avewright/proofwiki-math huggingface dataset:
Theorem: Number of Fibonacci Numbers with Same Number of Decimal Digits

Let $n$ be an integer such that $n > 1$.
When expressed in decimal notation, there are either $4$ or $5$ Fibonacci numbers with $n$ digits.

Proof:
Let $F_m$ be an $n$-digit Fibonacci number.
Then $F_m \ge 10^{n - 1} \ge 10$.
We have:
{{begin-eqn}}
{{eqn | l = F_m
      | r = F_{m - 1} + F_{m - 2}
      | c = {{Defof|Fibonacci Numbers}}
}}
{{eqn | o = \le
      | r = 2 F_{m - 1}
      | c = as $F_{m - 2} \le F_{m - 1}$
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = F_{m + 5}
      | r = F_{m - 1} F_5 + F_m  F_6
      | c = Honsberger's Identity
}}
{{eqn | r = 5 F_{m - 1} + 8 F_m
      | c =
}}
{{eqn | o = >
      | r = 2 F_m + 8 F_m
      | c = $F_m \le 2 F_{m - 1}$
}}
{{eqn | o = \ge
      | r = 10^n
      | c =
}}
{{end-eqn}}
So $F_{m + 5}$ is a number of at least $n + 1$ digits.
Therefore there is at most $5$ Fibonacci numbers with $n$ digits: $F_m, F_{m + 1}, F_{m + 2}, F_{m + 3}, F_{m + 4}$.
Similarly, $10 \le F_m < 10^n$.
Suppose $F_m$ is the smallest Fibonacci numbers with $n$ digits.
Then $F_{m - 1} < 10^{n - 1}$.
Thus:
{{begin-eqn}}
{{eqn | l = F_{m + 3}
      | r = F_{m - 1} F_3 + F_m F_4
      | c = Honsberger's Identity
}}
{{eqn | r = 2 F_{m - 1} + 3 F_m
      | c =
}}
{{eqn | o = <
      | r = 8 F_{m - 1}
      | c = $F_m \le 2 F_{m - 1}$
}}
{{eqn | o = <
      | r = 10^n
      | c =
}}
{{end-eqn}}
So there is at least $4$ Fibonacci numbers with $n$ digits: $F_m, F_{m + 1}, F_{m + 2}, F_{m + 3}$.
Combining the arguments above, there are either $4$ or $5$ Fibonacci numbers with $n$ digits.
{{qed}}
Theorem: Prime Number divides Infinite Number of Fibonacci Numbers

Let $p$ be a prime number.
Then there exist an infinite number of Fibonacci numbers which are divisible by $p$.

Proof:
From Prime Number divides Fibonacci Number, either $F_{p - 1}$ or $F_{p + 1}$ is divisible by $p$.
Thus:
{{begin-eqn}}
{{eqn | l = p
      | o = \divides
      | r = F_{p \pm 1}
      | c = Prime Number divides Fibonacci Number
}}
{{eqn | q = \forall n \in \Z_{>0}
      | l = F_{p \pm 1}
      | o = \divides
      | r = F_{n \paren {p \pm 1} }
      | c = Divisibility of Fibonacci Number
}}
{{eqn | ll= \leadsto
      | q = \forall n \in \Z_{>0}
      | l = p
      | o = \divides
      | r = F_{n \paren {p \pm 1} }
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Prime Number divides Fibonacci Number

For $n \in \Z$, let $F_n$ denote the $n$th Fibonacci number.
Let $p$ be a prime number.
Then:
:$p \equiv \pm 1 \pmod 5 \implies p \divides F_{p - 1}$
:$p \equiv \pm 2 \pmod 5 \implies p \divides F_{p + 1}$
where $\divides$ denotes divisibility.
Thus in all cases, except where $p = 5$ itself:
:$p \divides F_{p \pm 1}$

Proof:
It is worth noting the one case where $p = 5$:
:$5 \divides F_5 = 5$
{{ProofWanted|Googling around suggests there is a proof based on the Law of Quadratic Reciprocity but I have not laid hands on it yet.}}
Theorem: Existence of Fibonacci Number Divisible by Number

Let $m \in \Z_{\ne 0}$ be an integer.
Then in the first $m^2$ Fibonacci numbers there exists at least one Fibonacci number which is divisible by $m$.

Proof:
Consider pairs of Fibonacci numbers $\tuple {F_i, F_{i + 1}}$ modulo $m$.
There are $m^2$ possible pairs of remainders.
Thus by Pigeonhole Principle, among the $\paren {m^2 + 1}$ pairs of $\tuple {F_i, F_{i + 1}}$ with $0 \le i \le m^2$, at least two of them are identical modulo $m$.
That is, there exists $0 \le i < j \le m^2$ such that:
:$\tuple {F_i, F_{i + 1}} \equiv \tuple {F_j, F_{j + 1}} \pmod m$
It is trivial to then show by induction that:
:$F_{i - k} \equiv F_{j - k} \pmod m$
for any integer $k$.
In particular:
:$F_{j - i} \equiv F_{i - i} = F_0 = 0 \pmod m$
so $m \divides F_{j - i}$.
Note that $0 < j - i \le m^2$.
Hence the result.
{{qed}}
Theorem: Fibonacci Prime has Prime Index except for 3

Let $F_n$ denote the $n$th Fibonacci number.
Let $F_n$ be a prime number.
Then, apart from $F_4 = 3$, $n$ is a prime number.

Proof:
Let $F_n$ be a prime number.
{{AimForCont}} $n$ is a composite number greater than $4$.
Then $n = r s$ for some $1 < r, s < n$.
Except for the case where $n = 4$, at least one of $r$ and $s$ is greater than $2$.
From Divisibility of Fibonacci Number:
:$F_r \divides F_n$
and:
:$F_s \divides F_n$
where $\divides$ denotes divisibility.
When $k > 2$ we have that $F_k > 1$.
Thus when $n$ is composite such that $n > 4$, $F_n$ has at least one proper divisor.
That is, $F_n$ is not prime.
Thus by Proof by Contradiction, $n$ cannot be composite.
The exception is when $n = 4$, as noted above in which case its only proper divisor is $2$.
But $F_2 = 1$, and so divisibility by $F_2$ does not preclude primality.
Indeed, $F_4 = 3$, which is prime.
{{qed}}
Theorem: Fibonacci Number with Prime Index is not necessarily Prime

Let $p \in \Z_{>0}$ be a prime number.
Let $F_p$ be the $p$th Fibonacci number.
Then $F_p$ is not itself necessarily prime.

Proof:
Proof by Counterexample:
:$F_{19} = 4181 = 37 \times 113$
{{qed}}
Theorem: Complex Riemann Integral is Contour Integral

Let $f: \R \to \C$ be a complex Riemann integrable function over some closed real interval $\closedint a b$.
Then:
:$\ds \int_a^b \map f t \rd t = \int_\CC \map f t \rd t$
where:
:the integral on the {{LHS}} is a complex Riemann integral
:the integral on the {{RHS}} is a contour integral
:$\CC$ is a straight line segment along the real axis, connecting $a$ to $b$.

Proof:
{{begin-eqn}}
{{eqn | l = \int_a^b \map f t \rd t
      | r = \int_a^b \map f {\map \theta t} \map {\theta'} t \rd t
      | c = Complex Integration by Substitution: $\map \theta t = t$, $\map {\theta'} t = 1$
}}
{{eqn | r = \int_\CC \map f t \rd t
      | c = {{Defof|Complex Contour Integral}}
}}
{{end-eqn}}
{{qed}}
Category:Integral Calculus
Category:Complex Contour Integrals
84vwhluqerhf0drtj37uy8i2gryfv6q
Theorem: Fibonacci Number is not Product of Two Smaller Fibonacci Numbers

Let $m, n \in \Z$ be integers.
Suppose $\size m, \size n \ge 3$.
Let $F_m$ and $F_n$ be the $m$th and $n$th Fibonacci numbers.
Then $F_m \times F_n$ is not a Fibonacci number.

Proof:
From Honsberger's Identity:
:$F_n = F_{k - 1} F_{n - k + 2} + F_{k - 2} F_{n - k + 1}$
for $2 \le k \le n$.
{{AimForCont}} $F_n = F_m F_k$ for some $m, k \ge 3$.
Then:
{{begin-eqn}}
{{eqn | l = F_m
      | r = \dfrac {F_n} {F_k}
      | c = 
}}
{{eqn | r = \dfrac {F_{k - 1} F_{n - k + 2} + F_{k - 2} F_{n - k + 1} } {F_k}
      | c = 
}}
{{eqn | r = \paren {\dfrac {F_{k - 1} } {F_{k - 1} + F_{k - 2} } } F_{n - k + 2} + \paren {\dfrac {F_{k - 2} } {F_{k - 1} + F_{k - 2} } } F_{n - k + 1} 
      | c = 
}}
{{end-eqn}}
The {{RHS}} is a weighted mean of $2$ consecutive Fibonacci numbers.
Thus:
:$F_{n - k + 2} < F_m < F_{n - k + 1}$
which cannot happen.
The result follows by Proof by Contradiction.
{{qed}}
Theorem: Transformation of P-Norm

Let $p, q \ge 1$ be real numbers.
Let ${\ell^p}_\R$ denote the $p$-sequence space on $\R$.
Let $\norm {\mathbf x}_p$ denote the $p$-norm of $\mathbf x$.
Let $\mathbf x = \sequence {x_n} \in {\ell^{p q} }_\R$.
Suppose further that $\mathbf x^p = \sequence { {x_n}^p} \in {\ell^q}_\R$.
Then:
:$\norm {\mathbf x^p}_q = \paren {\norm {\mathbf x}_{p q} }^p$

Proof:
{{begin-eqn}}
{{eqn | l = \norm {\mathbf x^p}_q
      | r = \paren {\sum_{n \mathop = 0}^\infty \size { {x_n}^p}^q}^{1 / q}
      | c = {{Defof|P-Norm|$p$-Norm}}
}}
{{eqn | r = \paren {\sum_{n \mathop = 0}^\infty \size {x_n}^{p q} }^{1 / q}
      | c = Power of Power
}}
{{eqn | r = \paren {\paren {\sum_{n \mathop = 0}^\infty \size {x_n}^{p q} }^{1 / p q} }^p
      | c = Power of Power
}}
{{eqn | r = \paren {\norm {\mathbf x}_{p q} }^p
      | c = {{Defof|P-Norm|$p$-Norm}}
}}
{{end-eqn}}
{{qed}}
Category:Functional Analysis
Category:Norm Theory
Category:P-Sequence Metrics
Category:P-Norms
h8bx5fa5kmkw1ry9h27wzpli6f26fud
Theorem: Set is Subset of Finite Suprema Set

Let $\struct {S, \preceq}$ be an ordered set.
Let $X$ be a subset of $S$.
Then $X \subseteq \map {\mathrm {finsups} } X$
where $\map {\mathrm {finsups} } X$ denotes finite suprema set of $X$.

Proof:
Let $x \in X$.
By Supremum of Singleton:
:$\set x$ admits a supremum and $\sup \set x = x$
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 suprema set:
:$x \in \map {\mathrm {finsups} } X$
{{qed}}
Theorem: Lower Closure of Subset is Subset of Lower Closure

Let $\struct {S, \preceq}$ be an ordered set.
Let $X, Y$ be subsets of $S$.
Then
:$X \subseteq Y \implies X^\preceq \subseteq Y^\preceq$
where $X^\preceq$ is the lower closure of $X$.

Proof:
Let $X \subseteq Y$.
Let $x \in X^\preceq$.
By definition of lower closure of subset:
:$\exists y \in X: x \preceq y$
By definition of subset:
:$y \in Y$
Thus by definition of lower closure of subset:
:$x \in Y^\preceq$
{{qed}}
Theorem: Finite Suprema Set and Lower Closure is Smallest Ideal

Let $L = \struct {S, \vee, \preceq}$ be a join semilattice.
Let $X$ be a subset of $S$.
Then $X \subseteq \map {\operatorname {finsups} } X^\preceq$ and
:for every ideal $I$ in $L$: $X \subseteq I \implies \map {\operatorname {finsups} } X^\preceq \subseteq I$
where
:$\map {\operatorname {finsups} } X$ denotes the finite suprema set of $X$,
:$X^\preceq$ denotes the lower closure of $X$.

Proof:
By Set is Subset of Finite Suprema Set:
:$X \subseteq \map {\operatorname {finsups} } X$
By Lower Closure of Subset is Subset of Lower Closure:
:$X^\preceq \subseteq \map {\operatorname {finsups} } X^\preceq$
By Set is Subset of Lower Closure:
:$X \subseteq X^\preceq$
Thus by Subset Relation is Transitive:
:$X \subseteq \map {\operatorname {finsups} } X^\preceq$
Let $I$ be a ideal in $L$ such that
:$X \subseteq I$
Let $x \in \map {\operatorname {finsups} } X^\preceq$
By definition of lower closure of subset:
:$\exists y \in \map {\operatorname {finsups} } X: x \preceq y$
By definition of finite suprema set:
:$\exists A \in \map {\operatorname {Fin} } X: y = \sup A \land A$ admits a supremum
where $\map {\operatorname {Fin} } X$ denotes the set of all finite subsets of $S$.
By Subset Relation is Transitive:
:$A \subseteq I$
By Directed in Join Semilattice with Finite Suprema:
:$A \ne \O \implies \sup A \in I$
By Supremum of Empty Set is Smallest Element and Bottom in Ideal:
:$A = \O \implies \sup A = \bot \in I$
where $\bot$ denotes the smallest element of $S$.
So
:$y \in I$
Thus by definition of lower section:
:$x \in I$
{{qed}}
Theorem: Set is Subset of Lower Closure

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

Proof:
Let $x \in X$.
By definition of reflexivity:
:$x \preceq x$
Thus by definition of lower closure:
:$x \in X^\preceq$
{{qed}}
Theorem: Maximum Volume of Unit Radius Sphere in Fractional Dimensions

The maximum volume of a unit sphere in $x$-dimensional Euclidean space for real $x$ occurs when $x$ is given as:
:$x = 5 \cdotp 25694 \, 64048 \, 60 \ldots$
{{OEIS|A074455}}
The corresponding volume at that dimension is given by:
:$V = 5 \cdotp 27776 \, 80211 \, 13400 \, 997 \ldots$
{{OEIS|A074454}}

Proof:
{{ProofWanted|Requires significant work to even define the concepts.}}
Theorem: Even Perfect Number is Sum of Successive Odd Cubes except 6

Let $n$ be an even perfect number such that $n \ne 6$.
Then:
:$\ds n = \sum_{k \mathop = 1}^m \paren {2 k - 1}^3 = 1^3 + 3^3 + \cdots + \paren {2 m - 1}^3$
for some $m \in \Z_{>0}$.
That is, every even perfect number apart from $6$ is the sum of the sequence of the first $m$ odd cubes, for some $m$.

Proof:
From Sum of Sequence of Odd Cubes:
:$1^3 + 3^3 + 5^3 + \cdots + \paren {2 m − 1}^3 = m^2 \paren {2 m^2 − 1}$
By the Theorem of Even Perfect Numbers:
:$n = 2^{p - 1} \paren {2^p - 1}$
for some $p$, where $2^p - 1$ is prime.
From Primes of form Power Less One, it follows that $p$ is itself prime.
Let $p$ be an odd prime.
Then we have that $p \ge 3$.
It follows that $p - 1$ is even.
Hence $\dfrac {p - 1} 2$ is an integer.
Thus we see:
:$2^{p - 1} = \paren {2^{\frac {p - 1} 2} }^2$
Setting $m = 2^{\frac {p - 1} 2}$:
:$m^2 = 2^{p - 1}$
and so:
:$2 m^2 = 2^p$
and it follows that:
:$\ds n = \sum_{k \mathop = 1}^{2^{\frac {p - 1} 2} } \paren {2 k - 1}^3$
and hence the result.
{{qed|lemma}}
Let $p$ be an even prime.
That is:
:$p = 2$
This leads to the perfect number $n = 6$, where we have:
:$6 = 2^1 \paren {2^2 - 1}$
However, as $p - 1$ is now odd, it is no longer the case that $\dfrac {p - 1} 2$ is an integer.
At this point the formula fails to work.
{{qed}}
Theorem: Rational Power is of Exponential Order Epsilon

Let $r = \dfrac p q$ be a rational number, with $p, q \in \Z: q \ne 0, r > 0$.
Then: 
:$t \mapsto t^r$
is of exponential order $\epsilon$ for any  $\epsilon > 0$ arbitrarily small in magnitude.

Proof:
Write $t^r = t^{p/q}$, and set $t > 1$.
{{begin-eqn}}
{{eqn | l  = t^{p/q}
      | o = <
      | r = K e^{a t}
      | c = an Ansatz
}}
{{eqn | ll = \impliedby
      | l = t^p
      | o = <
      | r = \paren {K e^{a t} }^q
      | c = Rational Power is Strictly Increasing
}}
{{eqn | r = K^q e^{q a t}
      | c = Exponential of Product
}}
{{eqn | r =  K' e^{a' t}
      | c = $K^q = K', q a = a'$
}}
{{end-eqn}}
Recall from Polynomial is of Exponential Order Epsilon, $t^p < K'e^{a't}$ for ''any'' $a' > 0$, arbitrarily small in magnitude.
Therefore the inequality $t^{p/q} < K e^{a t}$ has solutions of the same nature.
{{qed}}
Category:Exponential Order
e4d45d6d8d4cdizv1nso1gtcjr75utz
Theorem: Sum of Sequence of Odd Cubes

:$\ds \sum_{r \mathop = 1}^n \paren {2 r - 1}^3 = 1^3 + 3^3 + 5^3 + \dotsb + \paren {2 n − 1}^3 = n^2 \paren {2 n^2 − 1}$

Proof:
{{begin-eqn}}
{{eqn | n = d
      | l = S_n
      | o = :=
      | r = \sum_{r \mathop = 1}^n r^3
      | c = 
}}
{{eqn | r = 1^3 + 2^3 + 3^3 + \cdots + \paren {n - 1}^3 + n^3
      | c = 
}}
{{eqn | n = d
      | l = O_n
      | o = :=
      | r = \sum_{r \mathop = 1}^n \paren {2 r - 1}^3
      | c = 
}}
{{eqn | r = 1^3 + 3^3 + 5^3 + \cdots + \paren {n - 1}^3 + \paren {2 n - 1}^3
      | c = 
}}
{{eqn | n = t
      | l = O_n
      | r = S_{2 n} - \paren {2^3 + 4^3 + 6^3 + \cdots + \paren {2 n - 2}^3 + \paren {2 n}^3}
      | c = 
}}
{{eqn | r = S_{2 n} - 2^3 \paren {1^3 + 2^3 + 3^3 + \cdots + \paren {n - 1}^3 + n^3}
      | c = 
}}
{{eqn | r = S_{2 n} - \sum_{r \mathop = 1}^n \paren {2 r}^3
      | c = 
}}
{{eqn | r = S_{2 n} - 8 S_n
      | c = 
}}
{{eqn | r = \frac {\paren {2 n}^2 \paren {2 n + 1}^2} 4 - 8 \frac {n^2 \paren {n + 1}^2} 4
      | c = Sum of Sequence of Cubes
}}
{{eqn | r = n^2 \paren {2 n + 1}^2 - 2 n^2 \paren {n + 1}^2
      | c = simplifying
}}
{{eqn | r = n^2 \paren {\paren {2 n + 1}^2 - 2 \paren {n + 1}^2}
      | c = factoring
}}
{{eqn | r = n^2 \paren {4 n^2 + 4 n + 1 - 2 n^2 - 4 n - 2}
      | c = multiplying out
}}
{{eqn | r = n^2 \paren {2 n^2 - 1}
      | c = simplifying
}}
{{end-eqn}}
Therefore:
:$\ds \forall n \in \Z_{> 0}: \sum_{r \mathop = 1}^n \paren {2 r - 1}^2 = n^2 \paren {2 n^2 − 1}$
{{qed}}
Theorem: Bottom in Ideal

Let $\struct {S, \preceq}$ be a bounded below ordered set.
Let $I$ be a ideal in $S$.
Then $\bot \in I$
where $\bot$ denotes the smallest element of $S$.

Proof:
By definition of ideal in ordered set:
:$I$ is non-empty and lower.
By definition of non-empty set:
:$\exists x: x \in I$
By definition of smallest element:
:$\bot \preceq x$
Thus by definition of lower section:
:$\bot \in I$
{{qed}}
Theorem: Real Power is of Exponential Order Epsilon

Let: 
:$f: \hointr 0 \to \to \R: t \mapsto t^r$
be $t$ to the power of $r$, for $r \in \R, r > -1$.
Then $f$ is of exponential order $\epsilon$ for any  $\epsilon > 0$ arbitrarily small in magnitude.

Proof:
For $t > 0$, $t^r$ is continuous.
At $t = 0$, defining $0^r = 0$, the function is continuous from the right:
{{begin-eqn}}
{{eqn | l = \lim_{t \mathop \to 0^+} t^r
      | r = \lim_{t \mathop \to 0^+} \map \exp {r \ln t}
}}
{{eqn | r = \map \exp {\lim_{t \mathop \to 0^+} r \ln t}
      | c = Exponential Function is Continuous
}}
{{eqn | r = 0
      | c = Logarithm Tends to Negative Infinity, Exponential Tends to Zero and Infinity
}}
{{end-eqn}}
Fix $t > 1$.
By the Axiom of Archimedes, there is a natural number $n$ such that $n > r$.
By Real Power Function on Base Greater than One is Strictly Increasing, $t^r < t^n$.
Recall from Polynomial is of Exponential Order Epsilon, $t^n < K e^{at}$ for ''any'' $a > 0$, arbitrarily small in magnitude.
Therefore the inequality $t^r < K e^{at}$ has solutions of the same nature.
{{qed}}
Category:Exponential Order
k6lpt5klkwg2bylpky9rx0nswn6ryhe
Theorem: Finite Suprema Set and Lower Closure is Ideal

Let $P = \struct {S, \vee, \preceq}$ be a join semilattice.
Let $X$ be a non-empty subset of $S$.
Then
:$\map {\operatorname{finsups} } X^\preceq$ is ideal in $P$.
where
:$\map {\operatorname{finsups} } X$ denotes the finite suprema set of $X$,
:$X^\preceq$ denotes the lower closure of $X$.

Proof:
By Finite Suprema Set and Lower Closure is Smallest Ideal:
:$X \subseteq \map {\operatorname{finsups} } X^\preceq$
By definition of non-empty set:
:$\map {\operatorname{finsups} } X^\preceq$ is a non-empty set.
We will prove that
:$\map {\operatorname{finsups} } X$ is directed.
Let $x, y \in \map {\operatorname{fininfs} } X$
By definition of finite suprema set:
:$\exists A \in \map {\operatorname {Fin} } X: x = \sup A \land A$ admits a supremum
and
:$\exists B \in \map {\operatorname {Fin} } X: y = \sup B \land B$ admits an supremum
where $\map {\operatorname {Fin} } X$ 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
:$C \in \map {\operatorname {Fin} } X$
By Existence of Non-Empty Finite Suprema in Join Semilattice:
:$C \ne \O \implies C$ admits a supremum.
By Union is Empty iff Sets are Empty:
:$C = \O \implies A = \O$
So
:$C$ admits a supremum.
By definition of finite suprema set:
:$\sup C \in \map {\operatorname{finsups} } X$
By Set is Subset of Union:
:$A \subseteq C$ and $B \subseteq C$
Thus by Supremum of Subset:
:$x \preceq \sup C$ and $y \preceq \sup C$
Hence $\map {\operatorname{finsups} } X$ is directed.
{{qed|lemma}}
By Directed iff Lower Closure Directed:
:$\map {\operatorname{finsups} } X^\preceq$ is directed.
By Lower Closure is Lower Section:
:$\map {\operatorname{finsups} } X^\preceq$ is lower.
Hence $\map {\operatorname{finsups}} X^\preceq$ is ideal in $P$.
{{qed}}
Theorem: Set of 3 Integers each Divisor of Sum of Other Two

There exists exactly one set of distinct coprime positive integers such that each is a divisor of the sum of the other two:
:$\set {1, 2, 3}$

Proof:
We note that if $\set {a, b, c}$ is such a set, then $\set {k a, k b, k c}$ satisfy the same properties trivially.
Hence the specification that $\set {a, b, c}$ is a coprime set.
We have that:
:$5 \times 1 = 2 + 3$ so $1 \divides 2 + 3$
:$2 \times 2 = 1 + 3$ so $2 \divides 1 + 3$
:$1 \times 3 = 1 + 2$ so $3 \divides 1 + 2$
It remains to be shown that this is the only such set.
We are to find all the sets $\set {a, b, c}$ such that:
:$a \divides b + c$
:$b \divides a + c$
:$c \divides a + b$
where $a \ne b$, $a \ne c$ and $b \ne c$.
{{WLOG}}, suppose $a < b < c$.
Since $2 c > a + b > 0$ and $c \divides a + b$, we must have $a + b = c$.
Then it follows that:
:$a \divides \paren {b + a + b}$
:$b \divides \paren {a + a + b}$
which reduces to:
:$a \divides 2 b$
:$b \divides 2 a$
Suppose $b$ is odd.
Then by Euclid's Lemma, we would have $b \divides a$.
By Absolute Value of Integer is not less than Divisors, this gives $b \le a$, which is a contradiction.
Thus $b$ is even.
Suppose $a$ is even.
Then $a, b, c$ are all even.
So $\gcd \set {a, b, c} \ne 1$, which is a contradiction.
Therefore it must be the case that $a$ is odd.
Then by Euclid's Lemma, we have:
:$a \divides \dfrac b 2$
and:
:$\dfrac b 2 \divides a$
By Absolute Value of Integer is not less than Divisors, this gives:
:$\dfrac b 2 = a$
Because $\gcd \set {a, b, c} = 1$, we must have $a = 1$.
Hence the set $\set {1, 2, 3}$ is obtained.
{{qed}}
Theorem: Only Number which is Sum of 3 Factors is 6

The only positive integer which is the sum of exactly $3$ of its distinct coprime divisors is $6$.

Proof:
Let $n$ be such a positive integer with corresponding divisors $a, b, c$ such that:
:$a + b + c = n$
We note that the set $\set {k a, k b, k c}$ satisfy the same properties trivially as divisors of $k n$.
Hence the specification that $\set {a, b, c}$ is a coprime set.
{{WLOG}}, suppose $a < b < c$.
Since $a, b, c$ are strictly positive, $n \ne c$.
Suppose $\dfrac n c \ge 3$.
Then:
{{begin-eqn}}
{{eqn | l = n
      | r = a + b + c
}}
{{eqn | o = <
      | r = c + c + c
}}
{{eqn | r = n
}}
{{end-eqn}}
which is a contradiction.
Hence:
{{begin-eqn}}
{{eqn | l = \frac n c
      | r = 2
}}
{{eqn | ll = \leadsto
      | l = a + b + c
      | r = 2 c
}}
{{eqn | ll = \leadsto
      | l = a + b
      | r = c
}}
{{eqn | ll = \leadsto
      | l = n
      | r = 2 a + 2 b
}}
{{end-eqn}}
Since $a, b$ are divisors of $n$:
:$a \divides \paren {2 a + 2 b}$
:$b \divides \paren {2 a + 2 b}$
which reduces to:
:$a \divides 2 b$
:$b \divides 2 a$
Suppose $b$ is odd.
Then by Euclid's Lemma, we would have $b \divides a$.
By Absolute Value of Integer is not less than Divisors, this gives $b \le a$, which is a contradiction.
Thus $b$ is even.
Suppose $a$ is even.
Then $a, b, c$ are all even.
So $\gcd \set {a, b, c} \ne 1$, which is a contradiction.
Therefore it must be the case that $a$ is odd.
Then by Euclid's Lemma, we have:
:$a \divides \dfrac b 2$
and:
:$\dfrac b 2 \divides a$
By Absolute Value of Integer is not less than Divisors, this gives:
:$\dfrac b 2 = a$
Because $\gcd \set {a, b, c} = 1$, we must have $a = 1$.
Hence the set $\set {1, 2, 3}$ is obtained.
{{qed}}
Theorem: Digital Root of 3 Consecutive Numbers ending in Multiple of 3

Let $n$, $n + 1$ and $n + 2$ be positive integers such that $n + 2$ is a multiple of $3$.
Let $m = n + \paren {n + 1} + \paren {n + 2}$.
Then the digital root of $m$ is $6$.

Proof:
Let $n + 2$ be expressed as $3 r$ for some positive integer $r$.
Then:
{{begin-eqn}}
{{eqn | l = m
      | r = \paren {3 r - 2} + \paren {3 r - 1} + 3 r
      | c = 
}}
{{eqn | r = 9 r - 3
      | c = 
}}
{{eqn | r = 9 \paren {r - 1} + 6
      | c = 
}}
{{end-eqn}}
The result follows from Digital Root of Number equals its Excess over Multiple of 9.
{{qed}}
Theorem: Characteristics of Regular 4-Dimensional Polytopes

The $4$-dimensional regular polytopes have the following characteristics:
{| border="1"
|-
! Name
! No. of cells
! No. of faces
! No. of edges
! No. of vertices
! Dual
|-
| align="right" | Pentatope
| align="right" | $5$
| align="right" | $10$
| align="right" | $10$
| align="right" | $5$
| align="right" | Self-dual
|-
| align="right" | Tesseract
| align="right" | $8$
| align="right" | $24$
| align="right" | $32$
| align="right" | $16$
| align="right" | $16$-cell
|-
| align="right" | $16$-cell
| align="right" | $16$
| align="right" | $32$
| align="right" | $24$
| align="right" | $8$
| align="right" | Tesseract
|-
| align="right" | $24$-cell
| align="right" | $24$
| align="right" | $96$
| align="right" | $96$
| align="right" | $24$
| align="right" | Self-dual
|-
| align="right" | $120$-cell
| align="right" | $120$
| align="right" | $720$
| align="right" | $1200$
| align="right" | $600$
| align="right" | $600$-cell
|-
| align="right" | $600$-cell
| align="right" | $600$
| align="right" | $1200$
| align="right" | $720$
| align="right" | $120$
| align="right" | $120$-cell
|}

Proof:
{{ProofWanted|Much background work to be covered}}
Theorem: Complex Power is of Exponential Order Epsilon

Let: 
:$f: \hointr 0 \to \to \C: t \mapsto t^\phi$
be $t$ to the power of $\phi$, for $\phi \in \C$, defined on its principal branch.
Let $\map \Re \phi > -1$.
Then $f$ is of exponential order $\epsilon$ for any  $\epsilon > 0$ arbitrarily small in magnitude.

Proof:
{{begin-eqn}}
{{eqn | l = \size {t^\phi}
      | r = t^{\map \Re \phi}
      | c = Modulus of Positive Real Number to Complex Power is Positive Real Number to Power of Real Part
}}
{{end-eqn}}
The result follows from Real Power is of Exponential Order Epsilon.
{{qed}}
Category:Exponential Order
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