avewright/proofwiki-math huggingface dataset:
Theorem: Divisor Sum of Non-Square Semiprime

Let $n \in \Z_{>0}$ be a semiprime with distinct prime factors $p$ and $q$.
Then:
:$\map {\sigma_1} n = \paren {p + 1} \paren {q + 1}$
where $\map {\sigma_1} n$ denotes the divisor sum function.

Proof:
From Divisor Sum of Integer:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$26 = 2 \times 13$
Hence:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {26}
      | r = \frac {2^2 - 1} {2 - 1} \times \frac {13^2 - 1} {13 - 1}
      | c = 
}}
{{eqn | r = \frac 3 1 \times \frac {168} {12}
      | c = 
}}
{{eqn | r = 3 \times 14
      | c = 
}}
{{eqn | r = 42
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Divisor Sum of Non-Square Semiprime

Let $n \in \Z_{>0}$ be a semiprime with distinct prime factors $p$ and $q$.
Then:
:$\map {\sigma_1} n = \paren {p + 1} \paren {q + 1}$
where $\map {\sigma_1} n$ denotes the divisor sum function.

Proof:
We have that:
:$26 = 2 \times 13$
and so by definition is a semiprime whose prime factors are distinct.
Hence:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {26}
      | r = \paren {2 + 1} \paren {13 + 1}
      | c = Divisor Sum of Non-Square Semiprime
}}
{{eqn | r = 3 \times 14
      | c = 
}}
{{eqn | r = 42
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Divisor Sum of Non-Square Semiprime

Let $n \in \Z_{>0}$ be a semiprime with distinct prime factors $p$ and $q$.
Then:
:$\map {\sigma_1} n = \paren {p + 1} \paren {q + 1}$
where $\map {\sigma_1} n$ denotes the divisor sum function.

Proof:
From Divisor Sum of Integer:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$94 = 2 \times 47$
Hence:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {94}
      | r = \frac {2^2 - 1} {2 - 1} \times \frac {47^2 - 1} {47 - 1}
      | c = 
}}
{{eqn | r = \frac 3 1 \times \frac {46 \times 48} {46}
      | c = Difference of Two Squares
}}
{{eqn | r = 3 \times 48
      | c = 
}}
{{eqn | r = 3 \times \paren {2^4 \times 3}
      | c = 
}}
{{eqn | r = 2^4 \times 3^2
      | c = 
}}
{{eqn | r = \paren {2^2 \times 3}^2
      | c = 
}}
{{eqn | r = 12^2
      | c = 
}}
{{eqn | r = 144
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Divisor Sum of Non-Square Semiprime

Let $n \in \Z_{>0}$ be a semiprime with distinct prime factors $p$ and $q$.
Then:
:$\map {\sigma_1} n = \paren {p + 1} \paren {q + 1}$
where $\map {\sigma_1} n$ denotes the divisor sum function.

Proof:
We have that:
:$94 = 2 \times 47$
and so by definition is a semiprime whose prime factors are distinct.
Hence:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {94}
      | r = \paren {2 + 1} \paren {47 + 1}
      | c = Divisor Sum of Non-Square Semiprime
}}
{{eqn | r = 3 \times 48
      | c = 
}}
{{eqn | r = 3 \times \paren {2^4 \times 3}
      | c = 
}}
{{eqn | r = 2^4 \times 3^2
      | c = 
}}
{{eqn | r = \paren {2^2 \times 3}^2
      | c = 
}}
{{eqn | r = 12^2
      | c = 
}}
{{eqn | r = 144
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Divisor Sum of Non-Square Semiprime

Let $n \in \Z_{>0}$ be a semiprime with distinct prime factors $p$ and $q$.
Then:
:$\map {\sigma_1} n = \paren {p + 1} \paren {q + 1}$
where $\map {\sigma_1} n$ denotes the divisor sum function.

Proof:
As $p$ and $q$ are distinct prime numbers, it follows that $p$ and $q$ are coprime.
Thus by Divisor Sum Function is Multiplicative:
:$\map {\sigma_1} n = \map {\sigma_1} p \map {\sigma_1} q$
From Divisor Sum of Prime Number:
:$\map {\sigma_1} p = \paren {p + 1}$
:$\map {\sigma_1} q = \paren {q + 1}$
Hence the result.
{{qed}}
Theorem: Divisor Sum of Non-Square Semiprime

Let $n \in \Z_{>0}$ be a semiprime with distinct prime factors $p$ and $q$.
Then:
:$\map {\sigma_1} n = \paren {p + 1} \paren {q + 1}$
where $\map {\sigma_1} n$ denotes the divisor sum function.

Proof:
A semiprime with distinct prime factors is a square-free integer.
By Divisor Sum of Square-Free Integer:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} p_i + 1$
Hence the result.
{{qed}}
Theorem: Integers which are Divisor Sum for 3 Integers

The sequence of integers which are the divisor sum of $3$ different integers begins:
:$24, 42, 48, 60, 84, 90, \ldots$
{{OEIS|A007372}}

Proof:
For a given $n$, to determine every $m$ such that $\map {\sigma_1} m = n$ can be determined by evaluating the divisor sum of all integers up to $n - 1$.
It is hence noted:
{{begin-eqn}}
{{eqn | l = 24
      | r = \map {\sigma_1} {14}
      | c = {{DSFLink|14}}
}}
{{eqn | r = \map {\sigma_1} {15}
      | c = {{DSFLink|15}}
}}
{{eqn | r = \map {\sigma_1} {23}
      | c = Divisor Sum of Prime Number: $23$ is prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 42
      | r = \map {\sigma_1} {20}
      | c = {{DSFLink|20}}
}}
{{eqn | r = \map {\sigma_1} {26}
      | c = {{DSFLink|26}}
}}
{{eqn | r = \map {\sigma_1} {41}
      | c = Divisor Sum of Prime Number: $41$ is prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 48
      | r = \map {\sigma_1} {33}
      | c = {{DSFLink|33}}
}}
{{eqn | r = \map {\sigma_1} {35}
      | c = {{DSFLink|35}}
}}
{{eqn | r = \map {\sigma_1} {47}
      | c = Divisor Sum of Prime Number: $47$ is prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 60
      | r = \map {\sigma_1} {24}
      | c = {{DSFLink|24}}
}}
{{eqn | r = \map {\sigma_1} {38}
      | c = {{DSFLink|38}}
}}
{{eqn | r = \map {\sigma_1} {59}
      | c = Divisor Sum of Prime Number: $59$ is prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 84
      | r = \map {\sigma_1} {44}
      | c = {{DSFLink|44}}
}}
{{eqn | r = \map {\sigma_1} {65}
      | c = {{DSFLink|65}}
}}
{{eqn | r = \map {\sigma_1} {83}
      | c = Divisor Sum of Prime Number: $83$ is prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 90
      | r = \map {\sigma_1} {40}
      | c = {{DSFLink|40}}
}}
{{eqn | r = \map {\sigma_1} {58}
      | c = {{DSFLink|58}}
}}
{{eqn | r = \map {\sigma_1} {89}
      | c = Divisor Sum of Prime Number: $89$ is prime
}}
{{end-eqn}}
{{qed}}
Theorem: Divisor Sum of Square-Free Integer

Let $n$ be an integer such that $n \ge 2$.
Let $n$ be square-free.
Let the prime decomposition of $n$ be:
:$\ds n = \prod_{1 \mathop \le i \mathop \le r} p_i =  p_1 p_2 \cdots p_r$
Let $\map {\sigma_1} n$ be the divisor sum of $n$.
That is, let $\map {\sigma_1} n$ be the sum of all positive divisors of $n$.
Then:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} p_i + 1$

Proof:
From Divisor Sum of Integer:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$66 = 2 \times 3 \times 11$
Hence:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {66}
      | r = \frac {2^2 - 1} {2 - 1} \times \frac {3^2 - 1} {3 - 1} \times \frac {11^2 - 1} {11 - 1}
      | c = 
}}
{{eqn | r = \frac 3 1 \times \frac 8 2 \times \frac {120} {10}
      | c = 
}}
{{eqn | r = 3 \times 4 \times 12
      | c = 
}}
{{eqn | r = 3 \times 2^2 \times \paren {2^2 \times 3}
      | c = 
}}
{{eqn | r = 2^4 \times 3^2
      | c = 
}}
{{eqn | r = \paren {2^2 \times 3}^2
      | c = 
}}
{{eqn | r = 12^2
      | c = 
}}
{{eqn | r = 144
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Divisor Sum of Square-Free Integer

Let $n$ be an integer such that $n \ge 2$.
Let $n$ be square-free.
Let the prime decomposition of $n$ be:
:$\ds n = \prod_{1 \mathop \le i \mathop \le r} p_i =  p_1 p_2 \cdots p_r$
Let $\map {\sigma_1} n$ be the divisor sum of $n$.
That is, let $\map {\sigma_1} n$ be the sum of all positive divisors of $n$.
Then:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} p_i + 1$

Proof:
We have that:
:$66 = 2 \times 3 \times 11$
Hence:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {66}
      | r = \paren {2 + 1} \paren {3 + 1} \paren {11 + 1}
      | c = Divisor Sum of Square-Free Integer
}}
{{eqn | r = 3 \times 4 \times 12
      | c = 
}}
{{eqn | r = 3 \times 2^2 \times \paren {2^2 \times 3}
      | c = 
}}
{{eqn | r = 2^4 \times 3^2
      | c = 
}}
{{eqn | r = \paren {2^2 \times 3}^2
      | c = 
}}
{{eqn | r = 12^2
      | c = 
}}
{{eqn | r = 144
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Divisor Sum of Square-Free Integer

Let $n$ be an integer such that $n \ge 2$.
Let $n$ be square-free.
Let the prime decomposition of $n$ be:
:$\ds n = \prod_{1 \mathop \le i \mathop \le r} p_i =  p_1 p_2 \cdots p_r$
Let $\map {\sigma_1} n$ be the divisor sum of $n$.
That is, let $\map {\sigma_1} n$ be the sum of all positive divisors of $n$.
Then:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} p_i + 1$

Proof:
From Divisor Sum of Integer:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$70 = 2 \times 5 \times 7$
Hence:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {70}
      | r = \frac {2^2 - 1} {2 - 1} \times \frac {5^2 - 1} {5 - 1} \times \frac {7^2 - 1} {7 - 1}
      | c = 
}}
{{eqn | r = \frac 3 1 \times \frac {24} 4 \times \frac {48} {6}
      | c = 
}}
{{eqn | r = 3 \times 6 \times 8
      | c = 
}}
{{eqn | r = 3 \times \paren {3 \times 2} \times 2^3
      | c = 
}}
{{eqn | r = 3^2 \times 2^4
      | c = 
}}
{{eqn | r = \paren {2^2 \times 3}^2
      | c = 
}}
{{eqn | r = 12^2
      | c = 
}}
{{eqn | r = 144
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Divisor Sum of Square-Free Integer

Let $n$ be an integer such that $n \ge 2$.
Let $n$ be square-free.
Let the prime decomposition of $n$ be:
:$\ds n = \prod_{1 \mathop \le i \mathop \le r} p_i =  p_1 p_2 \cdots p_r$
Let $\map {\sigma_1} n$ be the divisor sum of $n$.
That is, let $\map {\sigma_1} n$ be the sum of all positive divisors of $n$.
Then:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} p_i + 1$

Proof:
We have that:
:$70 = 2 \times 5 \times 7$
Hence:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {70}
      | r = \paren {2 + 1} \paren {5 + 1} \paren {7 + 1}
      | c = Divisor Sum of Square-Free Integer
}}
{{eqn | r = 3 \times 6 \times 8
      | c = 
}}
{{eqn | r = 3 \times \paren {3 \times 2} \times 2^3
      | c = 
}}
{{eqn | r = 3^2 \times 2^4
      | c = 
}}
{{eqn | r = \paren {2^2 \times 3}^2
      | c = 
}}
{{eqn | r = 12^2
      | c = 
}}
{{eqn | r = 144
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Divisor Sum of Square-Free Integer

Let $n$ be an integer such that $n \ge 2$.
Let $n$ be square-free.
Let the prime decomposition of $n$ be:
:$\ds n = \prod_{1 \mathop \le i \mathop \le r} p_i =  p_1 p_2 \cdots p_r$
Let $\map {\sigma_1} n$ be the divisor sum of $n$.
That is, let $\map {\sigma_1} n$ be the sum of all positive divisors of $n$.
Then:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} p_i + 1$

Proof:
We have that the Divisor Sum Function is Multiplicative.
From the definition of prime number, each of the prime factors of $n$ is coprime to all other divisors of $n$.
From Divisor Sum of Prime Number, we have:
:$\map {\sigma_1} {p_i} = p_i + 1$
Hence the result.
{{qed}}
Theorem: Divisor Sum of Square-Free Integer

Let $n$ be an integer such that $n \ge 2$.
Let $n$ be square-free.
Let the prime decomposition of $n$ be:
:$\ds n = \prod_{1 \mathop \le i \mathop \le r} p_i =  p_1 p_2 \cdots p_r$
Let $\map {\sigma_1} n$ be the divisor sum of $n$.
That is, let $\map {\sigma_1} n$ be the sum of all positive divisors of $n$.
Then:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} p_i + 1$

Proof:
From Divisor Sum of Integer:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where each of the $k_i$s are equal to $1$;
Hence:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^2 - 1} {p_i - 1}$
But from Difference of Two Squares:
:$\dfrac {p_i^2 - 1} {p_i - 1} = \dfrac {\paren {p_i + 1} \paren {p_i i 1} } {p_i - 1} = p_i + 1$
Hence the result.
{{qed}}
Theorem: Mapping Assigning to Element Its Lower Closure is Isomorphism

Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $I = \struct {\map {\mathit {Ids} } L, \precsim}$ be an inclusion ordered set
where:
:$\map {\mathit {Ids} } L$ denotes the set of all ideals in $L$
:$\mathord \precsim = \mathord \subseteq \cap \paren {\map {\mathit {Ids} } L \times \map {\mathit {Ids} } L}$
Let $P = \struct {\map K I, \precsim'}$ be an ordered subset of $I$
where:
:$\map K I$ denotes the compact subset of $I$.
Let $f: S \to \map K I$ be a mapping such that:
:$\forall x \in S: \map f x = x^\preceq$
Then $f$ is an order isomorphism between $L$ and $P$.

Proof:
By definition:
:$\forall x \in S: x^\preceq$ is a principal ideal.
By Compact Element iff Principal Ideal:
:$\forall x \in S: x^\preceq$ is a compact element in $I$.
By definition of compact subset:
:$\forall x \in S: x^\preceq \in \map K I$
Then $f$ is well-defined.
We will prove that:
:$f$ is an order embedding.
That means:
:$\forall x, y \in S: x \preceq y \iff \map f x \precsim' \map f y$
Let $x, y \in S$.
We will prove as lemma that:
:$x^\preceq \subseteq y^\preceq \implies x \preceq y$
Assume that:
:$x^\preceq \subseteq y^\preceq$
By definition of reflexivity:
:$x \preceq x$
By definition of lower closure of element:
:$x \in x^\preceq$
By definition of subset:
:$x \in y^\preceq$
Thus by definition of lower closure of element:
:$x \preceq y$
{{qed|lemma}}
:$x \preceq y$
{{iff}}:
:$x^\preceq \subseteq y^\preceq$ by lemma and Lower Closure is Increasing
{{iff}}:
:$\map f x \subseteq \map f y$ by definition of $f$
{{iff}}:
:$\map f x \precsim \map f y$ by definition of $\precsim$
{{iff}}:
:$\map f x \precsim' \map f y$ by definition of ordered subset.
{{qed|lemma}}
We will prove that:
:$\forall a \in \map K I: \exists y \in S: a = \map f y$
Let $a \in \map K I$.
By definition of compact subset:
:$a$ is a compact element in $I$.
By Compact Element iff Principal Ideal:
:$a$ is a principal ideal.
By definition of principal ideal:
:$\exists y \in S: a = y^\preceq$
Thus by definition of $f$:
:$a = \map f y$
{{qed|lemma}}
Thus $f$ is a surjection.
Hence $f$ is an order isomorphism.
{{qed}}
Theorem: Difference Triangle for Sequence of Fifth Powers

The difference triangle for the sequence of fifth powers ends on the fifth line with instances of $5!$, where $!$ denotes factorial.

Proof:
{{ProofWanted|Needs some difference calculus results first.}}
Theorem: Ideals form Arithmetic Lattice

Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below lattice.
Let $I = \struct {\map {\mathit {Ids} } L, \precsim}$ be an inclusion ordered set, where:
:$\map {\mathit {Ids} } L$ denotes the set of all ideals in $L$
:$\mathord \precsim = \mathord \subseteq \cap \paren {\map {\mathit {Ids} } L \times \map {\mathit {Ids} } L}$
Then $I$ is an arithmetic lattice.

Proof:
We will prove that:
:$\forall x, y \in \map K I: \exists z \in \map K I: z \precsim x \land z \precsim y \land \forall v \in \map K I: v \precsim x \land v \precsim y \implies v \precsim z$
Let $x, y \in \map K I$.
By definition of compact subset:
:$x$ and $y$ are compact elements in $I$.
By Compact Element iff Principal Ideal and definition of principal ideal:
:$\exists a \in S: x = a^\preceq$
and
:$\exists b \in S: y = b^\preceq$
By Intersection of Semilattice Ideals is Ideal:
:$x \cap y \in \map {\mathit {Ids} } L$
Define $z = x \cap y$.
By Meet in Set of Ideals:
:$z = x \wedge_I y$
We will prove that
:$\paren {a \wedge b}^\preceq \subseteq z$
Let $c \in \paren {a \wedge b}^\preceq$
By definition of lower closure of element:
:$c \preceq a \wedge b$
By Meet Precedes Operands:
:$a \wedge b \preceq a$ and $a \wedge b \preceq b$
By definition of transitivity:
:$c \preceq a$ and $c \preceq b$
By definition of lower closure of element:
:$c \in x$ and $c \in y$
Thus by definition of intersection:
:$c \in z$
{{qed|lemma}}
We will prove that:
:$z \subseteq \paren {a \wedge b}^\preceq$
Let $c \in z$.
By definition of intersection:
:$c \in x$ and $c \in y$
By definition of lower closure of element:
:$c \preceq a$ and $c \preceq b$
By definitions of infimum and lower bound:
:$c \preceq a \wedge b$
Thus by definition of lower closure of element:
:$c \in \paren {a \wedge b}^\preceq$
{{qed|lemma}}
By definition of set equality:
:$z = \paren {a \wedge b}^\preceq$
By definition:
:$z$ is a principal ideal.
By Compact Element iff Principal Ideal:
:$z$ is a compact element in $I$.
Thus by definition of compact subset:
:$z \in \map K I$
Thus by Meet Precedes Operands:
:$z \precsim x$ and $z \precsim y$
Let $v \in \map K I$ such that
:$v \precsim x$ and $v \precsim y$
By definition of $\precsim$:
:$v \subseteq x$ and $v \subseteq y$
By Intersection is Largest Subset:
:$v \subseteq z$
Thus by definition of $\precsim$:
:$v \precsim z$
{{qed|lemma}}
By definition:
:$\map K I$ forms a meet semilattice.
By Compact Subset is Join Subsemilattice:
:$\map K I$ forms a join semilattice.
By definition:
:$\map K I$ forms a lattice.
By Ideals form Algebraic Lattice:
:$I$ is an algebraic lattice.
Thus by Arithmetic iff Compact Subset form Lattice in Algebraic Lattice:
:$I$ is an arithmetic lattice.
{{qed}}
Theorem: Sums of Partial Sequences of Squares

Let $n \in \Z_{>0}$.
Consider the odd number $2 n + 1$ and its square $\paren {2 n + 1}^2 = 2 m + 1$.
Then:
:$\ds \sum_{j \mathop = 0}^n \paren {m - j}^2 = \sum_{j \mathop = 1}^n \paren {m + j}^2$
That is:
:the sum of the squares of the $n + 1$ integers up to $m$
equals:
:the sum of the squares of the $n$ integers from $m + 1$ upwards.

Proof:
First we express $m$ in terms of $n$:
{{begin-eqn}}
{{eqn | l = \paren {2 n + 1}^2
      | r = 4 n^2 + 4 n + 1
      | c = 
}}
{{eqn | r = 2 \paren {2 n^2 + 2 n} + 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = m
      | r = 2 n^2 + 2 n
}}
{{end-eqn}}
We have:
{{begin-eqn}}
{{eqn | o =
      | r = \sum_{j \mathop = 1}^n \paren {m + j}^2 - \sum_{j \mathop = 0}^n \paren {m - j}^2
}}
{{eqn | r = \sum_{j \mathop = 1}^n \paren {\paren {m + j}^2 - \paren {m - j}^2} - m^2
}}
{{eqn | r = \sum_{j \mathop = 1}^n 2 j \paren {2 m} - m^2
      | c = Difference of Two Squares
}}
{{eqn | r = 4 m \times \frac {n \paren {n + 1} } 2 - m^2
      | c = Closed Form for Triangular Numbers
}}
{{eqn | r = m \paren {2 n^2 + 2 n} - m^2
}}
{{eqn | r = m^2 - m^2
}}
{{eqn | r = 0
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Power of n equalling (n - 1)! + 1

There is exactly one solution to the equation in the integers:
:$\paren {n - 1}! + 1 = n^k$
for $k > 1$, and that is:
:$n = 5$
:$k = 2$

Proof:
We have that:
{{begin-eqn}}
{{eqn | l = \paren {1 - 1}! + 1
      | r = 0! + 1
      | c = 
}}
{{eqn | r = 1 + 1
      | c = Factorial of Zero
}}
{{eqn | r = 2
      | c = not a power of $1$
}}
{{eqn | l = \paren {2 - 1}! + 1
      | r = 1! + 1
      | c = 
}}
{{eqn | r = 1 + 1
      | c = Examples of Factorials
}}
{{eqn | r = 2
      | c = 
}}
{{eqn | r = 2^1
      | c = $k = 1$
}}
{{eqn | l = \paren {3 - 1}! + 1
      | r = 2! + 1
      | c = 
}}
{{eqn | r = 2 + 1
      | c = Examples of Factorials
}}
{{eqn | r = 3
      | c = 
}}
{{eqn | r = 3^1
      | c = $k = 1$
}}
{{eqn | l = \paren {4 - 1}! + 1
      | r = 3! + 1
      | c = 
}}
{{eqn | r = 6 + 1
      | c = Examples of Factorials
}}
{{eqn | r = 7
      | c = not a power of $4$
}}
{{eqn | l = \paren {5 - 1}! + 1
      | r = 4! + 1
      | c = 
}}
{{eqn | r = 24 + 1
      | c = Examples of Factorials
}}
{{eqn | r = 25
      | c = 
}}
{{eqn | r = 5^2
      | c = 
}}
{{end-eqn}}
Let $n > 5$.
Suppose $n$ is composite.
By Divisibility of n-1 Factorial by Composite n:
:$n \divides \paren {n - 1}!$
and thus:
:$n \nmid \paren {n - 1}! + 1$
showing that $\paren {n - 1}! + 1$ is not a power of $n$.
Suppose $n$ is prime.
Further suppose:
:$\exists k \in \N: \paren {n - 1}! + 1 = n^k$
Then:
{{begin-eqn}}
{{eqn | l = \paren {n - 1}!
      | r = n^k - 1
}}
{{eqn | r = \paren {n - 1} \sum_{i \mathop = 0}^{k - 1} n^i
      | c = Sum of Geometric Sequence
}}
{{eqn | ll = \leadsto
      | l = \paren {n - 2}!
      | r = \sum_{i \mathop = 0}^{k - 1} n^i
}}
{{eqn | o = \equiv
      | r = \sum_{i \mathop = 0}^{k - 1} 1^i
      | rr = \pmod {n - 1}
      | c = Congruence of Powers
}}
{{eqn | o = \equiv
      | r = k
      | rr = \pmod {n - 1}
}}
{{end-eqn}}
By Divisibility of n-1 Factorial by Composite n, since $n - 1 \ne 4$ and is composite:
:$n - 1 \divides \paren {n - 2}!$
thus $k$ must be a multiple of $n - 1$.
However:
{{begin-eqn}}
{{eqn | l = n^{n - 1}
      | o = >
      | r = n \paren {n - 1} \paren {n - 2} \cdots \paren 2
}}
{{eqn | r = n!
}}
{{eqn | o = >
      | r = 2 \paren {n - 1}!
}}
{{eqn | o = >
      | r = \paren {n - 1}! + 1
}}
{{eqn | o = >
      | r = 1
}}
{{eqn | r = n^0
}}
{{end-eqn}}
which shows that no multiple of $n - 1$ can satisfy our equation.
Hence there is no solution for $n > 5$.
{{qed}}
Theorem: Ideals form Algebraic Lattice

Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $I = \struct {\map {\operatorname{Ids} } L, \precsim}$ be an inclusion ordered set
where
:$\map {\operatorname{Ids} } L$ denotes the set of all ideals in $L$
:$\mathord \precsim = \mathord \subseteq \cap \paren {\map {\operatorname{Ids} } L \times \map {\operatorname{Ids} } L}$
Then $I$ is an algebraic lattice.

Proof:
By definition of subset:
:$\map {\operatorname{Ids} } L \subseteq \powerset S$
where $\powerset S$ denotes the power set of $S$.
Define:
:$P = \struct {\powerset S, \precsim'}$
where:
:$\mathord \precsim' = \mathord\subseteq \cap \paren {\powerset S \times \powerset S}$
By Ideals are Continuous Lattice Subframe of Power Set:
:$I$ is an continuous lattice subframe of $P$.
By Lattice of Power Set is Algebraic:
:$P$ is an algebraic lattice.
Thus by Continuous Lattice Subframe of Algebraic Lattice is Algebraic Lattice:
:$I$ is an algebraic lattice.
{{qed}}
Theorem: Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite

Let
:$\map P x : \closedint a b \to \R$ 
:$\map h x : \closedint a b \to \R$.
Let $\map h x$ be continuously differentiable $\forall x \in \closedint a b$.
Suppose:
:$\forall x \in \closedint a b: \map P x > 0$
Then:
:$\ds \forall \map h x : \map h a = \map h b = 0 : \int_a^b \paren {P h'^2 + Q h^2} \rd x > 0$
{{iff}} the interval $\closedint a b$ contains no points conjugate to $a$.

Proof:
=== Necessary Condition ===
Let $\map \omega x : \closedint a b \to \R$ be a continuously differentiable mapping.
Then:
{{begin-eqn}}
{{eqn | l = 0
      | r = \bigintlimits {\omega h^2} a b
      | c = Boundary Conditions for $h$
}}
{{eqn | r = \int_a^b \map {\frac \d {\d x} } {\omega h^2} \rd x
      | c = Fundamental theorem of calculus
}}
{{end-eqn}}   
Let $\omega$ be a solution to the following equation:
:$\map P {Q + \omega'} = \omega^2$
Then:
{{begin-eqn}}
{{eqn | l = P h'^2 + Q h^2 + \map {\frac \d {\d x} } {\omega h^2}
      | r = P h'^2 + 2 h h' \omega + \paren {Q + \omega'} h^2
}}
{{eqn | r = P h'^2 + 2 h h' \omega + \frac {\omega^2} P h^2
}}
{{eqn | r = P \paren {h' + \frac \omega P h}^2
}}
{{eqn | o = \ge
      | r = 0
}}
{{end-eqn}}
In other words:
{{begin-eqn}}
{{eqn | l = \int_a^b \paren {P h'^2 + Q h^2} \rd x
      | r = \int_a^b \paren {P h'^2 + Q h^2 + \map {\frac \d {\d x} } {\omega h^2} } \rd x
}}
{{eqn | r = \int_a^b P \paren {h' + \frac \omega P h}^2 \rd x
}}
{{eqn | o = \ge
      | r = 0
}}
{{end-eqn}}
Now we will show, that this expression never vanishes.
{{AimForCont}} that it does vanish.
Then:
:$h' + \dfrac \omega P h = 0$
By Existence-Uniqueness Theorem for First-Order Differential Equation:
:$\forall x \in \closedint a b: \map h a = 0 \implies \map h x = 0$
This implies an infinite number of conjugate points.
However, the theorem assumes no conjugate points.
This is a contradiction.
Hence:
:$\forall x \in \openint a b: \map h x \ne 0$
and:
:$P \paren {h' + \dfrac {\omega h} P}^2 > 0$
Thus, a functional formed by a definite integral of positive definite function is positive definite.
{{qed|lemma}}
Theorem: Square Cullen Numbers

The numbers:
:$1, 9, 25$
are Cullen numbers which are also square.

Proof:
We have:
{{begin-eqn}}
{{eqn | l = 1
      | r = 0 \times 2^0 + 1
}}
{{eqn | l = 9
      | r = 2 \times 2^2 + 1
}}
{{eqn | l = 25
      | r = 3 \times 2^3 + 1
}}
{{end-eqn}}
{{Qed}}