avewright/proofwiki-math huggingface dataset:
Theorem: Difference between Two Squares equal to Repunit/Corollary 1

{{begin-eqn}}
{{eqn | l = 6^2 - 5^2
      | r = 11
      | c = 
}}
{{eqn | l = 56^2 - 45^2
      | r = 1111
      | c = 
}}
{{eqn | l = 556^2 - 445^2
      | r = 111 \, 111
      | c = 
}}
{{eqn | o = :
      | c = 
}}
{{end-eqn}}
and in general for integer $n$:
:$R_{2 n} = {\underbrace {55 \ldots 56}_{\text {$n - 1$ $5$'s} } }^2 - {\underbrace {44 \ldots 45}_{\text {$n - 1$ $4$'s} } }^2$
that is:
:$\ds \sum_{k \mathop = 0}^{2 n - 1} 10^k = \paren {\sum_{k \mathop = 1}^{n - 1} 5 \times 10^k + 6}^2 - \paren {\sum_{k \mathop = 1}^{n - 1} 4 \times 10^k + 5}^2$

Proof:
From Difference between Two Squares equal to Repunit, $R_{2 n} = x^2 - y^2$ exactly when $R_{2 n} = a b$ where $x = \dfrac {a + b} 2$ and $y = \dfrac {a - b} 2$.
By the Basis Representation Theorem
{{begin-eqn}}
{{eqn | l = R_{2n}
      | r = \sum_{k \mathop = 0}^{2 n - 1} 10^k
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^{n - 1} 10^k + \sum_{k \mathop = n}^{2 n - 1} 10^k
      | c = splitting the summation into two
}}
{{eqn | r = \sum_{k \mathop = 0}^{n - 1} 10^k + 10^n \sum_{k \mathop = 0}^{n - 1} 10^k
      | c = factoring $10^n$ out of the second part
}}
{{eqn | r = \paren {10^n + 1} \sum_{k \mathop = 0}^{n - 1} 10^k
      | c = 
}}
{{eqn | r = \underbrace {100 \ldots 01}_{\text {$n - 1$ $0$'s} } \times \underbrace {111 \ldots 1}_{\text {$n$ $1$'s} }
      | c = 
}}
{{end-eqn}}
Thus, let:
:$a = \paren {10^n + 1}$
:$b = \ds \sum_{k \mathop = 0}^{n - 1}$
So:
{{begin-eqn}}
{{eqn | l = a + b
      | r = \sum_{k \mathop = 0}^n 10^k + 1
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 1}^n 10^k + 2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {a + b} 2
      | r = \sum_{k \mathop = 1}^n \frac {10^k} 2 + 1
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 1}^n 5 \times 10^{k - 1} + 1
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^{n - 1} 5 \times 10^k + 1
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 1}^{n - 1} 5 \times 10^k + 6
      | c = 
}}
{{eqn | r = \underbrace {55 \ldots 50}_{\text {$n - 1$ $5$'s} } + 6
      | c = 
}}
{{eqn | r = \underbrace {55 \ldots 56}_{\text {$n - 1$ $5$'s} }
      | c = 
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = a - b
      | r = \paren {10^n + 1} - \sum_{k \mathop = 0}^{n - 1} 10^k
      | c = 
}}
{{eqn | r = \paren {\sum_{k \mathop = 0}^{n - 1} 9 \times 10^k + 1} + 1 - \sum_{k \mathop = 0}^{n - 1} 10^k
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^{n - 1} 8 \times 10^k + 2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {a - b} 2
      | r = \sum_{k \mathop = 0}^{n - 1} 4 \times 10^k + 1
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 1}^{n - 1} 4 \times 10^k + 5
      | c = 
}}
{{eqn | r = \underbrace {44 \ldots 45}_{\text {$n - 1$ $4$'s} }
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Difference between Two Squares equal to Repunit/Corollary 2

{{begin-eqn}}
{{eqn | l = 6^2 - 5^2
      | r = 11
      | c = 
}}
{{eqn | l = 56^2 - 45^2
      | r = 1111
      | c = 
}}
{{eqn | l = 5056^2 - 5045^2
      | r = 111 \, 111
      | c = 
}}
{{eqn | o = :
      | c = 
}}
{{end-eqn}}
and in general for integer $n$:
:$R_{2 n} = {\underbrace{5050 \ldots 56}_{n - 1 \ 5 \text{'s} } }^2 - {\underbrace{5050 \ldots 45}_{n - 1 \ 5 \text{'s} } }^2$
that is:
:$\ds \sum_{k \mathop = 0}^{2 n - 1} 10^k = \left({\sum_{k \mathop = 1}^{n - 1} 5 \times 10^{2 k - 1} + 6}\right)^2 - \left({\sum_{k \mathop = 1}^{n - 1} 5 \times 10^{2 k - 1} - 5}\right)^2$

Proof:
From Difference between Two Squares equal to Repunit, $R_{2n} = x^2 - y^2$ exactly when $R_{2n} = a b$ where $x = \dfrac {a + b} 2$ and $y = \dfrac {a - b} 2$.
By the Basis Representation Theorem
{{begin-eqn}}
{{eqn | l = R_{2n}
      | r = \sum_{0 \mathop \le k \mathop < 2 n} 10^k
      | c = 
}}
{{eqn | r = \sum_{\substack {0 \mathop \le k \mathop < 2 n \\ k \text { even} } } 10^k + \sum_{\substack {0 \mathop \le k \mathop < 2 n \\ k \text { odd} } } 10^k
      | c = 
}}
{{eqn | r = \sum_{\substack {0 \mathop \le k \mathop < 2 n \\ k \text { even} } } 10^k + 10 \times \sum_{\substack {0 \mathop \le k \mathop < 2 n \\ k \text { even} } } 10^k
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^{n - 1} 10^{2 k} + 10 \times \sum_{k \mathop = 0}^n 10^{2 k}
      | c = change of indices
}}
{{eqn | r = \sum_{k \mathop = 0}^{n - 1} 11 \times 10^{2 k} 
      | c = 
}}
{{eqn | r = 11 \sum_{k \mathop = 0}^{n - 1} 10^{2 k}
      | c = 
}}
{{eqn | r = 11 \times \underbrace {10101 \ldots 01}_{n \ 1 \text{'s} }
      | c = 
}}
{{end-eqn}}
Thus, let:
:$a = \ds \sum_{k \mathop = 0}^{n - 1} \times 10^{2 k}$
:$b = 11$
So:
{{begin-eqn}}
{{eqn | l = a + b
      | r = \sum_{k \mathop = 0}^{n - 1} 10^{2 k} + 11
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 1}^{n - 1} 10^{2 k} + 12
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {a + b} 2
      | r = \sum_{k \mathop = 1}^{n - 1} \frac {10^{2 k} } 2 + 6
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 1}^{n - 1} 5 \times 10^{2 k - 1} + 6
      | c = 
}}
{{eqn | r = \underbrace {5050 \ldots 50}_{n - 1 \ 5 \text{'s} } + 6
      | c = 
}}
{{eqn | r = \underbrace {5050 \ldots 56}_{n - 1 \ 5 \text{'s} }
      | c = 
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = a - b
      | r = \sum_{k \mathop = 0}^{n - 1} \times 10^{2 k} - 11
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 1}^{n - 1} 10^{2 k} - 10
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {a - b} 2
      | r = \sum_{k \mathop = 1}^{n - 1} \frac {10^{2 k} } 2 - 5
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 1}^{n - 1} 5 \times 10^{2 k - 1} - 5
      | c = 
}}
{{eqn | r = \underbrace {5050 \ldots 50}_{n - 1 \ 5 \text{'s} } - 5
      | c = 
}}
{{eqn | r = \underbrace {5050 \ldots 45}_{n - 1 \ 5 \text{'s} }
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Category:Difference between Two Squares equal to Repunit
pop2gcap2gzk5j21keeru4ok1e9svqh
Theorem: Sixth Power as Sum of 7 Sixth Powers

The smallest known integer whose $6$th power can be expressed as the sum of $7$ smaller $6$th powers is $1141$:
:$1141^6 = 74^6 + 234^6 + 402^6 + 474^6 + 702^6 + 894^6 + 1077^6$

Proof:
{{begin-eqn}}
{{eqn | r = 74^6 + 234^6 + 402^6 + 474^6 + 702^6 + 894^6 + 1077^6
      | o = 
      | c = 
}}
{{eqn | r = 164 \, 206 \, 490 \, 176
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = 164 \, 170 \, 508 \, 913 \, 216
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = 4 \, 220 \, 426 \, 278 \, 476 \, 864
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = 11 \, 341 \, 488 \, 324 \, 787 \, 776
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = 119 \, 680 \, 300 \, 997 \, 734 \, 464
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = 510 \, 534 \, 520 \, 424 \, 456 \, 256
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = 1 \, 560 \, 609 \, 404 \, 742 \, 322 \, 089
      | c = 
}}
{{eqn | r = 2 \, 206 \, 550 \, 475 \, 483 \, 180 \, 841
      | c = 
}}
{{eqn | r = 1141^6
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Square Numbers which are Sum of Sequence of Odd Cubes

The sequence of square numbers which can be expressed as the sum of a sequence of odd cubes from $1$ begins:
:$1, 1225, 1 \, 413 \, 721, 1 \, 631 \, 432 \, 881, \dotsc$
{{OEIS|A046177}}
The sequence of square roots of this sequence is:
:$1, 35, 1189, 40 \, 391, \dotsc$
{{OEIS|A046176}}

Proof:
We have that:
{{begin-eqn}}
{{eqn | l = 1225
      | r = 35^2
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 1}^5 \paren {2 k - 1}^3 = 1^3 + 3^3 + 5^3 + 7^3 + 9^3
      | c = 
}}
{{eqn | l = 1 \, 413 \, 721
      | r = 1189^2
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 1}^{29} \paren {2 k - 1}^3 = 1^3 + 3^3 + 5^3 + \dotsb + 55^3 + 57^3
      | c = 
}}
{{end-eqn}}
From Sum of Sequence of Odd Cubes we have:
:$\ds \sum_{j \mathop = 1}^n \paren {2 j - 1}^3 = 1^3 + 3^3 + 5^3 + \dotsb + \paren {2 n − 1}^3 = n^2 \paren {2 n^2 − 1}$
Thus we need to find all $n$ such that $2 n^2 − 1$ is square.
This corresponds to the Pellian equation $x^2 - 2 y^2 = -1$, which has the positive integral solutions:
:<nowiki>$\begin {array} {r|r} x & y \\ \hline
1 & 1 \\
7 & 5 \\
41 & 29 \\
239 & 169 \\
1393 & 985 \\
\end {array}$</nowiki>
and so on.
By substituting $y = n$ and $x = \sqrt {2 n^2 - 1}$, we see that the products of $x$ and $y$ values give the required sequence of square roots.
{{qed}}
Theorem: Retract of Injective Space is Injective

Let $T = \struct {S, \tau}$ be an injective topological space.
Let $R = \struct {Z, \tau'}$ be a retract of $T$.
Then $R$ is injective.

Proof:
By definition of retract:
:there exists a continuous retraction $r: S \to Z$ of $T$.
Let $\YY = \struct {Y, \sigma}$ be a topological space.
Let $f: Y \to Z$ be a continuous mapping.
Let $\XX = \struct {X, \sigma'}$ such that
:$\YY$ is topological subspace of $\XX$.
By Inclusion Mapping is Continuous:
:$i_Z$ is continuous $\paren {R \to T}$
where $i_Z$ denotes the inclusion mapping from $Z$ in $S$.
By Composite of Continuous Mappings is Continuous:
:$i_Z \circ f: Y \to S$ is continuous.
By definition of injective space:
:there exists a continuous mapping $g: X \to S: g \restriction Y = i_Z \circ f$
Define $h := r \circ g$
By Composite of Continuous Mappings is Continuous:
:$h$ is continuous.
We will prove that
:$h \restriction Y = f$
By definition of topological subspace:
:$Y \subseteq X$ and $Z \subseteq S$
Thus by definitions of composition of mappings and restriction of mapping:
:$h \restriction Y: Y \to Z$ and $f: Y \to Z$
Let $y \in Y$.
By Restriction of Composition is Composition of Restriction:
:$h \restriction Y = r \circ \paren {g \restriction Y}$
By definition of mapping:
:$\map f y \in Z$
Thus
{{begin-eqn}}
{{eqn | l = \map {\paren {h \restriction Y} } y
      | r = \map r {\map {\paren {i_Z \circ f} } y}
      | c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = \map r {\map {i_Z} {\map f y} }
      | c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = \map r {\map f y}
      | c = {{Defof|Inclusion Mapping}}
}}
{{eqn | r = \map f y
      | c = {{Defof|Retraction (Topology)}}
}}
{{end-eqn}}
{{qed}}
Theorem: Numbers equal to Sum of Squares of two Parts

Integers that can be split into two parts whose squares add up to it include:
:$1233 = 12^2 + 33^2$
:$8833 = 88^2 + 33^2$
{{expand|Need to establish the parameters of this}}

Proof:
{{ProofWanted|Need to establish exactly what is to be proved}}
Theorem: Triples of Consecutive Sphenic Numbers

The sequence of triplets of consecutive sphenic numbers starts:
:$\tuple {1309, 1310, 1311}, \tuple {1885, 1886, 1887}, \tuple {2013, 2014, 2015}, \ldots$
{{OEIS|A066509|order = first}}
{{OEIS|A248202|order = middle}}

Proof:
Note that there cannot be quadruplets of such numbers, since one of the quadruplets must be divisible by $4$, making it non-sphenic.
We have:
{{begin-eqn}}
{{eqn | l = 1309
      | r = 7 \times 11 \times 17
}}
{{eqn | l = 1310
      | r = 2 \times 5 \times 131
}}
{{eqn | l = 1311
      | r = 3 \times 19 \times 23
}}
{{eqn | l = 1885
      | r = 5 \times 13 \times 29
}}
{{eqn | l = 1886
      | r = 2 \times 23 \times 41
}}
{{eqn | l = 1887
      | r = 3 \times 17 \times 37
}}
{{eqn | l = 2013
      | r = 3 \times 11 \times 61
}}
{{eqn | l = 2014
      | r = 2 \times 19 \times 53
}}
{{eqn | l = 2015
      | r = 5 \times 13 \times 31
}}
{{end-eqn}}
hence each number above is sphenic.
{{qed}}
Theorem: Cube Number as Sum of Three Consecutive Odd Squares

:$1331 = 11^3 = 19^2 + 21^2 + 23^2$
No other such sequence of $3$ consecutive odd squares has the same property.

Proof:
{{begin-eqn}}
{{eqn | l = 19^2 + 21^2 + 23^2
      | r = 361 + 441 + 529
      | c = 
}}
{{eqn | r = 1331
      | c = 
}}
{{end-eqn}}
Any sequence of $3$ consecutive odd integers that have squares that sum to a cube would satisfy:
:$m^3 = \paren {n - 2}^2 + n^2 + \paren {n + 2}^2$
where $n$ is the middle number of the sequence, with $m, n \in \Z$.
Expanding the {{RHS}}:
{{begin-eqn}}
{{eqn | l = m^3
      | r = \paren {n - 2}^2 + n^2 + \paren {n + 2}^2
}}
{{eqn | r = n^2 - 4 n + 4 + n^2 + n^2 + 4 n + 4
      | c = Square of Sum, Square of Difference
}}
{{eqn | r = 3 n^2 + 8
}}
{{end-eqn}}
Substituting $x = 3 m$ and $y = 9 n$:
{{begin-eqn}}
{{eqn | l = \paren {\frac x 3}^3
      | r = 3 \paren {\frac y 9}^2 + 8
}}
{{eqn | ll= \leadsto
      | l = \frac {x^3} {27}
      | r = \frac {y^2} {27} + 8
}}
{{eqn | ll= \leadsto
      | l = y^2
      | r = x^3 - 216
}}
{{end-eqn}}
which is an elliptic curve.
According to [https://www.lmfdb.org/EllipticCurve/Q/576/e/3 LMFDB], this elliptic curve has exactly $5$ lattice points:
:$\tuple {6, 0}, \tuple {10, \pm 28}, \tuple {33, \pm 189}$
which correspond to these values of $n$:
:$0, \pm \dfrac {28} 9, \pm 21$
$\pm \dfrac {28} 9$ are not integers.
The other possibilities are:
:the trivial $0 = \paren {-2}^2 + 0^2 + 2^2$
:the less trivial $11^3 = \paren {-23}^2 + \paren {-21}^2 + \paren {-19}^2$
The first is not a set of consecutive odd squares.
The second gives the same set of consecutive odd squares.
Hence there are no more solutions.
{{qed}}
Theorem: Smallest Triplet of Consecutive Integers Divisible by Cube

The smallest sequence of triplets of consecutive integers each of which is divisible by a cube greater than $1$ is:
:$\tuple {1375, 1376, 1377}$

Proof:
We will show that:
{{begin-eqn}}
{{eqn | l = 1375
      | r = 11 \times 5^3
      | c = 
}}
{{eqn | l = 1376
      | r = 172 \times 2^3
      | c = 
}}
{{eqn | l = 1377
      | r = 51 \times 3^3
      | c = 
}}
{{end-eqn}}
is the smallest such triplet.
Each number in such triplets of consecutive integers is divisible by a cube of some prime number.
Only $2, 3, 5, 7, 11$ are less than $\sqrt [3] {1377}$.
Since the numbers involved are small, we can check the result by brute force.
For general results one is encouraged to use the Chinese Remainder Theorem.
Theorem: Smallest Quadruplet of Consecutive Integers Divisible by Cube

The smallest sequence of quadruplets of consecutive integers each of which is divisible by a cube greater than $1$ is:
:$\tuple {22 \, 624, 22 \, 625, 22 \, 626, 22 \, 627}$

Proof:
{{begin-eqn}}
{{eqn | l = 22 \, 624
      | r = 2828 \times 2^3
      | c = 
}}
{{eqn | l = 22 \, 625
      | r = 181 \times 5^3
      | c = 
}}
{{eqn | l = 22 \, 626
      | r = 838 \times 3^3
      | c = 
}}
{{eqn | l = 22 \, 627
      | r = 17 \times 11^3
      | c = 
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown there is no smaller such quadruplet.}}
Theorem: Riemann Zeta Function as a Multiple Integral

For $n \in \Z_{> 0}$, the Riemann zeta function is given by:
:$\ds \map \zeta n = \int_{\closedint 0 1^n} \frac 1 {1 - \prod_{i \mathop = 1}^n x_i} \prod_{i \mathop = 1}^n \rd x_i$
where $\closedint 0 1^n$ denotes the Cartesian $n$th power of the closed real interval $\closedint 0 1$.

Proof:
{{begin-eqn}}
{{eqn | l = \int_{\closedint 0 1^n} \frac 1 {1 - \prod_{i \mathop = 1}^n x_i} \prod_{i \mathop = 1}^n \rd x_i 
      | r = \int_{\closedint 0 1^n} \sum_{j \mathop = 1}^\infty \paren {\prod_{i \mathop = 1}^n x_i}^{j - 1} \prod_{i \mathop = 1}^n \rd x_i 
      | c = Sum of Infinite Geometric Sequence
}}
{{eqn | r = \sum_{j \mathop = 1}^\infty \int_{\closedint 0 1^n}\prod_{i \mathop = 1}^n {x_i}^{j - 1} \prod_{i \mathop = 1}^n \rd x_i 
      | c = Integral of Series of Positive Measurable Functions, Fubini's Theorem
}}
{{eqn | r = \sum_{j \mathop = 1}^\infty \prod_{i \mathop = 1}^n \int_0^1 x^{j - 1}_i \rd x_i 
      | c = Fubini's Theorem
}}
{{eqn | r = \sum_{j \mathop = 1}^\infty \prod_{i \mathop = 1}^n \intlimits {\frac {x^j} j } 0 1
      | c = Integral of Power
}}
{{eqn | r = \sum_{j \mathop = 1}^\infty \prod_{i \mathop = 1}^n \frac 1 j
      | c = 
}}
{{eqn | r = \sum_{j \mathop = 1}^\infty \frac 1 {j^n}
}}
{{eqn | r = \map \zeta n
      | c = {{Defof|Riemann Zeta Function}}
}}
{{end-eqn}}
{{qed}}
Category:Riemann Zeta Function
Category:Analytic Number Theory
fe19gzhjlg5worf5r3c8gw3le9iukjs
Theorem: Closed Form for Hexagonal Pyramidal Numbers

The closed-form expression for the $n$th hexagonal pyramidal number is:
:$S_n = \dfrac {n \paren {n + 1} \paren {4 n - 1} } 6$

Proof:
{{begin-eqn}}
{{eqn | l = S_n
      | r = \sum_{k \mathop = 1}^n H_k
      | c = {{Defof|Hexagonal Pyramidal Number}}
}}
{{eqn | r = \sum_{k \mathop = 1}^n k \paren {2 k - 1}
      | c = Closed Form for Hexagonal Numbers
}}
{{eqn | r = 3 \sum_{k \mathop = 1}^n 2 k^2 - \sum_{k \mathop = 1}^n k
      | c = 
}}
{{eqn | r = \frac {2 n \paren {n + 1} \paren {2 n + 1} } 6 - \sum_{k \mathop = 1}^n k
      | c = Sum of Sequence of Squares
}}
{{eqn | r = \frac {2 n \paren {n + 1} \paren {2 n + 1} } 6 - \dfrac {n \paren {n + 1} } 2
      | c = Closed Form for Triangular Numbers
}}
{{eqn | r = \frac {2 n \paren {n + 1} \paren {2 n + 1} - 3 n \paren {n + 1} } 6
      | c = 
}}
{{eqn | r = \frac {n \paren {n + 1} \paren {4 n + 2 - 3} } 6
      | c = 
}}
{{eqn | r = \frac {n \paren {n + 1} \paren {4 n - 1} } 6
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Closed Forms
Category:Pyramidal Numbers
1b9e5x8004owfzmnt6dfbmrdwpbcibr
Theorem: Tetrahedral and Triangular Numbers

The only positive integers which are simultaneously tetrahedral and triangular are:
:$1, 10, 120, 1540, 7140$

Proof:
{{begin-eqn}}
{{eqn | l = 1
      | r = \dfrac {1 \paren {1 + 1} \paren {1 + 2} } 6
      | c = Closed Form for Tetrahedral Numbers
}}
{{eqn | r = \dfrac {1 \times \paren {1 + 1} } 2
      | c = Closed Form for Triangular Numbers
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 10
      | r = \dfrac {3 \paren {3 + 1} \paren {3 + 2} } 6
      | c = Closed Form for Tetrahedral Numbers
}}
{{eqn | r = \dfrac {4 \times \paren {4 + 1} } 2
      | c = Closed Form for Triangular Numbers
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 120
      | r = \dfrac {8 \paren {8 + 1} \paren {8 + 2} } 6
      | c = Closed Form for Tetrahedral Numbers
}}
{{eqn | r = \dfrac {15 \times \paren {15 + 1} } 2
      | c = Closed Form for Triangular Numbers
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 1540
      | r = \dfrac {20 \paren {20 + 1} \paren {20 + 2} } 6
      | c = Closed Form for Tetrahedral Numbers
}}
{{eqn | r = \dfrac {55 \times \paren {55 + 1} } 2
      | c = Closed Form for Triangular Numbers
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 7140
      | r = \dfrac {34 \paren {34 + 1} \paren {34 + 2} } 6
      | c = Closed Form for Tetrahedral Numbers
}}
{{eqn | r = \dfrac {119 \times \paren {119 + 1} } 2
      | c = Closed Form for Triangular Numbers
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown that these are the only such instances.}}
Theorem: Restriction of Composition is Composition of Restriction

Let $X, Y, Z$ be sets.
Let $f: X \to Y$ and $g: Y \to Z$ be mappings.
Let $S \subseteq X$.
Then:
:$\paren {g \circ f} \restriction S = g \circ \paren {f \restriction S}$

Proof:
By definitions of composition of mappings and restriction of mapping:
:$\paren {g \circ f} \restriction S: S \to Z$ and $g \circ \paren {f \restriction S}: S \to Z$
Let $s \in S$.
By definition of restriction of mapping:
:$\map {\paren {\paren {g \circ f} \restriction S} } s = \map {\paren {g \circ f} } s$
Thus
{{begin-eqn}}
{{eqn | l = \map {\paren {g \circ \paren {f \restriction S} } } s
      | r = \map g {\map {\paren {f \restriction S} } s}
      | c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = \map g {\map f s}
      | c = {{Defof|Restriction of Mapping}}
}}
{{eqn | r = \map {\paren {\paren {g \circ f} \restriction S} } s
      | c = {{Defof|Composition of Mappings}}
}}
{{end-eqn}}
{{qed}}
Theorem: One-Digit Number is Harshad

Let $n$ be a $1$-digit positive integer.
Then $n$ is a harshad number.

Proof:
By definition, a harshad number is divisible by the sum of its digits base $10$.
Let $n$ be a $1$-digit positive integer.
The sum of the digits of $n$ is trivially $n$.
The result follows from Integer Divides Itself.
{{qed}}
Category:Harshad Numbers
kxykvvanwxcz0ibl71hfbpqgniu0v7p
Theorem: Smallest Fermat Pseudoprime to Bases 2, 3 and 5

The smallest Fermat pseudoprime to bases $2$, $3$ and $5$ is $1729$.

Proof:
{{ProofWanted|We have the list of Poulet numbers and Fermat pseudoprimes base $3$, but not of base $5$. Once we get that list, we can find the numbers on the list for both.}}
Theorem: 1782 is 3 Times Sum of all 2-Digit Numbers from its Digits

$1782$ equals $3$ multiplied by the sum of all the $2$-digit integers that can be formed from its digits.

Proof:
The number of $2$-digit integers that can be formed from the digits of $1782$ equals the number of $2$-permutations of $\set {1, 7, 8, 2}$.
That is:
:$\set {17, 18, 12, 71, 78, 72, 81, 87, 82, 21, 27, 28}$
Hence:
:$17 + 18 + 12 + 71 + 78 + 72 + 81 + 87 + 82 + 21 + 27 + 28 = 594 = \dfrac {1782} 3$
{{qed}}
Theorem: Triple of Consecutive Happy Numbers

The smallest triple of consecutive integers all of which are happy is:
:$\left({1880, 1881, 1882}\right)$

Proof:
{{begin-eqn}}
{{eqn | o = 
      | r = 1880
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 1^2 + 8^2 + 8^2 + 0^2
      | r = 1 + 64 + 64 + 0
      | c = 
}}
{{eqn | r = 129
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 1^2 + 2^2 + 9^2
      | r = 1 + 4 + 81
      | c = 
}}
{{eqn | r = 86
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 8^2 + 6^2
      | r = 64 + 36
      | c = 
}}
{{eqn | r = 100
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 1^2 + 0^2 + 0^2
      | r = 1 + 0 + 0
      | c = 
}}
{{eqn | r = 1
      | c = and so $1880$ is happy
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | o = 
      | r = 1881
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 1^2 + 8^2 + 8^2 + 1^2
      | r = 1 + 64 + 64 + 1
      | c = 
}}
{{eqn | r = 130
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 1^2 + 3^2 + 0^2
      | r = 1 + 9 + 0
      | c = 
}}
{{eqn | r = 10
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 1^2 + 0^2
      | r = 1 + 0
      | c = 
}}
{{eqn | r = 1
      | c = and so $1881$ is happy
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | o = 
      | r = 1882
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 1^2 + 8^2 + 8^2 + 2^2
      | r = 1 + 64 + 64 + 4
      | c = 
}}
{{eqn | r = 133
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 1^2 + 3^2 + 3^2
      | r = 1 + 9 + 9
      | c = 
}}
{{eqn | r = 19
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 1^2 + 9^2
      | r = 1 + 81
      | c = 
}}
{{eqn | r = 82
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 8^2 + 2^2
      | r = 64 + 4
      | c = 
}}
{{eqn | r = 68
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 6^2 + 8^2
      | r = 36 + 64
      | c = 
}}
{{eqn | r = 100
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 1^2 + 0^2 + 0^2
      | r = 1 + 0 + 0
      | c = 
}}
{{eqn | r = 1
      | c = and so $1882$ is happy
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown that this is the smallest such triple}}
Theorem: Numbers whose Digits are Unchanged when Subtracting Reversal

The following sequence consists of the integers which have the property that subtraction of their reversals results in anagrams of them:
:$954, 1980, 2961, 3870, 5823, 7641, 9108, 19980, 29880, 29961, 32760, \ldots$
{{OEIS|A121969}}

Proof:
{{begin-eqn}}
{{eqn | l = 954 - 459
      | r = 495
}}
{{eqn | l = 1980 - 0891
      | r = 1089
}}
{{eqn | l = 2961 - 1692
      | r = 1269
}}
{{eqn | l = 3870 - 0783
      | r = 3087
}}
{{eqn | l = 5823 - 3285
      | r = 2538
}}
{{eqn | l = 7641 - 1467
      | r = 6174
}}
{{eqn | l = 9108 - 8019
      | r = 1089
}}
{{eqn | l = 19 \, 980 - 08 \, 991
      | r = 10 \, 989
}}
{{eqn | l = 29 \, 880 - 08 \, 892
      | r = 20 \, 988
}}
{{eqn | l = 29 \, 961 - 16 \, 992
      | r = 12 \, 969
}}
{{eqn | l = 32 \, 760 - 06 \, 732
      | r = 26 \, 037
}}
{{end-eqn}}
{{ProofWanted|that there are no more in between these}}