nyuuzyou/wikis huggingface dataset:
Gamma Function of Zero

== Theorem ==
Let $\Gamma$ denotes the Gamma function
Then:
:$\map \Gamma 0$ is not defined.
== Proof ==
But $\dfrac 1 0$ is not defined.
Hence the result.
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: The Gamma Function: $33 \ \text{(d)}$
Category:Examples of Gamma Function Values
Gardner's Conjecture on Divisibility of Sum of Even Amicable Pairs

== Famous False Conjecture ==
The sum of every amicable pair of even integers is divisible by $9$.
For example:
... and so on.
== Refutation ==
Counterexamples are rare, but they exist:
For example:
== Historical Note ==
== Sources ==
* 
* : $220$
* : $220$
Category:Amicable Pairs
Category:Famous False Conjectures
Gardner's Conjecture on Divisibility of Sum of Even Amicable Pairs/Historical Note

== Historical Note on Gardner's Conjecture on Divisibility of Sum of Even Amicable Pairs ==
 made this conjecture in $1968$ after noticing that all amicable pairs of even integers that were known at that time had this property.
The counterexample given was provided by , having originally been discovered by .
== Sources ==
* : $220$
* : $220$
Category:Historical Notes
Gauge Transformation/Examples/Magnetic Field

== Example of Gauge Transformation ==
Consider the magnetic fields $\mathbf E$ and $\mathbf B$.
They can be written in terms of scalar potential $\phi$ and vector potential $\mathbf A$ as:
These may be changed by means of the gauge transformation:
The physical quantities $\mathbf E$ and $\mathbf B$ have not been changed, but the purely mathematical $\phi$ and $\mathbf A$ have.
== Sources ==
* 
Category:Examples of Gauge Transformations
Gauss's Continued Fraction

== Theorem ==
:$\ds \frac {\map F {\alpha, \beta + 1; \gamma + 1; x} } {\map F {\alpha, \beta; \gamma; x} } = \cfrac 1 {1 - \cfrac {a x} {1 - \cfrac {b x} {1 - \cfrac {c x} {1 - \cfrac {d x} {1 - \cfrac {\ddots} {\ddots } } } } } }$
where:
:$\map F {a, b; c; x}$ is the Gaussian hypergeometric function of $x$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } { c^{\overline k} } \dfrac {x^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
:$\begin{array}{rc|rc}
a = & \dfrac {\alpha \paren {\gamma - \beta} } {\gamma \paren {\gamma + 1} } & b = & \dfrac {\paren {\beta + 1} \paren {\gamma - \alpha + 1} } {\paren {\gamma + 1} \paren {\gamma + 2} } \\
c = & \dfrac {\paren {\alpha + 1} \paren {\gamma - \beta + 1} } {\paren {\gamma + 2} \paren {\gamma + 3} } & d = & \dfrac {\paren {\beta + 2} \paren {\gamma - \alpha + 2} } {\paren {\gamma + 3} \paren {\gamma + 4} } \\
e = & \dfrac {\paren {\alpha + 2} \paren {\gamma - \beta + 2} } {\paren {\gamma + 4} \paren {\gamma + 5} } & f = & \dfrac {\paren {\beta + 3} \paren {\gamma - \alpha + 3} } {\paren {\gamma + 5} \paren {\gamma + 6} } \\
g = & \dfrac {\paren {\alpha + 3} \paren {\gamma - \beta + 3} } {\paren {\gamma + 6} \paren {\gamma + 7} } & h = & \dfrac {\paren {\beta + 4} \paren {\gamma - \alpha + 4} } {\paren {\gamma + 7} \paren {\gamma + 8} } \\
\end{array}$
:
== Proof ==
:First, we demonstate:
:$\ds \dfrac {\map F {\alpha, \beta + 1, \gamma + 1; x} } {\map F {\alpha, \beta, \gamma; x} } = \dfrac 1 {1 - \dfrac {\alpha \paren {\gamma - \beta} x } {\gamma \paren {\gamma + 1} } \dfrac {\map F {\alpha + 1, \beta + 1, \gamma + 2; x} } {\map F {\alpha, \beta + 1, \gamma + 1; x} } }$
:Next, we demonstate:
:$\ds \dfrac {\map F {\alpha + 1, \beta, \gamma + 1; x} } {\map F {\alpha, \beta, \gamma; x} } = \dfrac 1 {1 - \dfrac {\beta \paren {\gamma - \alpha} x } {\gamma \paren {\gamma + 1} } \dfrac {\map F {\alpha + 1, \beta + 1, \gamma + 2; x} } {\map F {\alpha + 1, \beta, \gamma + 1; x} } }$
Therefore:
== Sources ==
* : Chapter Disquisitiones generales circa seriem infinitam $1 + \dfrac {\alpha \beta} \gamma x + $ etc.
Category:Continued Fractions
Category:Gaussian Hypergeometric Function
Gauss's Digamma Theorem

== Theorem == 
Let $\dfrac p q$ be a positive rational number with $p < q$.
Then: 
:$\ds \map \psi {\dfrac p q} = -\gamma - \ln 2 q - \dfrac \pi 2 \map \cot {\dfrac p q \pi} + 2 \sum_{n \mathop = 1}^{\ceiling {q / 2} - 1} \map \cos {\dfrac {2 \pi p n} q} \map \ln {\map \sin {\dfrac {\pi n} q} }$
where:
:$\psi$ is the digamma function
:$\cot$ is the cotangent function
:$\ln$ is the natural logarithm. 
== Proof == 
We now note:
The term $\ds -q \sum_{n \mathop = 0}^\infty \paren {\frac {t^{p + n q} } {p + n q} }$ is $q$ multiplied by the Power Series Expansion: $\ds \map \ln {1 - t} = -\sum_{n \mathop = 1}^\infty \dfrac {t^n} n$, but only when $n \equiv p \pmod q$
From Simpson's Dissection: 
:$\ds \sum_{n \mathop = 0}^\infty a_{n q + p} x^{n q + p} = \dfrac 1 q \sum_{j \mathop = 0}^{q - 1} \omega^{- j p} \map f {\omega^j x}$
where $\omega = e^{2 i \pi/ q}$ is a primitive $q$th root of unity.
Since the $\map \psi {\dfrac p q}$ is a real number, we only want the real part of the natural logarithm.
This requires us to determine the modulus of the argument.
Plugging this back into our main result:
== Sources ==
* 
* : Chapter $1$. The Gamma and Beta Functions
* 
Category:Digamma Function
Gauss's Hypergeometric Theorem

== Theorem ==
Let $a, b, c \in \C$.
Let $c \notin \Z_{\le 0}$.
Let $\map \Re {c - a - b} > 0$.
Then:
:$\map F {a, b; c; 1} = \dfrac {\map \Gamma c \map \Gamma {c - a - b} } {\map \Gamma {c - a} \map \Gamma {c - b} }$
where:
:$\map F {a, b; c; 1}$ is the Gaussian hypergeometric function of $1$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } { c^{\overline k} } \dfrac {1^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
:$\map \Gamma {n + 1} = n!$ is the Gamma function.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof 1 ==
== Proof 2 ==
== Examples ==
== Also see ==
* Dixon's Hypergeometric Theorem
* Kummer's Hypergeometric Theorem
* Properties of Generalized Hypergeometric Function
== Sources ==
* : $\S 31$: Hypergeometric Functions: Miscellaneous Properties: $31.14$
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
* 
Category:Gauss's Hypergeometric Theorem
Category:Gaussian Hypergeometric Function
Category:Gamma Function
Gauss's Hypergeometric Theorem/Corollary 1

== Corollary to Gauss's Hypergeometric Theorem ==
Let $\map \Re {1 - a} > 0$.
Let $c \notin \Z_{\le 0}$ and $c \ne 1$.
Then:
:$\ds \sum_{k \mathop = 0}^\infty \dfrac {a^{\overline k} } {\paren {c - 1 + k} k!} = \dfrac {\map \Gamma {c - 1} \map \Gamma {1 - a} } {\map \Gamma {c - a} }$
== Proof ==
Set $b = c - 1$ in Gauss's Hypergeometric Theorem
Before substitution:
After substitution:
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
Category:Gauss's Hypergeometric Theorem
Gauss's Hypergeometric Theorem/Corollary 2

== Corollary to Gauss's Hypergeometric Theorem ==
Let $\map \Re {a - 1} < 0$.
Then:
:$\ds \dfrac 1 a + \dfrac {a } {\paren {a + 1} 1!} + \dfrac {a \paren {a + 1} } {\paren {a + 2} 2!} + \dfrac {a \paren {a + 1} \paren {a + 2} } {\paren {a + 3} 3!} + \cdots = \dfrac {\pi} {\map \sin {\pi a } } $
== Proof ==
Set $c - 1 = a$ in Corollary 1:
Before substitution:
After substitution:
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
Category:Gauss's Hypergeometric Theorem
Gauss's Hypergeometric Theorem/Examples

== Examples of Use of Gauss's Hypergeometric Theorem==
=== Example: $\map F {1, 2; 4; 1}$ ===
=== Example: $\map F {1, 1; \dfrac 5 2; 1}$ ===
=== Example: $\map F {\dfrac 1 2, \dfrac 1 2; \dfrac 3 2; 1}$ ===
=== Example: $\dfrac {10} 3 \map F {\dfrac 3 {10}, \dfrac 3 {10}; \dfrac {13} {10}; 1}$ ===
Category:Examples of Use of Gauss's Hypergeometric Theorem
Gauss's Hypergeometric Theorem/Examples/2F1(0.5,0.5;1.5;1)

== Example of Use of Gauss's Hypergeometric Theorem ==
:$1 + \paren {\dfrac 1 {2^1 \times 3 \times 1!} } + \paren {\dfrac {1 \times 3} {2^2 \times 5 \times 2!} } + \paren {\dfrac {1 \times 3 \times 5} {2^3 \times 7 \times 3!} } + \cdots = \dfrac \pi 2$
== Proof ==
From Gauss's Hypergeometric Theorem:
:$\map F {a, b; c; 1} = \dfrac {\map \Gamma c \map \Gamma {c - a - b} } {\map \Gamma {c - a} \map \Gamma {c - b} }$
where:
:$\map F {a, b; c; 1}$ is the Gaussian hypergeometric function: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } { c^{\overline k} } \dfrac {1^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
:$\map \Gamma {n + 1} = n!$ is the Gamma function.
We have:
and:
Therefore:
:$1 + \paren {\dfrac 1 {2^1 \times 3 \times 1!} } + \paren {\dfrac {1 \times 3} {2^2 \times 5 \times 2!} } + \paren {\dfrac {1 \times 3 \times 5} {2^3 \times 7 \times 3!} } + \cdots = \dfrac \pi 2$
Category:Examples of Use of Gauss's Hypergeometric Theorem‎
Category:Formulas for Pi
Gauss's Hypergeometric Theorem/Examples/2F1(1,1;2.5;1)

== Example of Use of Gauss's Hypergeometric Theorem ==
:$1 + \dfrac 2 5 + \paren {\dfrac {2 \times 4} {5 \times 7} } + \paren {\dfrac {2 \times 4 \times 6} {5 \times 7 \times 9} } + \cdots = 3$
== Proof ==
From Gauss's Hypergeometric Theorem:
:$\map F {a, b; c; 1} = \dfrac {\map \Gamma c \map \Gamma {c - a - b} } {\map \Gamma {c - a} \map \Gamma {c - b} }$
where:
:$\map F {a, b; c; 1}$ is the Gaussian hypergeometric function of $1$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } { c^{\overline k} } \dfrac {1^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
:$\map \Gamma {n + 1} = n!$ is the Gamma function.
We have:
and:
Therefore:
:$1 + \dfrac 2 5 + \paren {\dfrac {2 \times 4} {5 \times 7} } + \paren {\dfrac {2 \times 4 \times 6} {5 \times 7 \times 9} } + \cdots = 3$
Category:Examples of Use of Gauss's Hypergeometric Theorem‎
Gauss's Hypergeometric Theorem/Examples/2F1(1,2;4;1)

== Example of Use of Gauss's Hypergeometric Theorem ==
:$1 + \dfrac 2 4 + \paren {\dfrac {2 \times 3} {4 \times 5} } + \paren {\dfrac {2 \times 3 \times 4} {4 \times 5 \times 6} } + \cdots = 3$
== Proof ==
From Gauss's Hypergeometric Theorem:
:$\map F {a, b; c; 1} = \dfrac {\map \Gamma c \map \Gamma {c - a - b} } {\map \Gamma {c - a} \map \Gamma {c - b} }$
where:
:$\map F {a, b; c; 1}$ is the Gaussian hypergeometric function: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } { c^{\overline k} } \dfrac {1^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
:$\map \Gamma {n + 1} = n!$ is the Gamma function.
We have:
and:
Therefore:
:$1 + \dfrac 2 4 + \paren {\dfrac {2 \times 3} {4 \times 5} } + \paren {\dfrac {2 \times 3 \times 4} {4 \times 5 \times 6} } + \cdots = 3$
Category:Examples of Use of Gauss's Hypergeometric Theorem‎
Gauss's Hypergeometric Theorem/Examples/Corollary 2 2pi over phi

== Example of Use of Gauss's Hypergeometric Theorem ==
:$\paren {\dfrac 1 {10^{-1} \times 3 \times 0!} } + \paren {\dfrac 3 {10^0 \times 13 \times 1!} } + \paren {\dfrac {3 \times 13} {10^1 \times 23 \times 2!} } + \paren {\dfrac {3 \times 13 \times 23} {10^2 \times 33 \times 3!} } + \cdots = \dfrac {2 \pi} \phi$
== Proof ==
From Gauss's Hypergeometric Theorem: Corollary 2:
:$\ds \dfrac 1 a + \dfrac a {\paren {a + 1} 1!} + \dfrac {a \paren {a + 1} } {\paren {a + 2} 2!} + \dfrac {a \paren {a + 1} \paren {a + 2} } {\paren {a + 3} 3!} + \cdots = \dfrac {\pi} {\map \sin {\pi a } }$
Let $a = \dfrac 3 {10}$.
On the :
On the :
Therefore:
:$\paren {\dfrac 1 {10^{-1} \times 3 \times 0!} } + \paren {\dfrac 3 {10^0 \times 13 \times 1!} } + \paren {\dfrac {3 \times 13} {10^1 \times 23 \times 2!} } + \paren {\dfrac {3 \times 13 \times 23} {10^2 \times 33 \times 3!} } + \cdots = \dfrac {2\pi} \phi$
Category:Examples of Use of Gauss's Hypergeometric Theorem‎
Category:Formulas for Pi
Gauss's Hypergeometric Theorem/Historical Note

== Historical Note on Gauss's Hypergeometric Theorem ==
The proof shown above is a more detailed version of a proof by Srinivasa Ramanujan.
Based on Ramanujan's Notebook, as transcribed in Chapter $10$ of Berndt's book, Ramanujan's proof goes as follows:
:Assume that $n$ and $x$ are integers with $n \ge 0$ and $n + x \ge 0$. 
:Expanding $\paren {1 + u}^{y + n}$ and $\paren {1 + \dfrac 1 u}^x$ in their formal binomial series and taking their product, we find that, if $a_n$ is the coefficient of $u^n$:
:On the other hand, expanding $\paren {1 + u}^{x + y + n}$ in its binomial series and dividing by $u^x$, we find that:
:Comparing $(8.2)$ and $(8.3)$, we deduce $(8.1)$.
As can be seen, Ramanujan jumped over several intermediate steps in $(8.2)$, but his assertions were all correct.
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
Category:Historical Notes
Gauss's Hypergeometric Theorem/Proof 1

== Theorem ==
== Proof ==
Let $x, y, n \in \C$ be complex numbers such that $\map \Re {x + y + n + 1} > 0$.
Let $u \in \C$ be a complex number such that $\cmod u < 1$.
Expanding the product of $\paren {1 + u}^{y + n}$ and $\paren {\dfrac {1 + u} u}^x$:
The coefficient $a_n$ of $u^n$ of the product $\dfrac {\paren {1 + u}^{x + y + n} } {u^x}$ above can be determined by setting $k = k + n$ in the series with $\dbinom {y + n} k$:
Now expand $\paren {1 + u}^{x + y + n}$ and divide by $u^x$:
The coefficient $a_n$ of $u^n$ of the product $\dfrac {\paren {1 + u}^{x + y + n} } {u^x}$ above is:
Equating coefficients gives us:
Letting $a = -x$, $b = -y$ and $c = n + 1$, we obtain:
== Historical Note ==
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
Category:Gauss's Hypergeometric Theorem
Category:Examples of Equating Coefficients
Gauss's Hypergeometric Theorem/Proof 2

== Theorem ==
== Proof ==
From Euler's Integral Representation of Hypergeometric Function, we have:
:$\ds \map F {a, b; c; x} = \dfrac {\map \Gamma c } {\map \Gamma b \map \Gamma {c - b} } \int_0^1 t^{b - 1} \paren {1 - t}^{c - b - 1} \paren {1 - x t}^{- a} \rd t$
Where $a, b, c \in \C$.
and $\size x < 1$
and $\map \Re c > \map \Re b > 0$.
Since Euler's Integral Representation only applies where $\size x < 1$, we will determine the limit of the integral as $x \to 1$.
Therefore:
Category:Gauss's Hypergeometric Theorem
Gauss's Integral Form of Digamma Function

== Theorem == 
Let $z$ be a complex number with a positive real part.
Then: 
:$\ds \map \psi z = \int_0^\infty \paren {\frac {e^{-t} } t - \frac {e^{-z t} } {1 - e^{-t} } } \rd t$
where $\psi$ is the digamma function.
== Proof == 
From Extension of Harmonic Number to Non-Integer Argument, we have:
:$\map H x = \gamma + \dfrac {\map {\Gamma'} {x + 1} } {\map \Gamma {x + 1} }$
which is equivalent to:
:$\ds \map \psi z = -\gamma + H_{z - 1}$
We aim to demonstrate:
:$\ds \int_0^\infty \paren {\frac {e^{-t} } t - \frac {e^{-z t} } {1 - e^{-t} } } \rd t = -\gamma + H_{z - 1}$
We have:
Let:
We also have:
Then:
== Also see ==
* Definite Integral to Infinity of Reciprocal of Exponential of x minus One minus Exponential of -x over x
* Dirichlet's Integral Form of Digamma Function
== Sources == 
* : $12.3$: Gauss's expression for the logarithmic derivate of the Gamma function as an infinite integral
Category:Digamma Function
Category:Complex Analysis
Category:Definite Integrals
Gauss's Law

== Physical Law ==
=== In the context of Maxwell's Equations ===
where:
:$\nabla \cdot \mathbf E$ denotes the divergence of the electric field $\mathbf E$
:$\rho$ denotes electric charge density
:$\varepsilon_0$ denotes the vacuum permittivity.
== Informal Explanation ==
== Also known as ==
Some sources present the name as Gauss' Law.
== Examples ==
Category:Gauss's Law (Electrostatics)
Category:Gauss's Law
Category:Electrostatics
Gauss's Law/Examples/Spherically Symmetric Body

== Examples of Use of Gauss's Law ==
Let $B$ be a spherical conducting body in space.
Let $B$ have an electric charge on it.
Then the electric field generated by $B$ is the same as an electric field generated by a point charge with the same charge as the total charge as $B$, placed at the center of $B$.
== Proof ==
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.4$ Gauss's Law
Category:Gauss's Law
Gauss's Law/Informal Explanation

== Gauss's Law Informal Explanation ==
Consider a point charge $q$ at the center of a sphere of radius $r$.
From Surface Area of Sphere, the area of the sphere is $4 \pi r^2$.
From Coulomb's Law of Electrostatics, the electric field at the surface of the sphere is of magnitude $\dfrac q {4 \pi \varepsilon_0 r^2}$.
Hence the product of the area and the magnitude of the electric field is:
:$4 \pi r^2 \times \dfrac q {4 \pi \varepsilon_0 r^2} = \dfrac q {\varepsilon_0}$
Hence the total magnitude of the electric field is independent of the radius.
=== Analogy with Light ===
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.4$ Gauss's Law
Category:Gauss's Law
Gauss's Law/Informal Explanation/Analogy with Light

== Gauss's Law Informal Explanation: Example ==
Consider a light bulb at the center of a spherical lampshade which is completely transparent.
All the light from the light bulb escapes.
That is, the total light emerging from the lampshade is independent of the radius of the lampshade.
In fact, the total light emerging from the lampshade is also independent of the shape of the lampshade.
All the light gets out, whether the lampshade is spherical or not.
Gauss's Law is an example of the same basic principle.
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.4$ Gauss's Law
Category:Gauss's Law
Gauss's Law (Electrostatics)

== Physical Law ==
:$\nabla \cdot \mathbf D = \rho$
where:
:$\nabla \cdot \mathbf D$ denotes the divergence of the electric displacement field $\mathbf D$
:$\mathbf D = \varepsilon_0 \mathbf E$ denotes the electric displacement field
:$\rho$ denotes electric charge density
:$\mathbf E$ denotes the electric field
:$\varepsilon_0$ denotes the vacuum permittivity.
== Also presented as ==
== Also see ==
* Definition:Maxwell's Equations
== Sources ==
* : Introduction: Electromagnetic Theory
* 
* 
* 
* 
Category:Gauss's Law (Electrostatics)
Category:Gauss's Law
Category:Electrostatics
Category:Maxwell's Equations
Gauss's Law (Electrostatics)/Also presented as

== Gauss's Law: Also known as == 
Gauss's law can also be seen presented as:
:$\nabla \cdot \mathbf E = \dfrac \rho {\varepsilon_0}$
where:
:$\nabla \cdot \mathbf E$ denotes the divergence of the electric field $\mathbf E$
:$\rho$ denotes electric charge density
:$\varepsilon_0$ denotes the vacuum permittivity.
Category:Gauss's Law (Electrostatics)
Category:Maxwell's Equations
Gauss's Lemma

* Gauss's Lemma (Number Theory): relating the number of least positive residues of a prime to its Legendre symbol;
* Gauss's Lemma (Polynomial Theory)
** Gauss's Lemma on Primitive Rational Polynomials: the product of two primitive polynomials is primitive.
** Gauss's Lemma on Primitive Polynomials over Ring: the same in a wider context.
** Gauss's Lemma on Irreducible Polynomials: concerning the reducibility of polynomials
** Gauss's Lemma on Unique Factorization Domains
Gauss's Lemma (Number Theory)

== Lemma ==
Let $p$ be an odd prime.
Let $a \in \Z: a \not \equiv 0 \pmod p$.
Let $S = \set {a, 2 a, 3 a, \ldots, \dfrac {p - 1} 2 a}$.
Let $n$ denote the number of elements of $S$ whose least positive residue modulo $p$ is greater than $\dfrac p 2$.
Then:
:$\paren {\dfrac a p} = \paren {-1}^n$
where $\paren {\dfrac a p}$ is the Legendre symbol.
== Proof ==
First note that no two elements of $s$ are congruent modulo $p$:
Let $r a \equiv s a \pmod p$ for some $r, s$ such that $1 \le s \le r \le \dfrac {p - 1} 2$.
As $a \perp p$, we have that $r a \equiv s a \implies r \equiv s \pmod p$ from Cancellability of Congruences.
Now $r \equiv s \pmod p$ can happen only when $r = s$ as $0 \le r - s \le \dfrac {p - 1} 2$.
Next, no element of $S$ is congruent modulo $p$ to $0$.
This is because $r a \equiv 0 \pmod p$ when $a \not \equiv 0$ requires $r \equiv 0$ which doesn't happen.
Now, we create $S'$ from $S$ by replacing each element of $S$ by its least positive residue modulo $p$.
Arranging $S'$ into increasing order, we get:
:$S' = \set {b_1, b_2, \ldots, b_m, c_1, c_2, \ldots, c_n}$
where $b_m < \dfrac p 2 < c_1$ and $m + n = \dfrac {p - 1} 2$.
As $p$ is an odd prime, $\dfrac p 2$ is not an integer so neither $b_m$ nor $c_1$ can be equal to $\dfrac p 2$.
Now we let $S' ' = \set {b_1, b_2, \ldots, b_m, p - c_1, p - c_2, \ldots, p - c_n}$
We need to show that $S' ' = \set {1, 2, 3, \ldots, \dfrac {p - 1} 2}$.
Now:
:all the elements of $S' '$ are positive, as all the $c_j < p$
:the largest one is either $b_m$ or $p - c_1$
:there are $\dfrac {p - 1} 2$ of them
:they are all less than $\dfrac p 2$.
So $S' '$ contains $\dfrac {p - 1} 2$ positive integers all less than $\dfrac p 2$.
So all we need to do is show that they are distinct.
As all the elements of $S'$ are distinct, the $m$ elements $b_1, b_2, \ldots, b_m$ are distinct.
Similarly, so are the $n$ elements $p - c_1, p - c_2, \ldots, p - c_n$.
So, we suppose $b_i = p - c_j$ for some $i, j$ and try to derive a contradiction.
So:
:$p = b_i + c_j \equiv r a + s a \pmod p$
for some $r, s$ such that $1 \le r, s \le \dfrac {p - 1} 2$.
But from:
:$p \equiv \paren {r + s} a \pmod p$ and $a \not \equiv 0 \pmod p$
we apply Euclid's Lemma to get $r + s \equiv 0 \pmod p$.
But this is impossible since $2 \le r + s \le p - 1$.
This establishes that $S' ' = \set {1, 2, 3, \ldots, \dfrac {p - 1} 2}$.
Now, we multiply all the elements of $S' '$ together:
Now, from GCD with Prime:
:$\gcd \set {p, \paren {\dfrac {p - 1} 2} !} = 1$
So from Cancellability of Congruences the term $\paren {\dfrac {p - 1} 2} !$ can be cancelled from both sides of the above congruence:
:$1 \equiv \paren {-1}^n a^{\paren {\frac {p - 1} 2} } \pmod p$
Finally, from the definition of the Legendre symbol:
:$\paren {\dfrac a p} \equiv a^{\paren {\frac {p - 1} 2} } \pmod p$
Multiplying both sides of the congruence by $\paren {-1}^n$:
:$\paren {\dfrac a p} = \paren {-1}^n$
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $47 \ \text {a)}$
Category:Legendre Symbol
Gauss's Lemma (Polynomial Theory)

== Theorem ==
Gauss's lemma on polynomials may refer to any of the following statements.
== Product of primitive polynomials is primitive ==
== Gauss's Lemma on Primitive Polynomials over Ring ==
== Content is Multiplicative ==
== Statement on irreducible polynomials ==
== Polynomial ring is UFD ==
Category:Polynomial Theory
Category:Gauss's Lemma (Polynomial Theory)