l$ as $n \to \infty$.
Then it is '''not''' necessarily the case that:
:$(1): \quad \forall n \in \N: x_n > a \implies l > a$
:$(2): \quad \forall n \in \N: x_n < b \implies l < b$

Proof:
Take the examples:
:$(1): \quad \sequence {x_n} = \dfrac 1 n$
:$(2): \quad \sequence {y_n} = -\dfrac 1 n$
Then :
:$\forall n \in \N_{>0}: \dfrac 1 n > 0, -\dfrac 1 n < 0$
From Sequence of Reciprocals is Null Sequence, we have
:$x_n \to 0$
:$y_n \to 0$
as $n \to \infty$.
However, it is clearly '''false''' that $0 > 0$ and $0 < 0$.
{{qed}}
Theorem: Squares of 23...3

The following pattern holds:
{{begin-eqn}}
{{eqn | l = 3^2
      | r = 9
}}
{{eqn | l = 23^2
      | r = 529
}}
{{eqn | l = 233^2
      | r = 54 \, 289
}}
{{eqn | l = 2333^2
      | r = 5 \, 442 \, 889
}}
{{eqn | l = 23333^2
      | r = 544 \, 428 \, 889
}}
{{end-eqn}}
and so on.

Proof:
{{ProofWanted|Simple but tedious.}}
Theorem: Smallest Fourth Power as Sum and Difference of Fourth Powers

The smallest $4$th power that can be expressed as the sum of $2$ $4$th powers minus a $3$rd is:
:$2401 = 7^4 = 227^4 + 157^4 - 239^4$
with all numbers less than $10^4$.

Proof:
{{begin-eqn}}
{{eqn | o = 
      | r = 227^4 + 157^4 - 239^4
      | c = 
}}
{{eqn | r = 2 \, 655 \, 237  \, 841 + 607 \, 573 \, 201 - 3 \, 262 \, 808 \, 641
      | c = 
}}
{{eqn | r = 2401
      | c = 
}}
{{eqn | r = 7^4
      | c = 
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown that this is the smallest such.}}
Theorem: Zsigmondy's Theorem for Sums

Let $a > b > 0$ be coprime positive integers.
Let $n \ge 1$ be a (strictly) positive integer.
Then there is a prime number $p$ such that
:$p$ divides $a^n + b^n$
:$p$ does not divide $a^k + b^k$ for all $k < n$
with the following exception:
:$n = 3$, $a = 2$, $b = 1$

Proof:
By Zsigmondy's Theorem, there exists a prime divisor $p$ of $a^{2 n} - b^{2 n}$ which does not divide $a^k - b^k$ for all $k < 2 n$ unless:
:$n = 1$ and $a + b$ is a power of $2$
:$n = 3$, $a = 2$, $b = 1$
In particular, $p$ does not divide $a^{2 k} - b^{2 k} = \paren {a^k - b^k} \paren {a^k + b^k}$ for $k < n$.
It remains to check the case $n = 1$ and $a + b$ a power of $2$.
We have to show that $a^2 + b^2$ has an odd prime divisor.
Since $a$ and $b$ are coprime, both $a$ and $b$ are odd.
By Square Modulo 4, $a^2 + b^2 \equiv 2 \pmod 4$.
Because $a > b > 0$, $a^2 + b^2 > 2$.
But $4 \divides 2^k$ for $k > 1$.
Thus $a^2 + b^2$ is not a power of $2$.
Hence $a^2 + b^2$ has an odd prime divisor.
{{qed}}
{{Namedfor|Karl Zsigmondy|cat = Zsigmondy}}
Category:Number Theory
3spcigpduvdts8wuzbojjnubqlu3z3v
Theorem: Multiplicative Order of Roots of Cyclotomic Polynomial Modulo Prime

Let $n \ge 1$ be a natural number.
Let $\Phi_n$ be the $n$-th cyclotomic polynomial.
Let $p$ be a prime number.
Let $n = p^\alpha q$ where $\alpha = \map {\nu_p} n$ is the $p$-adic valuation of $n$.
Let $a \in \Z$ with $\map {\Phi_n} a \equiv 0 \pmod p$.
Then the order of $a$ modulo $p$ is $q$:
:$\map {\operatorname{ord}_p} a = q$.

Proof:
By Product of Cyclotomic Polynomials, $p \mid \map {\Phi_n} a \mid a^n-1$.
Thus $a$ is coprime to $p$.
By Fermat's Little Theorem, $1 \equiv a^n \equiv a^q \pmod p$.
Thus $\map {\operatorname{ord}_p} a \le q$.
{{AimForCont}} $\map {\operatorname{ord}_p} a = k < q$.
By Product of Cyclotomic Polynomials, $p \mid \map {\Phi_d} a$ for some $d \mid k$.
Then $a$ is a double root of $x^q - 1$ modulo $p$.
By Double Root of Polynomial is Root of Derivative, $q \equiv 0 \pmod p$.
This is a contradiction, thus $k = q$.
{{qed}}
Category:Cyclotomic Polynomials
44cx0cgq3ygvencbo00gumox6jil9xi
Theorem: Lower Topology is Unique

Let $T_1 = \left({S, \preceq, \tau_1}\right)$ and $T_2 = \left({S, \preceq, \tau_2}\right)$ be relational structures with lower topologies.
Then:
: $\tau_1 = \tau_2$

Proof:
Define:
: $B := \left\{ {\complement_S \left({x^\succeq}\right): x \in S}\right\}$
where $x^\succeq$ denotes the upper closure of $x$.
Thus:
{{begin-eqn}}
{{eqn | l = \tau_1
      | r = \tau \left({B}\right)
      | c = {{Defof|Topology Generated by Synthetic Sub-Basis}}
}}
{{eqn | r = \tau_2
      | c = {{Defof|Topology Generated by Synthetic Sub-Basis}}
}}
{{end-eqn}}
{{qed}}
Theorem: Homogeneous Cyclotomic Polynomial is Symmetric

Let $n>1$ be a natural number.
Let $\Phi_n(x,y)$ be the $n$th homogeneous cyclotomic polynomial.
Then $\Phi_n(x,y) = \Phi_n(y,x)$, that is, $\Phi_n(x,y)$ is symmetric.

Proof:
{{ProofWanted}}
Category:Cyclotomic Polynomials
1klkb4mge55s34ezlu50w1to0q0ktsa
Theorem: Cyclotomic Polynomial of Index Power of Two

Let $n \ge 1$ be a natural number.
Then the $2^n$th cyclotomic polynomial is:
:$\map {\Phi_{2^n} } x = x^{2^{n - 1} } + 1$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\Phi_{2^n} } x
      | r = \prod_{\zeta} \paren {x - \zeta}
      | c = where the product runs over all primitive complex $2^n$th roots of unity
}}
{{eqn | r = \prod_{\substack {1 \mathop \le k \mathop \le 2^n \\ \gcd \set {k, 2^n} = 1} } \paren {x - \map \exp {\frac {2 \pi i k} {2^n} } }
      | c = Condition for Complex Root of Unity to be Primitive
}}
{{eqn | r = \prod_{\substack {1 \mathop \le k \mathop \le 2^n \\ k \text { odd} } } \paren {x - \map \exp {\frac {k i \pi} {2^{n - 1} } } }
      | c = 
}}
{{eqn | r = \prod_{k \mathop = 0}^{2^{n - 1} - 1} \paren {x - \map \exp {\frac {\paren {2 k + 1} i \pi} {2^{n - 1} } } }
      | c = by writing each odd number as $2 k + 1$
}}
{{eqn | r = x^{2^{n - 1} } + 1
      | c = Factorisation of $z^n + 1$
}}
{{end-eqn}}
{{qed}}
Category:Cyclotomic Polynomials
5f6f3syoexhclaxk8bbaw0tlv0k0h4y
Theorem: Trivial Estimate for Cyclotomic Polynomials

Let $n \ge 1$ be a natural number.
Let $\Phi_n$ be the $n$th cyclotomic polynomial.
Let $\phi$ be the Euler totient function.
Let $z \in \C$ be a complex number.
Then:
:$\size {\size z - 1}^{\map \phi n} \le \size {\map {\Phi_n} z} \le \paren {\size z + 1}^{\map \phi n}$
where:
:the first inequality becomes an equality only if:
:::$n = 1$ and $z \in \R_{\ge 0}$
::or:
:::$n = 2$ and $z \in \R_{\le 0}$
:the second inequality becomes an equality only if:
:::$n = 1$ and $z \in \R_{\le 0}$
::or:
:::$n = 2$ and $z \in \R_{\ge 0}$

Proof:
{{ProofWanted}}
Category:Cyclotomic Polynomials
8irnibmc93ig6ztlrhmeb6bb0y6sfpt
Theorem: 2520 equals Sum of 4 Divisors in 6 Ways

The number $2520$ can be expressed as the sum of $4$ of its divisors in $6$ different ways:
{{begin-eqn}}
{{eqn | l = 2520
      | r = 1260 + 630 + 504 + 126
}}
{{eqn | r = 1260 + 630 + 421 + 210
}}
{{eqn | r = 1260 + 840 + 360 + 60
}}
{{eqn | r = 1260 + 840 + 315 + 105
}}
{{eqn | r = 1260 + 840 + 280 + 140
}}
{{eqn | r = 1260 + 840 + 252 + 168
}}
{{end-eqn}}
This is the maximum possible number of ways it is possible to express an integer as the sum of $4$ of its divisors.

Proof:
We apply 1 can be Expressed as Sum of 4 Distinct Unit Fractions in 6 Ways:
{{:1 can be Expressed as Sum of 4 Distinct Unit Fractions in 6 Ways}}
Find the maximum powers of the primes in each equation, and choose the largest that appears:
{{begin-eqn}}
{{eqn | l = 42
      | r = 2^1 \times 3^1 \times 7^1
}}
{{eqn | l = 24
      | r = 2^3 \times 3^1
}}
{{eqn | l = 18
      | r = 2^1 \times 3^2
}}
{{eqn | l = 30
      | r = 2^1 \times 3^1 \times 5^1
}}
{{eqn | l = 20
      | r = 2^2 \times 5^1
}}
{{eqn | l = 12
      | r = 2^2 \times 3^1
}}
{{end-eqn}}
Therefore the smallest number would be:
:$2^3 \times 3^2 \times 5^1 \times 7^1 = 2520$
{{qed}}
Theorem: Complete List of Special Highly Composite Numbers

There are exactly $6$ special highly composite numbers:
:$1, 2, 6, 12, 60, 2520$
{{OEIS|A106037}}

Proof:
We have the following:
:$1$ is a Special Highly Composite Number
:$2$ is a Special Highly Composite Number
:$6$ is a Special Highly Composite Number
:$12$ is a Special Highly Composite Number
:$60$ is a Special Highly Composite Number
:$2520$ is a Special Highly Composite Number
By inspection of the sequence of highly composite numbers:
:$1, 2, 4, 6, 12, 24, 36, 48, 60, 120, 180, 240, 360, 720, 840, 1260, 1680, 2520, \ldots$
it can be seen that there are no more special highly composite numbers less than $2520$.
Let $n > 18$.
From Ratio between Consecutive Highly Composite Numbers Greater than 2520 is Less than 2, the $n$th highly composite number does not divide the $n+1$th.
Hence the $n$th highly composite number is not a special highly composite number.
The result follows.
{{qed}}
Theorem: Lifting The Exponent Lemma for Sums for p=2

Let $x, y \in \Z$ be integers with $x + y \ne 0$.
Let $n \ge 1$ be an odd natural number.
Let:
:$2 \divides x + y$
where $\divides$ denotes divisibility.
Then:
:$\map {\nu_2} {x^n + y^n} = \map {\nu_2} {x + y}$
where $\nu_2$ denotes $2$-adic valuation.
</onlyinclude>

Proof:
This follows from the Lifting The Exponent Lemma for p=2 with $y$ replaced by $-y$.
{{qed}}
Theorem: Prime Decomposition of Highly Composite Number

Let $n$ be a highly composite number.
Let the prime decomposition of $n$ be expressed as:
:$n = \ds \prod_{k \mathop \in \N} {p_k}^{r_k}$
where $p_k$ denotes the $k$th prime.
Then the sequence $\sequence {r_k}$ is decreasing.
That is:
:$\forall k \in \N: r_k \ge r_{k + 1}$

Proof:
Let $n = \ds \prod_{k \mathop \in \N} {p_k}^{r_k}$ be highly composite.
By definition of divisor count function:
:$\map {\sigma_0} n = \ds \prod_{k \mathop \in \N} \paren {r_k + 1}$
{{AimForCont}} $r_{l + 1} > r_l$ for some $l \in \N$.
Consider $m \in \Z$ whose prime decomposition of $n$ is expressed as:
:$m = \ds \prod_{k \mathop \in \N} {p_k}^{s_k}$
where:
:$\forall j < l: s_j = r_j$
:$\forall j > l + 1: s_j = r_j$
:$s_l = r_{l + 1}$
:$s_{l + 1} = r_l$
We have that:
{{begin-eqn}}
{{eqn | l = \map {\sigma_0} m
      | r = \prod_{k \mathop \in \N} \paren {s_k + 1}
      | c = {{Defof|Divisor Count Function}}
}}
{{eqn | r = \prod_{k \mathop \in \N} \paren {r_k + 1}
      | c = as all the $s$s are the same as all the $r$s, but in a different order
}}
{{eqn | r = \map {\sigma_0} n
      | c = {{Defof|Divisor Count Function}}
}}
{{end-eqn}}
Thus:
:$\map {\sigma_0} m = \map {\sigma_0} n$
Now we have that:
:$r_l < r_{l + 1}$
and so:
:$s_l > s_{l + 1}$
and so:
{{begin-eqn}}
{{eqn | l = \frac { {p_l}^{r_l} {p_{l + 1} }^{r_{l + 1} } } { {p_l}^{s_l} {p_{l + 1} }^{s_{l + 1} } }
      | r = \frac { {p_l}^{r_l} {p_{l + 1} }^{r_{l + 1} } } { {p_l}^{r_{l + 1} } {p_{l + 1} }^{r_l} }
      | c = 
}}
{{eqn | r = \frac { {p_l}^c} { {p_{l + 1}^c} }
      | c = where $c = r_{l + 1} - r_l$
}}
{{end-eqn}}
Then:
:$p_{l + 1} > p_l$
and so:
:${p_{l + 1} }^c > {p_l}^c$
from which it follows that:
:$m < n$
while:
:$\map {\sigma_0} m = \map {\sigma_0} n$
But $n$ is highly composite.
This means that if $\map {\sigma_0} m = \map {\sigma_0} n$, then $n \le m$.
This is contradicted by $n > m$.
Thus by Proof by Contradiction:
:$\forall k \in \N: r_k \ge r_{k + 1}$
{{qed}}
Theorem: Complement of Upper Closure of Element is Open in Lower Topology

Let $T = \struct {S, \preceq, \tau}$ be a relational structure with lower topology.
Let $x \in S$.
Then $\relcomp S {x^\succeq}$ is open and $x^\succeq$ is closed.

Proof:
Define $B := \set {\relcomp S {y^\succeq}: y \in S}$
By definition of lower topology:
:$B$ is sub-basis of $T$.
By definition of sub-basis:
:$B \subseteq \tau$
By definition of $B$:
:$\relcomp S {x^\succeq} \in B$
Thus by definition of subset:
:$\relcomp S {x^\succeq} \in \tau$
Thus by definition:
:$x^\succeq$ is closed.
{{qed}}
Theorem: Open Subset is Lower Section in Lower Topology

Let $T = \struct {S, \preceq, \tau}$ be a transitive relational structure with lower topology.
Let $A \subseteq S$ such that
:$A$ is open.
Then $A$ is a lower section of $S$.

Proof:
Define $B = \set {\relcomp S {x^\succeq}: x \in S}$
By definition of lower topology:
:$B$ is sub-basis of $T$.
By definitions of sub-basis and basis:
:$\ds \BB = \set {\bigcap \FF: \FF \subseteq B, \FF \text{ is finite} }$ is a basis.
By definition of basis:
:$\ds \tau \subseteq \set {\bigcup X: X \subseteq \BB}$
Let $x \in A$, $y \in S$ such that:
:$y \preceq x$
By definition of open set:
:$A \in \tau$
By definition of subset:
:$\exists Y \subseteq \BB: A = \bigcup Y$
By definition of union:
:$\exists Z \in Y: x \in Z$
By definition of subset:
:$\exists X \subseteq B: Z = \bigcap X \land X$ is finite.
We will prove that:
:$\forall Q \in X: y \in Q$
Let $Q \in X$.
By definition of subset:
:$Q \in B$
By definition of $B$:
:$\exists z \in S: Q = \relcomp S {z^\succeq}$
By definition of intersection:
:$x \in Q$
By Upper Closure is Upper Section:
:$z^\succeq$ is an upper section.
By Complement of Upper Section is Lower Section:
:$Q$ is a lower section.
Thus by definition of lower section:
:$y \in Q$
{{qed|lemma}}
By definition of intersection:
:$y \in Z$
Thus by definition of union:
:$y \in A$
{{qed}}
Theorem: Modified Kaprekar Process on 4-Digit Number terminates in 2538

Let $n$ be a $4$-digit number.
Let $n$ be operated on by the modified Kaprekar process.
The eventual result is always $2538$.

Proof:
{{ProofWanted|I find these processes dull. Does anyone else want to go into a detailed analysis of this and other similar?}}
Theorem: Fibonacci Number equal to Sum of Sequence of Cubes

The following Fibonacci number can be expressed as the sum of a sequence of cubes:
:$F_{18} = 2584 = 7^3 + 8^3 + 9^3 + 10^3$
{{expand|Any more?}}

Proof:
{{begin-eqn}}
{{eqn | l = 2584
      | r = 343 + 512 + 729 + 1000
      | c = 
}}
{{eqn | r = 7^3 + 8^3 + 9^3 + 10^3
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: 2601 as Sum of 3 Squares in 12 Different Ways

$2601$ can be expressed as the sum of $3$ squares in $12$ different ways.

Proof:
{{begin-eqn}}
{{eqn | l = 2601
      | r = 51^2
      | c = 
}}
{{eqn | r = 1^2 + 10^2 + 50^2
      | c = 
}}
{{eqn | r = 2^2 + 14^2 + 49^2
      | c = 
}}
{{eqn | r = 10^2 + 10^2 + 49^2
      | c = 
}}
{{eqn | r = 14^2 + 14^2 + 47^2
      | c = 
}}
{{eqn | r = 1^2 + 22^2 + 46^2
      | c = 
}}
{{eqn | r = 14^2 + 17^2 + 46^2
      | c = 
}}
{{eqn | r = 1^2 + 34^2 + 38^2
      | c = 
}}
{{eqn | r = 14^2 + 31^2 + 38^2
      | c = 
}}
{{eqn | r = 3^2 + 36^2 + 36^2
      | c = 
}}
{{eqn | r = 24^2 + 27^2 + 36^2
      | c = 
}}
{{eqn | r = 17^2 + 34^2 + 34^2
      | c = 
}}
{{eqn | r = 22^2 + 31^2 + 34^2
      | c = 
}}
{{end-eqn}}
That there are no more can be determined by exhaustion.
{{qed}}
Theorem: Reversal of Number Multiplied by 11

Let $n \in \N$ be a number for which, when written in decimal notation, no two adjacent digits total to more than $9$.
Let $n'$ denote the reversal of $n$.
Then $n \times 11$ is the reversal of $n' \times 11$.

Proof:
By Basis Representation Theorem, there exists one and only one sequence $\sequence {r_j}_{0 \mathop \le j \mathop \le t}$ such that:
:$(1): \quad \ds n = \sum_{k \mathop = 0}^t r_k 10^k$
:$(2): \quad \ds \forall k \in \closedint 0 t: r_k \in \N_{10}$
:$(3): \quad r_t \ne 0$
Since no two adjacent digits of $n$ total to more than $9$, we have:
:$\paren {r_i + r_{i - 1} } \in \N_{10}$ for $i = 1, 2, \dots, t$
Now:
{{begin-eqn}}
{{eqn | l = n \times 11
      | r = 11 \sum_{k \mathop = 0}^t r_k 10^k
}}
{{eqn | r = 10 \sum_{k \mathop = 0}^t r_k 10^k + \sum_{k \mathop = 0}^t r_k 10^k
}}
{{eqn | r = \sum_{k \mathop = 0}^t r_k 10^{k + 1} + \sum_{k \mathop = 0}^t r_k 10^k
}}
{{eqn | r = \sum_{k \mathop = 1}^{t + 1} r_{k - 1} 10^k + \sum_{k \mathop = 0}^t r_k 10^k
      | c = Translation of Index Variable of Summation
}}
{{eqn | r = r_0 + \sum_{k \mathop = 1}^t \paren {r_k + r_{k - 1} } 10^k + r_t 10^{t + 1}
}}
{{end-eqn}}
Since $r_0, r_t$ and each $r_k + r_{k - 1}$ is in $\N_{10}$, the above is the unique representation of $n \times 11$.
The reversal of $n$, $n'$, is given by:
:$\ds n = \sum_{k \mathop = 0}^t r_{t - k} 10^k$
Hence:
{{begin-eqn}}
{{eqn | l = n' \times 11
      | r = 11 \sum_{k \mathop = 0}^t r_{t - k} 10^k
}}
{{eqn | r = 10 \sum_{k \mathop = 0}^t r_{t - k} 10^k + \sum_{k \mathop = 0}^t r_{t - k} 10^k
}}
{{eqn | r = \sum_{k \mathop = 0}^t r_{t - k} 10^{k + 1} + \sum_{k \mathop = 0}^t r_{t - k} 10^k
}}
{{eqn | r = \sum_{k \mathop = 1}^{t + 1} r_{t - k + 1} 10^k + \sum_{k \mathop = 0}^t r_{t - k} 10^k
      | c = Translation of Index Variable of Summation
}}
{{eqn | r = r_t + \sum_{k \mathop = 1}^t \paren {r_{t - k} + r_{t - k + 1} } 10^k + r_0 10^{t + 1}
}}
{{end-eqn}}
Since $r_t, r_0$ and each $r_{t - k} + r_{t - k + 1}$ is in $\N_{10}$, the above is the unique representation of $n' \times 11$.
Its reversal is given by:
:$\ds r_0 + \paren {r_0 + r_1} 10^1 + \paren {r_1 + r_2} 10^2 + \dots + \paren {r_{t - 1} + r_t} 10^t + r_t 10^{t + 1}$
which is:
:$\ds r_0 + \sum_{k \mathop = 1}^t \paren {r_{k - 1} + r_k} 10^k + r_t 10^{t + 1}$
which can be seen to be equal to $n \times 11$.
{{qed}}
Theorem: Reduction Formula for Integral of Power of Tangent

For all $n \in \Z_{> 1}$:
Let:
:$I_n := \ds \int \tan^n x \rd x$
Then:
:$I_n = \dfrac {\tan^{n - 1} x} {n - 1} - I_{n - 2}$
is a reduction formula for $\ds \int \tan^n x \rd x$.

Proof:
{{begin-eqn}}
{{eqn | l = I_n
      | r = \int \tan^n x \rd x
      | c = by definition
}}
{{eqn | r = \int \tan^{n - 2} x \tan^2 x \rd x
}}
{{eqn | r = \int \tan^{n - 2} x \paren {\sec^2 x - 1} \rd x
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \int \tan^{n - 2} x \sec^2 x \rd x - \int \tan^{n - 2} x \rd x
      | c = Linear Combination of Primitives
}}
{{eqn | n = 1
      | r = \int \tan^{n - 2} x \sec^2 x \rd x - I_{n - 2}
      | c = by definition
}}
{{end-eqn}}
Let:
{{begin-eqn}}
{{eqn | l = t
      | r = \tan x
}}
{{eqn | ll= \leadsto
      | l = \frac {\d t} {\d x}
      | r = \sec^2 x
      | c = Derivative of Tangent Function
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \tan^{n - 2} x \sec^2 x \rd x
      | r = \int t^{n - 2} \rd t
      | c = Integration by Substitution
}}
{{eqn | r = \frac {t^{n - 1} } {n - 1}
      | c = Primitive of Power
}}
{{eqn | r = \frac {\tan^{n - 1} x} {n - 1} 
      | c = substituting back $t \to \map \tan x$
}}
{{eqn | ll= \leadsto
      | l = I_n
      | r = \frac {\tan^{n - 1} x} {n - 1} - I_{n - 2}
      | c = substituting for $\ds \int \tan^{n - 2} x \sec^2 x \rd x$ in $(1)$
}}
{{end-eqn}}
Hence the result.
{{qed}}