avewright/proofwiki-math huggingface dataset:
Theorem: Condition for Group Endomorphism to Commute with All Inner Automorphisms

Let $G$ be a group.
Let $\phi: G \to G$ be an endomorphism on $G$.
Let $\phi$ be such that:
:$\forall a \in G: \kappa_a \circ \phi = \phi \circ \kappa_a$
where:
:$\kappa_a$ denotes the inner automorphism of $G$ given by $a$
:$\circ$ denotes composition of mappings.
Then:
:$H = \set {x \in G: \map \phi {\map \phi x} = \map \phi x}$
is a normal subgroup of $G$.
Also, the quotient group $G / H$ is an abelian group.

Proof:
We have for all $a, g \in G$:
{{begin-eqn}}
{{eqn | l = a \map \phi g a^{-1}
      | r = \map {\kappa_a} {\map \phi g}
      | c = {{Defof|Inner Automorphism}}
}}
{{eqn | r = \map {\paren{\kappa_a \circ \phi} } g
      | c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = \map {\paren{\phi \circ \kappa_a} } g
}}
{{eqn | r = \map \phi {\map {\kappa_a} g}
      | c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = \map \phi {a g a^{-1} }
      | c = {{Defof|Inner Automorphism}}
}}
{{end-eqn}}
Let $x \in H$. 
Then by the above:
{{begin-eqn}}
{{eqn | l = \map \phi {\map \phi {a x a^{-1} } }
      | r = \map \phi {a \map \phi x a^{-1} }
      | c = 
}}
{{eqn | r = \map \phi a \map \phi {\map \phi x} \map \phi {a^{-1} }
      | c = {{Defof|Group Endomorphism}}
}}
{{eqn | r = \map \phi a \map \phi x \map \phi {a^{-1} }
      | c = $\map \phi x = \map \phi {\map \phi x}$
}}
{{eqn | r = \map \phi {a x a^{-1} }
      | c = {{Defof|Group Endomorphism}}
}}
{{end-eqn}}
Hence $a x a^{-1} \in H$, and thus $a H a^{-1} \subseteq H$.
By definition of a normal subgroup, $H$ is a normal subgroup of $G$.
It remains to show that the quotient group $G / H$ is an abelian group.
We need to show that for any $a, b \in G$, we have:
:$a H b H = b H a H$
that is,
:$a b H = b a H$
By Cosets are Equal iff Product with Inverse in Subgroup, this is equivalent to:
:$\paren {b a}^{-1} a b = a^{-1} b^{-1} a b \in H$
We have:
{{begin-eqn}}
{{eqn | l = \map \phi {\map \phi {a^{-1} b^{-1} a b} }
      | r = \map \phi {\map \phi {a^{-1} b^{-1} a} \map \phi b}
      | c = {{Defof|Group Endomorphism}}
}}
{{eqn | r = \map \phi {a^{-1} \map \phi {b^{-1} } a \map \phi b}
      | c = by commutativity condition above
}}
{{eqn | r = \map \phi {a^{-1} } \map \phi {\map \phi {b^{-1} } a \map \phi b}
      | c = {{Defof|Group Endomorphism}}
}}
{{eqn | r = \map \phi {a^{-1} } \map \phi {\map \phi b^{-1} a \map \phi b}
      | c = Group Homomorphism Preserves Inverses
}}
{{eqn | r = \map \phi {a^{-1} } \map \phi b^{-1} \map \phi a \map \phi b
      | c = by commutativity condition above
}}
{{eqn | r = \map \phi {a^{-1} } \map \phi {b^{-1} } \map \phi a \map \phi b
      | c = Group Homomorphism Preserves Inverses
}}
{{eqn | r = \map \phi {a^{-1} b^{-1} a b}
      | c = {{Defof|Group Endomorphism}}
}}
{{end-eqn}}
and thus $a^{-1} b^{-1} a b \in H$.
Hence the result.
{{qed}}
Theorem: External Direct Product Associativity/Sufficient Condition

Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Let $\circ_1$ and $\circ_2$ be associative.
Then $\circ$ is also associative.

Proof:
Let $\circ_1$ and $\circ_2$ be associative. 
{{begin-eqn}}
{{eqn | l = \paren {\tuple {s_1, t_1} \circ \tuple {s_2, t_2} } \circ \tuple {s_3, t_3}
      | r = \tuple {\paren {s_1 \circ_1 s_2} \circ_1 s_3, \paren {t_1 \circ_2 t_2} \circ_2 t_3}
      | c = {{Defof|Operation Induced by Direct Product}}
}}
{{eqn | r = \tuple {s_1 \circ_1 s_2 \circ_1 s_3, t_1 \circ_2 t_2 \circ_2 t_3}
      | c = {{Defof|Associative Operation}}
}}
{{eqn | r = \tuple {s_1 \circ_1 \paren {s_2 \circ_1 s_3}, t_1 \circ_2 \paren {t_2 \circ_2 t_3} }
      | c = {{Defof|Associative Operation}}
}}
{{eqn | r = \tuple {s_1, t_1} \circ \paren {\tuple {s_2, t_2} \circ \tuple {s_3, t_3} }
      | c = {{Defof|Operation Induced by Direct Product}}
}}
{{end-eqn}}
Thus $\circ$ is associative.
Theorem: External Direct Product Associativity/Necessary Condition

Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Let $\circ$ be associative.
Then $\circ_1$ and $\circ_2$ are both also associative.

Proof:
Let $\circ$ be associative. 
{{AimForCont}} it is not the case that both $\circ_1$ and $\circ_2$ are associative.
{{WLOG}}, suppose $\circ_1$ is not associative.
Hence:
{{begin-eqn}}
{{eqn | q = \exists s_1, s_2, s_3 \in S
      | l = \paren {s_1 \circ_1 s_2} \circ_1 s_3
      | o = \ne
      | r = s_1 \circ_1 \paren {s_2 \circ_1 s_3}
      | c = as $\circ_1$ is not associative
}}
{{eqn | ll= \leadsto
      | q = \forall t_1, t_2, t_3 \in T
      | l = \tuple {\paren {s_1 \circ_1 s_2}, s_3} \circ \tuple {\paren {t_1 \circ_2 t_2}, t_3}
      | o = \ne
      | r = \tuple {s_1 \circ_1 \paren {s_2 \circ_1 s_3}, t_1 \circ_2 \paren {t_2 \circ_2 t_3} }
      | c = Equality of Ordered Pairs
}}
{{eqn | ll= \leadsto
      | q = \forall t_1, t_2, t_3 \in T
      | l = \paren {\tuple {s_1, t_1} \circ \tuple {s_2, t_2} } \circ \tuple {s_3, t_3}
      | o = \ne
      | r = \tuple {s_1, t_1} \circ \paren {\tuple {s_2, t_2} \circ \tuple {s_3, t_3} }
      | c = {{Defof|Operation Induced by Direct Product}}
}}
{{end-eqn}}
This contradicts our assumption that $\circ$ is associative.
The same argument can be applied, {{mutatis}}, to the case where $\circ_2$ is not associative.
Hence, by Proof by Contradiction, $\circ_1$ and $\circ_2$ are both associative.
Theorem: External Direct Product Commutativity/Sufficient Condition

Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Let $\circ_1$ and $\circ_2$ be commutative operations.
Then $\circ$ is also a commutative operation.

Proof:
Let $\circ_1$ and $\circ_2$ be commutative operations.
{{begin-eqn}}
{{eqn | l = \tuple {s_1, t_1} \circ \tuple {s_2, t_2}
      | r = \tuple {s_1 \circ_1 s_2, t_1 \circ_2 t_2}
      | c = {{Defof|Operation Induced by Direct Product}}
}}
{{eqn | r = \tuple {s_2 \circ_1 s_1, t_2 \circ_2 t_1}
      | c = {{Defof|Commutative Operation}}
}}
{{eqn | r = \tuple {s_2, t_2} \circ \tuple {s_1, t_1}
      | c = {{Defof|Operation Induced by Direct Product}}
}}
{{end-eqn}}
Thus $\circ$ is commutative.
Theorem: External Direct Product Commutativity/Necessary Condition

Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Let $\circ$ be commutative.
Then $\circ_1$ and $\circ_2$ are both also commutative.

Proof:
Let $\circ$ be commutative. 
{{AimForCont}} it is not the case that both $\circ_1$ and $\circ_2$ are commutative.
{{WLOG}}, suppose $\circ_1$ is not commutative.
Hence:
{{begin-eqn}}
{{eqn | q = \exists s_1, s_2 \in S
      | l = s_1 \circ_1 s_2
      | o = \ne
      | r = s_2 \circ_1 s_1
      | c = as $\circ_1$ is not commutative
}}
{{eqn | ll= \leadsto
      | q = \forall t_1, t_2 \in T
      | l = \tuple {s_1 \circ_1 s_2} \circ \tuple {t_1 \circ_2 t_2}
      | o = \ne
      | r = \tuple {s_2 \circ_1 s_1} \circ \tuple {t_2 \circ_2 t_1}
      | c = Equality of Ordered Pairs
}}
{{eqn | ll= \leadsto
      | q = \forall t_1, t_2 \in T
      | l = \tuple {s_1, t_1} \circ \tuple {s_2, t_2}
      | o = \ne
      | r = \tuple {s_2, t_2} \circ \tuple {s_1, t_1}
      | c = {{Defof|Operation Induced by Direct Product}}
}}
{{end-eqn}}
This contradicts our assumption that $\circ$ is commutative.
The same argument can be applied, {{mutatis}}, to the case where $\circ_2$ is not commutative.
Hence, by Proof by Contradiction, $\circ_1$ and $\circ_2$ are both commutative.
Theorem: External Direct Product Identity/Sufficient Condition

Let $\struct {S \times T, \circ}$ be the external direct product of two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Let:
:$\struct {S, \circ_1}$ have identity element $e_S$
and:
:$\struct {T, \circ_2}$ have identity element $e_T$.
Then $\tuple {e_S, e_T}$ is the identity element for $\struct {S \times T, \circ}$.

Proof:
Let $e_S$ and $e_T$ be the identity elements of $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ respectively.
{{begin-eqn}}
{{eqn | l = \tuple {s, t} \circ \tuple {e_S, e_T}
      | r = \tuple {s \circ_1 e_S, t \circ_2 e_T}
      | c = 
}}
{{eqn | r = \tuple {s, t}
      | c = 
}}
{{eqn | l = \tuple {e_S, e_T} \circ \tuple {s, t}
      | r = \tuple {e_S \circ_1 s, e_T \circ_2 t}
      | c = 
}}
{{eqn | r = \tuple {s, t}
      | c = 
}}
{{end-eqn}}
Thus $\tuple {e_S, e_T}$ is the identity of $\struct {S \times T, \circ}$.
Theorem: External Direct Product Identity/Necessary Condition

Let $\struct {S \times T, \circ}$ be the external direct product of two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Let $\struct {S \times T, \circ}$ have an identity element $\tuple {e_S, e_T}$.
Then:
:$\struct {S, \circ_1}$ has an identity element $e_S$
and:
:$\struct {T, \circ_2}$ has an identity element $e_T$.

Proof:
Let $\tuple {e_S, e_T}$ be an identity of $\struct {S \times T, \circ}$.
Then we have:
{{begin-eqn}}
{{eqn | q = \forall \tuple {s, t} \in S \times T
      | l = \tuple {s, t} \circ \tuple {e_S, e_T}
      | r = \tuple {s, t}
      | c = {{Defof|Identity Element}}
}}
{{eqn | ll= \leadsto
      | l = \tuple {s \circ_1 e_S, t \circ_2 e_T}
      | r = \tuple {s, t}
      | c = {{Defof|External Direct Product}}
}}
{{eqn | ll= \leadsto
      | q = \forall s \in S, t \in T
      | l = s \circ_1 e_S
      | r = s
      | c = Equality of Ordered Pairs
}}
{{eqn | lo= \land
      | l = t \circ_2 e_T
      | r = t
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | q = \forall \tuple {s, t} \in S \times T
      | l = \tuple {e_S, e_T} \circ \tuple {s, t}
      | r = \tuple {s, t}
      | c = {{Defof|Identity Element}}
}}
{{eqn | ll= \leadsto
      | l = \tuple {e_S \circ_1 s, e_T \circ_2 t}
      | r = \tuple {s, t}
      | c = {{Defof|External Direct Product}}
}}
{{eqn | ll= \leadsto
      | q = \forall s \in S, t \in T
      | l = e_S \circ_1 s
      | r = s
      | c = Equality of Ordered Pairs
}}
{{eqn | lo= \land
      | l = e_T \circ_2 t
      | r = t
      | c = 
}}
{{end-eqn}}
Thus $e_S$ and $e_T$ are identity elements of $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ respectively.
Theorem: External Direct Product Inverses/Sufficient Condition

Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Let:
:$s^{-1}$ be an inverse of $s \in \struct {S, \circ_1}$
and:
:$t^{-1}$ be an inverse of $t \in \struct {T, \circ_2}$.
Then $\tuple {s^{-1}, t^{-1} }$ is an inverse of $\tuple {s, t} \in \struct {S \times T, \circ}$.

Proof:
Let:
:$e_S$ be the identity for $\struct {S, \circ_1}$
and:
:$e_T$ be the identity for $\struct {T, \circ_2}$.
Also let:
:$s^{-1}$ be the inverse of $s \in \struct {S, \circ_1}$
and
:$t^{-1}$ be the inverse of $t \in \struct {T, \circ_2}$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {s, t} \circ \tuple {s^{-1}, t^{-1} }
      | r = \tuple {s \circ_1 s^{-1}, t \circ_2 t^{-1} }
      | c = 
}}
{{eqn | r = \tuple {e_S, e_T}
      | c = 
}}
{{eqn | r = \tuple {s^{-1} \circ_1 s, t^{-1} \circ_2 t}
      | c = 
}}
{{eqn | r = \tuple {s^{-1}, t^{-1} } \circ \tuple {s, t}
      | c = 
}}
{{end-eqn}}
Thus the inverse of $\tuple {s, t}$ is $\tuple {s^{-1}, t^{-1} }$.
Theorem: External Direct Product Inverses/Necessary Condition

Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Let $\tuple {s^{-1}, t^{-1} }$ be an inverse of $\tuple {s, t} \in \struct {S \times T, \circ}$.
Then:
:$s^{-1}$ is an inverse of $s \in \struct {S, \circ_1}$
and:
:$t^{-1}$ is an inverse of $t \in \struct {T, \circ_2}$.

Proof:
Let $\tuple {e_S, e_T}$ be the identity element of $\struct {S \times T, \circ}$.
Let $\tuple {s^{-1}, t^{-1} }$ be an inverse element of $\tuple {s, t} \in \struct {S \times T, \circ}$.
Then we have:
{{begin-eqn}}
{{eqn | l = \tuple {s, t} \circ \tuple {s^{-1}, t^{-1} }
      | r = \tuple {e_S, e_T}
      | c = {{Defof|Inverse Element}}
}}
{{eqn | ll= \leadsto
      | l = \tuple {s \circ_1 s^{-1}, t \circ_2 t^{-1} }
      | r = \tuple {e_S, e_T}
      | c = {{Defof|External Direct Product}}
}}
{{eqn | ll= \leadsto
      | q = \forall s \in S, t \in T
      | l = s \circ_1 s^{-1}
      | r = e_S
      | c = Equality of Ordered Pairs
}}
{{eqn | lo= \land
      | l = t \circ_2 t^{-1}
      | r = e_T
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \tuple {s^{-1}, t^{-1} } \circ \tuple {s, t} 
      | r = \tuple {e_S, e_T}
      | c = {{Defof|Inverse Element}}
}}
{{eqn | ll= \leadsto
      | l = \tuple {s^{-1} \circ_1 s, t^{-1} \circ_2 t}
      | r = \tuple {e_S, e_T}
      | c = {{Defof|External Direct Product}}
}}
{{eqn | ll= \leadsto
      | l = s^{-1} \circ_1 s
      | r = e_S
      | c = Equality of Ordered Pairs
}}
{{eqn | lo= \land
      | l = t^{-1} \circ_2 t
      | r = e_T
      | c = 
}}
{{end-eqn}}
Thus $s^{-1}$ and $t^{-1}$ are inverse elements of $s \in \struct {S, \circ_1}$ and $t \in \struct {T, \circ_2}$ respectively.
Theorem: External Direct Product Closure/Sufficient Condition

Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be algebraic structures.
Let $\struct {S \times T, \circ}$ be the external direct product of $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be closed.
Then $\struct {S \times T, \circ}$ is also closed.

Proof:
Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be closed.
Let $\tuple {s_1, t_1} \in S \times T$ and $\tuple {s_2, t_2} \in S \times T$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {s_1, t_1} \circ \tuple {s_2, t_2}
      | r = \tuple {s_1 \circ_1 s_2, t_1 \circ_2 t_2}
      | c = {{Defof|External Direct Product}}
}}
{{eqn | o = \in
      | r = S \times T
      | c = as $S$ and $T$ are closed: $s_1 \circ_1 s_2 \in S, t_1 \circ_2 t_2 \in T$
}}
{{end-eqn}}
demonstrating that $\struct {S \times T, \circ}$ is closed.
Theorem: External Direct Product Closure/Necessary Condition

Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be algebraic structures.
Let $\struct {S \times T, \circ}$ be the external direct product of $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Let $\struct {S \times T, \circ}$ be a closed algebraic structure. 
Then $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ are also closed.

Proof:
Let $\struct {S \times T, \circ}$ be closed.
Let $s_1, s_2 \in S$ and $t_1, t_2 \in T$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {s_1 \circ_1 s_2, t_1 \circ_2 t_2}
      | r = \tuple {s_1, t_1} \circ \tuple {s_2, t_2}
      | c = {{Defof|External Direct Product}}
}}
{{eqn | o = \in
      | r = S \times T
      | c = as $\struct {S \times T, \circ}$ is closed
}}
{{eqn | ll= \leadsto
      | l = s_1 \circ_1 s_2
      | o = \in
      | r = S
      | c = {{Defof|External Direct Product}}
}}
{{eqn | lo= \land
      | l = t_1 \circ_2 t_2
      | o = \in
      | r = T
      | c = 
}}
{{end-eqn}}
demonstrating that $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ are closed.
Theorem: Green's Identities

Let $\struct {M, g}$ be a compact Riemannian manifold with or without boundary $\partial M$.
Let $\hat g$ be the induced metric on $\partial M$.
Let $\map {C^\infty} M$ be the smooth function space.
Let $u, v \in \map {C^\infty} M$ be smooth real functions on $M$.
Let $\rd V_g$ be the Riemannian volume form.
Let $\nabla^2$ be the Laplace-Beltrami operator.
Let $\grad$ be the gradient operator.
Let $N$ be the outward-pointing unit normal vector field on $\partial M$.
Let $\innerprod \cdot \cdot_g$ be the Riemannian metric {{WRT}} the metric $g$.
Then:
:$\ds \int_M u \nabla^2 v \rd V_g = \int_{\partial M} u N v \rd V_{\hat g} - \int_M \innerprod {\grad u} {\grad v}_g \rd V_g$
:$\ds \int_M \paren {u \nabla^2 v - v \nabla^2 u} \rd V_g = \int_{\partial M} \paren {u N v - v N u} \rd V_{\hat g}$
{{Research|There are also $\R^3$ identities, often presented separately, and there are 3 of them, the third being Laplacian of Green function}}

Proof:
{{ProofWanted}}
{{Namedfor|George Green|cat = Green}}
Theorem: Integer as Difference between Two Squares/Formulation 1

Let $n$ be a positive integer.
Then $n$ can be expressed as:
:$n = a^2 - b^2$
{{iff}} $n$ has at least two distinct divisors of the same parity that multiply to $n$.

Proof:
{{begin-eqn}}
{{eqn | l = n
      | r = a^2 - b^2
      | c = 
}}
{{eqn | r = \paren {a + b} \paren {a - b}
      | c = Difference of Two Squares
}}
{{end-eqn}}
Thus $n = p q$ where:
{{begin-eqn}}
{{eqn | n = 1
      | l = p
      | r = \paren {a + b}
      | c = 
}}
{{eqn | n = 2
      | l = q
      | r = \paren {a - b}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = p + q
      | r = 2 a
      | c = $(1) + (2)$
}}
{{eqn | l = p - q
      | r = 2 b
      | c = $(1) - (2)$
}}
{{eqn | ll= \leadsto
      | l = a
      | r = \dfrac {p + q} 2
      | c = 
}}
{{eqn | l = b
      | r = \dfrac {p - q} 2
      | c = 
}}
{{end-eqn}}
Thus for $a$ and $b$ to be integers, both $p$ and $q$ must be:
:distinct, otherwise $p = q$ and so $b = 0$
:either both even or both odd, otherwise both $p + q$ and $p - q$ will be odd, and so neither $\dfrac {p + q} 2$ nor $\dfrac {p - q} 2$ are defined in $\Z$.
Hence the result.
{{qed}}
Category:Square Numbers
Category:Difference of Two Squares
9733z9nz11in0djv1un5gnh1497a7h6
Theorem: Integer as Difference between Two Squares/Formulation 2

Any integer can be expressed as the difference of two squares {{iff}} that integer is NOT $n \equiv 2 \pmod 4$

Proof:
: Each integer will be in one of the  4 sets of residue classes modulo $4$:
: $n \equiv 0 \pmod 4$
: $n \equiv 1 \pmod 4$
: $n \equiv 2 \pmod 4$
: $n \equiv 3 \pmod 4$
{{begin-eqn}}
{{eqn | l = 4x
      | r = \paren{x + 1 }^2 - \paren{x - 1 }^2
      | c = 
}}
{{eqn | l = 4x + 1
      | r = \paren{2x + 1 }^2 - 4x^2
      | c = 
}}
{{eqn | l = 4x + 3
      | r = \paren{2x + 2 }^2 - \paren{2x + 1 }^2
      | c = 
}}
{{end-eqn}}
: For $n \equiv 2 \pmod 4$, it is impossible to represent such an integer as the difference of two squares.
Taking the squares of each of the residue classes, we have:
: $0^2 \equiv 0 \pmod 4$
: $1^2 \equiv 1 \pmod 4$
: $2^2 \equiv 0 \pmod 4$
: $3^2 \equiv 1 \pmod 4$
: Therefore, when taking the difference of two squares, $n \equiv 2 \pmod 4$ is never a result.
: $0 - 0 \equiv 0 \pmod 4$
: $1 - 1 \equiv 0 \pmod 4$
: $0 - 1 \equiv 3 \pmod 4$
: $1 - 0 \equiv 1 \pmod 4$
{{qed}}
Category:Square Numbers
Category:Difference of Two Squares
n1o1m5xgnu87toc12jr3hd2kp6bkj4n
Theorem: Integer as Sums and Differences of Consecutive Squares

Every integer $k$ can be represented in infinitely many ways in the form:
:$k = \pm 1^2 \pm 2^2 \pm 3^3 \pm \dots \pm m^2$
for some (strictly) positive integer $m$ and some choice of signs $+$ or $-$.

Proof:
First we notice that:
{{begin-eqn}}
{{eqn | o =
      | r = \paren {m + 1}^2 - \paren {m + 2}^2 - \paren {m + 3}^2 + \paren {m + 4}^2
}}
{{eqn | r = \paren {-1} \paren {m + 1 + m + 2} + \paren {m + 3 + m + 4}
      | c = Difference of Two Squares
}}
{{eqn | r = 4
}}
{{end-eqn}}
Now we prove a weaker form of the theorem by induction:
:Every ''positive'' integer $k$ can be represented in ''at least one way'' in the form:
::$k = \pm 1^2 \pm 2^2 \pm 3^3 \pm \dots \pm m^2$
:for some (strictly) positive integer $m$ and some choice of signs $+$ or $-$.
Theorem: Condition for Mapping between Structure and Cartesian Product of Substructures to be Bijection

Let $\struct {S, \circ}$ be an algebraic structure with $1$ operation.
Let $\struct {A, \circ {\restriction_A} }$ and $\struct {B, \circ {\restriction_B} }$ be closed algebraic substructures of $\struct {S, \circ}$, where $\circ {\restriction_A}$ and $\circ {\restriction_B}$ are the operations induced by the restrictions of $\circ$ to $A$ and $B$ respectively.
Let the mapping $\phi: A \times B \to S$ be defined as:
:$\forall \tuple {a, b} \in A \times B: \map \phi {a, b} = a \circ b$
where $A \times B$ denotes the Cartesian product of $A$ and $B$.
Then:
:$\phi$ is a bijection
{{iff}}:
:for all $s \in S$: there exists a unique $\tuple {a, b} \in A \times B$ such that $a \circ b = s$.

Proof:
First we establish that from Set of Finite Subsets under Induced Operation is Closed:
:$A \times B \subseteq S$
Thus:
:$\forall \tuple {a, b} \in A \times B: \exists s \in S: s = a \circ b = \map \phi {a, b}$
Thus $\phi$ is indeed a mapping.
Theorem: Mapping on Cartesian Product of Substructures is Restriction of Operation

Let $\struct {S, \circ}$ be an algebraic structure with $1$ operation.
Let $\struct {A, \circ {\restriction_A} }, \struct {B, \circ {\restriction_B} }$ be closed algebraic substructures of $\struct {S, \circ}$, where $\circ {\restriction_A}$ and $\circ {\restriction_B}$ are the operations induced by the restrictions of $\circ$ to $A$ and $B$ respectively.
Let the mapping $\phi: A \times B \to S$ be defined as:
:$\forall \tuple {a, b} \in A \times B: \map \phi {a, b} = a \circ b$
where $A \times B$ denotes the Cartesian product of $A$ and $B$.
Then $\phi$ is the restriction to $A \times B$ of the operation $\circ$ on $S \times S \to S$.

Proof:
{{Proofread}}
Suppose the mapping $\phi: A \times B \to S$ is defined as:
:$\forall \tuple {a, b} \in A \times B: \map \phi {a, b} = a \circ b$
where $A \times B$ denotes the Cartesian product of $A$ and $B$.
Then:
{{begin-eqn}}
{{eqn | q = \forall a \in A, b \in B
      | l = a \circ b  
      | r = \map \phi {a, b}
      | c = Definition of $\phi$
}}
{{eqn | r = \paren{\phi \cap \paren {A \times B} } \tuple{a, b}
      | c = {{Defof|Restriction/Mapping}}
}}
{{eqn | r = \phi {\restriction_{A \times B} } \tuple{a, b }
      | c = {{Defof|Restriction/Mapping}}
}}
{{eqn | r = a \mathbin {\circ {\restriction_{A \times B} } } b
      | c = {{Defof|Operation Induced by Restriction}}
}}
{{end-eqn}}
{{qed}}
Theorem: Codomain of Internal Direct Isomorphism is Subset Product of Factors

Let $\struct {S, \circ}$ be an algebraic structure with $1$ operation.
Let $\struct {A, \circ {\restriction_A} }$ and $\struct {B, \circ {\restriction_B} }$ be closed algebraic substructures of $\struct {S, \circ}$, where $\circ {\restriction_A}$ and $\circ {\restriction_B}$ are the operations induced by the restrictions of $\circ$ to $A$ and $B$ respectively.
Let $\struct {S, \circ}$ be the '''internal direct product of $A$ and $B$'''.
Then $S$ is the subset product of $A$ and $B$:
:$S = A \circ B$

Proof:
First we establish that from Set of Finite Subsets under Induced Operation is Closed:
:$A \times B \subseteq S$
From Condition for Mapping between Structure and Cartesian Product of Substructures to be Bijection:
:$\phi$ is a bijection
{{iff}}:
:for all $s \in S$: there exists a unique $\tuple {a, b} \in A \times B$ such that $a \circ b = s$.
By definition of internal direct product:
:the mapping $\phi: A \times B \to S$ defined as:
::$\forall a \in A, b \in B: \map \phi {a, b} = a \circ b$
:is an isomorphism from the '''(external) direct product''' $\struct {A, \circ {\restriction_A} } \times \struct {B, \circ {\restriction_B} }$ onto $\struct {S, \circ}$.
Thus $\phi$ is {{afortiori}} a bijection.
From Condition for Mapping between Structure and Cartesian Product of Substructures to be Bijection it follows that:
:for all $s \in S$: there exists a unique $\tuple {a, b} \in A \times B$ such that $a \circ b = s$.
That is:
:$S = \set {a \circ b: \tuple {a, b} \in A \times B}$
Hence $S = A \circ B$ by definition of subset product.
{{qed}}
Category:Subset Products
Category:Internal Direct Products
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Theorem: Left Operation is Entropic

The left operation is entropic:
:$\forall a, b, c, d: \paren {a \gets b} \gets \paren {c \gets d} = \paren {a \gets c} \gets \paren {b \gets d}$

Proof:
{{begin-eqn}}
{{eqn | l = \paren {a \gets b} \gets \paren {c \gets d}
      | r = a \gets c
      | c = {{Defof|Left Operation}}
}}
{{eqn | r = a
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \paren {a \gets c} \gets \paren {b \gets d}
      | r = a \gets b
      | c = {{Defof|Left Operation}}
}}
{{eqn | r = a
      | c = 
}}
{{eqn | r = \paren {a \gets b} \gets \paren {c \gets d}
      | c = {{apriori}}
}}
{{end-eqn}}
{{qed}}