nyuuzyou/wikis huggingface dataset:
Split Monomorphism is Monic

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $f: C \to D$ be a split monomorphism.
Then $f: C \rightarrowtail D$ is monic.
== Proof ==
Let $g: D \to C$ be a morphism such that:
:$g \circ f = \operatorname{id}_C$
which is guaranteed to exist by definition of split monomorphism.
Suppose that $x, y: B \to C$ are morphisms such that:
:$f \circ x = f \circ y$
Then necessarily also:
:$g \circ \paren {f \circ x} = g \circ \paren {f \circ y}$
and by Composition of Morphisms is Associative:
:$\paren {g \circ f} \circ x = \paren {g \circ f} \circ y$
We are given that $g \circ f = \operatorname{id}_C$.
Hence:
:${\operatorname{id}_C} \circ x = {\operatorname{id}_C} \circ y$
So by definition of identity morphism:
:$x = y$
Let $\mathbf C_1$ denotes the collection morphisms $f, g, h, \ldots$ between its objects for the metacategory $\mathbf C$.
We have that $f \in \mathbf C_1$ such that:
:$f \circ x = f \circ y \implies x = y$
for all morphisms $x, y \in \mathbf C_1$ for which these compositions are defined.
Hence the result by definition of monic.
The situation is illustrated by the following commutative diagram:
:$\begin{xy}
<-4em,0em>*+{B} = "B",
<0em,0em> *+{C} = "C",
<4em,0em> *+{D} = "D",
<4em,-4em>*+{C} = "C2",
"B"+/r.5em/+/^.25em/;"C"+/l.5em/+/^.25em/ **@{-} ?>*@{>} ?*!/_.6em/{x},
"B"+/r.5em/+/_.25em/;"C"+/l.5em/+/_.25em/ **@{-} ?>*@{>} ?*!/^.6em/{y},
"C";"D" **@{-} ?>*@{>} ?*!/_.6em/{f},
"C";"C2" **@{-} ?>*@{>} ?*!/^.6em/{\operatorname{id}_C},
"D";"C2" **@{-} ?>*@{>} ?*!/_.6em/{g},
\end{xy}$
== Sources ==
* : $\S 2.1.1$
Category:Category Theory
Sprague's Property of Root 2

== Theorem ==
Let $S = \sequence {s_n}$ be the sequence of fractions defined as follows:
Let the numerator of $s_n$ be:
:$\floor {n \sqrt 2}$
where $\floor x$ denotes the floor of $x$.
Let the denominators of the terms of $S$ be the (strictly) positive integers missing from the numerators of $S$:
:$S := \dfrac 1 3, \dfrac 2 6, \dfrac 4 {10}, \dfrac 5 {13}, \dfrac 7 {17}, \dfrac 8 {20}, \ldots$
Then the difference between the numerator and denominator of $s_n$ is equal to $2 n$.
== Proof ==
Denote the numerators of the terms of $S$ as $\sequence {N_n}$.
Denote the denominators of the terms of $S$ as $\sequence {D_n}$.
From the definition:
:$\sequence {N_n}$ is a Beatty sequence, where $\sequence {N_n} = \BB_{\sqrt 2} = \sequence{\floor{n \sqrt 2} }_{n \mathop \in \Z_{> 0} }$
:$\sequence {N_n}$ and $\sequence {D_n}$ are complementary Beatty sequences.
Then by Beatty's Theorem, $\sequence {D_n}$ is a Beatty sequence.
Define $\sequence {D_n} = \BB_y = \sequence{\floor{n y} }_{n \mathop \in \Z_{> 0} }$.
Then we have: 
So we have $s_n = \dfrac {N_n} {D_n} = \dfrac {\floor {n \sqrt 2} } {\floor {n \paren {2 + \sqrt 2} } }$.
The difference between the numerator and denominator of $s_n$ is $\floor {n \paren {2 + \sqrt 2} } - \floor {n \sqrt 2}$.
We have:
Hence the result.
== Sources ==
* 
* : $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
* : $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
Category:Square Root of 2
Category:Beatty Sequences
Square-Bracing Problem

== Classic Problem ==
Let there be an unlimited supply of rigid rods of unit length which may be joined together at their ends with a hinged joint.
Let $4$ of these rods be used to construct a square.
How many more of these rods are needed to make the square rigid?
== Solution ==
=== Non-Crossing Rods ===
=== Crossing Rods ===
== Sources ==
* 
Category:Square-Bracing Problem
Category:Classic Problems
Category:Geometry
Square-Summable Indexed Sets Closed Under Addition

== Theorem ==
Let $\family {a_i}_{i \mathop \in I}, \family {b_i}_{i \mathop \in I}$ be $I$-indexed families of real numbers.
Let:
where $\ds \sum$ denotes the generalized sums.
Then:
:$\ds \sum \set {\paren {a_i + b_i}^2: i \in I} < \infty$
== Proof ==
First we note that:
:$\forall i \in I: {a_i}^2 > 0$ and ${b_i}^2 > 0$
This will be used throughout implicitly.
There are two cases to consider: $a_i b_i < 0$ and $a_i b_i > 0$.
=== Case $1$: $a_i b_i < 0$ ===
We have:
=== Case $1$: $a_i b_i > 0$ ===
Suppose $a_i = b_i$.
Then:
Otherwise $a_i \ne b_i$.
, suppose ${a_i}^2 < {b_i}^2$.
Thus:
From both cases we see that:
:$\forall i \in \N: \paren {a_i + b_i}^2 < 3 \paren { {a_i}^2 + {b_i}^2}$
We note that:
Hence:
Hence the result.
Category:Generalized Sums
Square Divides Product of Multiples

== Theorem ==
Let $a, b, c, \in \Z$ be integers.
Let:
:$a \divides b, a \divides c$
where $\divides$ denotes divisibility.
Then:
:$a^2 \divides b c$
== Proof ==
We have that:
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $2 \ \text{(b)}$
Category:Divisors
Square Function is Even

== Theorem ==
The square function:
:$\map f x = x^2$
is an even function.
== Proof ==
Hence the result by definition of even function.
== Sources ==
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.3$ Functions of a Real Variable: $\text {(h)}$ Even and Odd Functions
* 
* 
Category:Square Function
Category:Examples of Even Functions
Square Inscribed in Circle is greater than Half Area of Circle

== Theorem ==
A square inscribed in a circle has an area greater than half that of the circle.
== Proof ==
:
Let $ABCD$ be a square inscribed in a circle.
Let $EFGH$ be a square circumscribed around the same circle.
We have that:
:$ABCD$ is twice the area of the triangle $ADB$.
:$EFGH$ is twice the area of the rectangle $EFBD$.
From Area of Rectangle, the area of $EFBD$ is $ED \cdot DB$.
From Area of Triangle in Terms of Side and Altitude, the area of $ADB$ is $\dfrac 1 2 \cdot ED \cdot DB$.
Thus:
:$EFGH = 2 \cdot ABCD$
But the area of $EFGH$ is greater than the area of the circle around which it is circumscribed.
Therefore half of the area of $EFGH$ is greater than half of the area of the circle around which it is circumscribed.
Therefore the area of $ABCD$ is greater than half of the area of the circle within which it is inscribed.
== Historical Note ==
This result is used in , specifically Book $\text{XII}$, in several places. However, it is never extracted and proved separately as a lemma; the result is merely stated without comment after the first time it is demonstrated.
On it has been determined as being an important enough result to be proved in its own right.
Category:Circles
Category:Squares
Square Matrices over Real Numbers under Multiplication form Monoid

== Theorem ==
Let $\map {\MM_\R} n$ be a $n \times n$ matrix space over the set of real numbers $\R$.
Then the set of all $n \times n$ real matrices $\map {\MM_\R} n$ under matrix multiplication (conventional) forms a monoid.
== Proof ==
:Matrix Multiplication over Order n Square Matrices is Closed.
:Matrix Multiplication is Associative.
:The Unit Matrix is Unity of Ring of Square Matrices.
== Sources ==
* : $\S 29$. Semigroups: definition and examples: $(4)$
Category:Matrix Algebra
Category:Examples of Monoids
Square Matrices with +1 or -1 Determinant under Multiplication forms Group

== Theorem ==
Let $n \in \Z_{>0}$ be a strictly positive integer.
Let $S$ be the set of square matrices of order $n$ of real numbers whose determinant is either $1$ or $-1$.
Let $\struct {S, \times}$ denote the algebraic structure formed by $S$ whose operation is (conventional) matrix multiplication.
Then $\struct {S, \times}$ is a group.
== Proof ==
Taking the group axioms in turn:
=== ===
Let $\mathbf A, \mathbf B \in S$.
By definition of matrix product, $\mathbf {A B}$ is a square matrix of order $n$.
From Determinant of Matrix Product:
:$\det \mathbf A \det \mathbf B = \map \det {\mathbf {A B} }$
Hence the determinant of $\mathbf {A B}$ is either $1$ or $-1$.
Thus $\mathbf {A B} \in S$ and so $\struct {S, \times}$ is closed.
=== ===
We have that Matrix Multiplication is Associative.
Thus $\times$ is associative on $\struct {S, \times}$.
=== ===
From Determinant of Unit Matrix, the unit matrix $\mathbf I$ is in $S$.
From Unit Matrix is Unity of Ring of Square Matrices, the unit matrix $\mathbf I$ serves as the identity element of $\struct {S, \times}$.
=== ===
Because the determinants of the elements of $S$ are $1$ or $-1$, they are by definition nonsingular.
We have that $\mathbf I$ is the identity element of $\struct {\R, \circ}$.
From the definition of nonsingular matrix, the inverse of any nonsingular matrix $\mathbf A$ is $\mathbf A^{-1}$.
All the group axioms are thus seen to be fulfilled, and so $\struct {S, \times}$ is a group.
== Sources ==
* : Chapter $1$: Definitions and Examples: Exercise $1 \ \text{(d)}$
Category:Examples of Groups
Category:Matrix Groups
Square Matrix/Examples/3 x 3

== Example of Square Matrix ==
The $3 \times 3$ square matrix is as follows:
$\quad \mathbf A = \begin {bmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33} \\
\end {bmatrix}$
== Sources ==
* : Chapter $2$: 'And you do addition?': $\S 2.2$: Functions on vectors: $\S 2.2.2$: $3 \times 3$ matrices
Category:Examples of Matrices
Square Matrix/Examples/Real 2 x 2

== Example of Square Matrix ==
A $2 \times 2$ real square matrix is an array of $4$ real numbers $p, q, r, s$ arranged as:
$\quad \mathbf A = \begin {bmatrix}
p & q \\
r & s
\end {bmatrix}$
== Sources ==
* : Chapter $1$: Integral Domains: $\S 3$. Definition of an Integral Domain: Example $3$
Category:Examples of Matrices
Square Matrix is Row Equivalent to Triangular Matrix

== Theorem ==
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$ over a commutative ring $R$.
Then $\mathbf A$ can be converted to an upper or lower triangular matrix by elementary row operations.
== Proof 1 ==
== Proof 2 ==
Category:Square Matrices
Category:Triangular Matrices
Category:Square Matrix is Row Equivalent to Triangular Matrix
Square Matrix is Row Equivalent to Triangular Matrix/Proof 1

== Theorem ==
== Proof ==
Let $\mathbf A$ be a square matrix of order $n$.
We proceed by induction on $n$, the number of rows of $\mathbf A$.
=== Basis for the Induction ===
For $n = 1$, we have a matrix of just one element, which is trivially diagonal, hence both upper and lower triangular.
This is the basis for the induction.
=== Induction Hypothesis ===
Fix $n \in \N$, and assume all $n \times n$-matrices can be upper triangularised by elementary row operations. 
If $R$ is a field, assume all $n \times n$-matrices can be upper triangularised by elementary row operations of type $2$.
This forms our induction hypothesis.
=== Induction Step ===
Let $\mathbf A = \sqbrk a_{n + 1}$ be a square matrix of order $n + 1$.
When the first column of $\mathbf A$ contains only zeroes, it is upper triangularisable the submatrix $\map {\mathbf A} {1; 1}$ is.
Each elementary row operation used in triangularisation process of the submatrix $\map {\mathbf A} {1; 1}$ will not change the zeros of the first column of $\mathbf A$.
So when $\map {\mathbf A} {1; 1}$ is upper triangularised, then $A$ will also be upper triangularised.
From the induction hypothesis, we conclude that $\mathbf A$ can be upper triangularised by elementary row operations.
Now suppose that its first column contains a non-zero value.
Suppose that $a_{1 1} = 0$.
Let $j$ be the smallest row index such that $a_{j 1} \ne 0$, and note that $j$ exists by assumption.
Now apply the following operation of type $2$:
:$r_1 \to r_1 + r_j$
As $a_{j 1} \ne 0$, this enforces $a_{1 1} \ne 0$, and we continue as in the case below.
Suppose $a_{1 1} \ne 0$. 
We use the following operations for all $j \in \set {2, \ldots, n + 1}$:
:$(1): \quad$ Put $c = a_{j 1}$.
:$(2): \quad r_j \to a_{1 1} r_j$
:$(3): \quad r_j \to r_j - c r_1$
This will put the first column to zero (except for the first element, $a_{1 1}$).
It follows that $\mathbf A$ can be upper triangularised the submatrix $\map {\mathbf A} {1; 1}$ can.
Again, the induction hypothesis renders $\mathbf A$ upper triangularisable by elementary row operations.
This completes the case distinction, and hence the result follows by induction.
To put the matrix $\mathbf A$ into lower triangular form, just do the same thing, but start with the last column and the last diagonal element $a_{n + 1 \; n + 1}$.
Category:Square Matrix is Row Equivalent to Triangular Matrix
Square Matrix is Row Equivalent to Triangular Matrix/Proof 2

== Theorem ==
== Proof ==
This proof assumes that $R$ is a field, which makes the triangulation process slightly quicker.
By this assumptions, all elements of $\mathbf A$ have multiplicative inverses.
Let $\mathbf A$ be a square matrix of order $n$.
We proceed by induction on $n$, the number of rows of $\mathbf A$.
=== Basis for the Induction ===
For $n = 1$, we have a matrix of just one element, which is trivially diagonal, hence both upper and lower triangular.
This is the basis for the induction.
=== Induction Hypothesis ===
Fix $n \in \N$, and assume all $n \times n$-matrices can be upper triangularised by elementary row operations. 
If $R$ is a field, assume all $n \times n$-matrices can be upper triangularised by elementary row operations of type $2$.
This forms our induction hypothesis.
=== Induction Step ===
Let $\mathbf A = \sqbrk a_{n + 1}$ be a square matrix of order $n + 1$.
When the first column of $\mathbf A$ contains only zeroes, it is upper triangularisable the submatrix $\map {\mathbf A} {1; 1}$ is.
Each elementary row operation used in triangularisation process of the submatrix $\map {\mathbf A} {1; 1}$ will not change the zeros of the first column of $\mathbf A$.
So when $\map {\mathbf A} {1; 1}$ is upper triangularised, then $\mathbf A$ will also be upper triangularised.
From the induction hypothesis, we conclude that $\mathbf A$ can be upper triangularised by elementary row operations.
Now suppose that its first column contains a non-zero value.
Suppose that $a_{1 1} = 0$.
Let $j$ be the smallest row index such that $a_{j 1} \ne 0$, and note that $j$ exists by assumption.
Now apply the following operation of type $2$:
:$r_1 \to r_1 + r_j$
As $a_{j 1} \ne 0$, this enforces $a_{1 1} \ne 0$, and we continue as in the case below.
Suppose $a_{1 1} \ne 0$. 
We use the following operations of type $2$:
:$\forall j \in \set {2, \ldots, n + 1}: r_j \to r_j - \dfrac {a_{j 1}} {a_{1 1}} r_1$
This will put the first column to zero (except for the first element, $a_{1 1}$).
It follows that $\mathbf A$ can be upper triangularised the submatrix $\map {\mathbf A} {1; 1}$ can.
Again, the induction hypothesis renders $\mathbf A$ upper triangularisable by elementary row operations.
This completes the case distinction, and hence the result follows by induction.
To put the matrix $\mathbf A$ into lower triangular form, just do the same thing, but start with the last column and the last diagonal element $a_{n + 1 \; n + 1}$.
Category:Square Matrix is Row Equivalent to Triangular Matrix
Square Matrix with Duplicate Columns has Zero Determinant

== Theorem ==
If two columns of a square matrix over a commutative ring $\struct {R, +, \circ}$ are identical, then its determinant is zero.
=== Corollary ===
== Proof ==
Let $\mathbf A$ be a square matrix over $R$ with two identical columns.
Let $\mathbf A^\intercal$ denote the transpose of $\mathbf A$.
Then $\mathbf A^\intercal$ has two identical rows.
Then:
== Also see ==
* Square Matrix with Duplicate Rows has Zero Determinant
== Sources ==
* : $\S 7.2$
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Proposition $1.13$
* 
* 
* 
Category:Matrix Algebra
Category:Determinants
Category:Square Matrix with Duplicate Columns has Zero Determinant
Square Matrix with Duplicate Columns has Zero Determinant/Corollary

== Corollary to Square Matrix with Duplicate Columns has Zero Determinant ==
If a square matrix has a zero column, its determinant is zero.
== Proof ==
If you add any column to a zero column, you get a square matrix with two identical columns.
From Multiple of Column Added to Column of Determinant, performing this operation does not change the value of the determinant.
So a square matrix with a zero column has the same determinant as that with two identical columns.
From Square Matrix with Duplicate Columns has Zero Determinant, that is zero.
== Also see ==
* 
== Sources ==
* : $\text{A}.2$: Linear algebra and determinants: Theorem $\text{A}.10 \ (1)$
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Fact $1.7 \ \text {(e)}$
Category:Square Matrix with Duplicate Columns has Zero Determinant
Square Matrix with Duplicate Rows has Zero Determinant

== Theorem ==
If two rows of a square matrix over a commutative ring $\struct {R, +, \circ}$ are the same, then its determinant is zero.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Square Matrix with Duplicate Columns has Zero Determinant
== Sources ==
* : $\S 7.2$
* 
* 
* 
Category:Matrix Algebra
Category:Determinants
Category:Square Matrix with Duplicate Rows has Zero Determinant
Square Matrix with Duplicate Rows has Zero Determinant/Corollary

== Corollary to Square Matrix with Duplicate Rows has Zero Determinant ==
If a square matrix has a zero row, its determinant is zero.
== Proof ==
If you add any row to a zero row, you get a square matrix with two identical rows.
From Multiple of Row Added to Row of Determinant, performing this operation does not change the value of the determinant.
So a square matrix with a zero row has the same determinant as that with two identical rows.
From Square Matrix with Duplicate Rows has Zero Determinant, that is zero.
== Also see ==
* 
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Fact $1.7 \ \text {(e)}$
Category:Square Matrix with Duplicate Rows has Zero Determinant
Square Matrix with Duplicate Rows has Zero Determinant/Proof 1

== Theorem ==
== Proof ==
The proof proceeds by induction over $n$, the order of the square matrix.
=== Basis for the Induction ===
Let $n = 2$, which is the smallest natural number for which a square matrix of order $n$ can have two identical rows.
Let $\mathbf A = \sqbrk a_2$ be a square matrix over $R$ with two identical rows.
Then, by definition of determinant:
:$\map \det {\mathbf A} = a_{1 1} a_{2 2} - a_{1 2} a_{2 1} = a_{1 1} a_{2 2} - a_{2 2} a_{1 1} = 0$
This is the basis for the induction.
=== Induction Hypothesis ===
Assume for $n \in \N_{\ge 2}$ that any square matrices of order $n$ over $R$ with two identical rows has determinant equal to $0$.
=== Induction Step ===
Let $\mathbf A$ be a square matrix of order $n+1$ over $R$ with two identical rows.
Let $i_1, i_2 \in \set {1, \ldots, n + 1}$ be the indices of the identical rows, and let $i_1 < i_2$.
Let $i \in \set {1, \ldots, n + 1}$.
Let $\map {\mathbf A} {i; 1}$ denote the submatrix obtained from $\mathbf A$ by removing row $i$ and column $1$.
If $i \ne i_1$ and $i \ne i_2$, then $\map {\mathbf A} {i; 1}$ still contains two identical rows.
By the induction hypothesis:
:$\map \det {\map {\mathbf A} {i; 1} } = 0$
Now consider the matrices $\map {\mathbf A} {i_1; 1}$ and $\map {\mathbf A} {i_2; 1}$.
Let $r_j$ denote row $j$ of $\map {\mathbf A} {i_1; 1}$.
If we perform the following sequence of $i_2 - i_1 - 1$ elementary row operations on $\map {\mathbf A} {i_1; 1}$:
:$r_{i_1} \leftrightarrow r_{i_1 + 1} \ ; \ r_{i_1 + 1} \leftrightarrow r_{i_1 + 2} \ ; \ \ldots \ ; \ r_{i_2 - 1} \leftrightarrow r_{i_2}$
we will transform $\map {\mathbf A} {i_1; 1}$ into $\map {\mathbf A} {i_2; 1}$.
From Determinant with Rows Transposed:
:$\map \det {\map {\mathbf A} {i_1; 1} } = \paren {-1}^{i_2 - i_1 - 1} \map \det {\map {\mathbf A} {i_2; 1} }$
Then:
== Sources ==
* : $\S 7.2$
Category:Square Matrix with Duplicate Rows has Zero Determinant
Square Matrix with Duplicate Rows has Zero Determinant/Proof 2

== Theorem ==
== Proof ==
Suppose that $\forall x \in R: x + x = 0 \implies x = 0$.
From Determinant with Rows Transposed, if you exchange two rows of a square matrix, the sign of its determinant changes.
If you exchange two identical rows of a square matrix, then the sign of its determinant changes from $D$, say, to $-D$.
But the matrix stays the same.
So $D = -D$ and so $D = 0$.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Proposition $1.9$
Category:Square Matrix with Duplicate Rows has Zero Determinant
Square Modulo 24 of Odd Integer Not Divisible by 3

== Theorem ==
Let $a \in \Z$ be an integer such that:
:$2 \nmid a$
:$3 \nmid a$
where $\nmid$ denotes non-divisibility.
Then:
:$a^2 \equiv 1 \pmod {24}$
That is:
:$24 \divides \paren {a^2 - 1}$
where $\divides$ denotes divisibility.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $7$
Category:Square Modulo 24 of Odd Integer Not Divisible by 3
Category:Odd Squares
Square Modulo 24 of Odd Integer Not Divisible by 3/Proof 1

== Theorem ==
== Proof ==
Let $a$ be as asserted.
We have that:
:$2 \nmid a$
From Odd Square Modulo 8:
:$a^2 \equiv 1 \pmod 8$
which means:
:$8 \divides a^2 - 1$
We also have that:
:$3 \nmid a$
From Square Modulo 3: Corollary 3:
:$3 \divides a^2 - 1$
We have from Coprime Integers: $3$ and $8$ that:
:$3 \perp 8$
where $\perp$ denotes coprimality.
As we have that:
:$8 \divides a^2 - 1$
and:
:$3 \divides a^2 - 1$
it follows from Product of Coprime Factors that:
:$24 \divides a^2 - 1$
Hence the result.
Category:Square Modulo 24 of Odd Integer Not Divisible by 3
Square Modulo 24 of Odd Integer Not Divisible by 3/Proof 2

== Theorem ==
== Proof ==
Let $a$ be as asserted.
We have that:
:$3 \nmid a$
which means that either of the following hold:
Suppose $a = 3 k_1 + 1$.
As $a$ is odd, it must be the case that $k_1$ is even.
Hence:
If $k$ is even, then:
:$\exists r \in \Z: k = 2 r$
Hence:
:$a^2 = 24 r \paren {3 k + 1} + 1$
and so:
:$24 \divides a^2 - 1$
If $k$ is odd, then $3 k + 1$ is even, and:
:$\exists r \in \Z: 3 k + 1 = 2 r$
Hence:
:$a^2 = 24 r k + 1$
and so:
:$24 \divides a^2 - 1$
Suppose $a = 3 k_2 + 2$.
As $a$ is odd, it must be the case that $k_2$ is also odd.
Hence:
If $k$ is even, then:
:$\exists r \in \Z: k = 2 r$
Hence:
and so:
:$24 \divides a^2 - 1$
If $k$ is odd, then $\paren {3 k + 5}$ is even, and:
:$\exists r \in \Z: 3 k + 5 = 2 r$
Hence:
and so again:
:$24 \divides a^2 - 1$
All cases have been accounted for.
Hence the result.
Category:Square Modulo 24 of Odd Integer Not Divisible by 3
Square Modulo 3

== Theorem ==
Let $x \in \Z$ be an integer.
Then one of the following holds:
where:
:$\divides$ denotes divisibility
:$\nmid$ denotes non-divisibility.
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
== Proof ==
Let $x$ be an integer.
Using Congruence of Powers throughout, we make use of:
: $x \equiv y \pmod 3 \implies x^2 \equiv y^2 \pmod 3$
There are three cases to consider:
: $(1): \quad x \equiv 0 \pmod 3$: we have $x^2 \equiv 0^2 \pmod 3 \equiv 0 \pmod 3$
: $(2): \quad x \equiv 1 \pmod 3$: we have $x^2 \equiv 1^2 \pmod 3 \equiv 1 \pmod 3$
: $(3): \quad x \equiv 2 \pmod 3$: we have $x^2 \equiv 2^2 \pmod 3 \equiv 1 \pmod 3$
== Sources ==
* : Chapter $2$: Some Properties of $\Z$: Exercise $2.9$
* : Chapter $2$: Divisibility Theory in the Integers: $2.1$ The Division Algorithm: Problems $2.1$: $3 \ \text{(b)}$
Category:Modulo Arithmetic
Category:Square Numbers
Category:Square Modulo 3