nyuuzyou/wikis huggingface dataset:
Pythagorean Triangle from Fibonacci Numbers

== Theorem ==
Take $4$ consecutive Fibonacci numbers:
:$F_n, F_{n + 1}, F_{n + 2}, F_{n + 3}$
Let:
:$a := F_n F_{n + 3}$
:$b := 2 F_{n + 1} F_{n + 2}$
:$c := F_{2 n + 3}$
Then:
:$a^2 + b^2 = c^2$
and:
:$\dfrac {a b} 2 = F_n \times F_{n + 1} \times F_{n + 2} \times F_{n + 3}$
That is, if the legs of a right triangle are the product of the outer terms and twice the inner terms, then:
:the hypotenuse is the Fibonacci number whose index is half the sum of the indices of the four given Fibonacci numbers.
:the area is the product of the four given Fibonacci numbers.
== Proof ==
By definition of Fibonacci numbers:
:$F_n = F_{n + 2} - F_{n + 1}$
and:
:$F_{n + 3} = F_{n + 2} + F_{n + 1}$
Then $a$ can be expressed as:
Now consider:
and it can be seen that the ordered triple:
:$\paren {{F_{n + 2} }^2 - {F_{n + 1} }^2, 2 F_{n + 1} F_{n + 2}, {F_{n + 2} }^2 + {F_{n + 1} }^2}$
is in the form of Solutions of Pythagorean Equation.
It remains to be shown that:
:${F_{n + 2} }^2 + {F_{n + 1} }^2 = F_{2 n + 3}$
We have:
and the proof is complete.
The area is trivial:
== Sources ==
* 
* : $5$
* : $5$
Category:Fibonacci Numbers
Category:Pythagorean Triangles
Pythagorean Triangle from Sum of Reciprocals of Consecutive Same Parity Integers

== Theorem ==
Let $a, b \in \Z_{>0}$ be (strictly) positive integers such that they are consecutively of the same parity.
Let $\dfrac p q = \dfrac 1 a + \dfrac 1 b$.
Then $p$ and $q$ are the legs of a Pythagorean triangle.
== Proof ==
Let $a$ and $b$ both be even.
Then:
From Solutions of Pythagorean Equation, $\tuple {n, n + 1}$ form the generator for the primitive Pythagorean triple $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ where $m = n + 1$.
The legs of the resulting primitive Pythagorean triangle are $2 n \paren {n + 1}$ and $\paren {n + 1}^2 - n^2$.
Let $a$ and $b$ both be odd.
Then:
From Solutions of Pythagorean Equation, $\tuple {2 n + 2, 1}$ form the generator for the primitive Pythagorean triple $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ where $m = 2 n + 2$ and $n = 1$.
The legs of the resulting primitive Pythagorean triangle are $2 \paren {2 n + 2}$ and $\paren {2 n + 2}^2 - 1$.
== Sources ==
* : $13$
* : $13$
Category:Pythagorean Triangles
Pythagorean Triangle whose Area is Half Perimeter

== Theorem ==
The $3-4-5$ triangle is the only Pythagorean triangle whose area is half its perimeter.
== Proof ==
Let $a, b, c$ be the lengths of the sides of a Pythagorean triangle $T$.
Thus $a, b, c$ form a Pythagorean triple.
By definition of Pythagorean triple, $a, b, c$ are in the form:
:$2 m n, m^2 - n^2, m^2 + n^2$
We have that $m^2 + n^2$ is always the hypotenuse.
Thus the area of $T$ is given by:
:$\AA = m n \paren {m^2 - n^2}$
The perimeter of $T$ is given by:
:$\PP = m^2 - n^2 + 2 m n + m^2 + n^2 = 2 m^2 + 2 m n$
We need to find all $m$ and $n$ such that $\PP = 2 \AA$. 
Thus:
As $m$ and $n$ are both (strictly) positive integers, it follows immediately that:
:$n = 1$
:$m - n = 1$
and so:
:$m = 2, n = 1$
and the result follows.
== Sources ==
* : $5$
* : $13$
* : $5$
* : $13$
Category:Pythagorean Triangles
Pythagorean Triangle whose Hypotenuse and Leg differ by 1

== Theorem ==
Let $P$ be a Pythagorean triangle whose sides correspond to the Pythagorean triple $T$.
Then:
: the hypotenuse of $P$ is $1$ greater than one of its legs
:
: the generator for $T$ is of the form $G = \tuple {n, n + 1}$ where $n \in \Z_{> 0}$ is a (strictly) positive integer.
== Proof ==
We have from Solutions of Pythagorean Equation that the set of all primitive Pythagorean triples is generated by:
:$\tuple {2 m n, m^2 - n^2, m^2 + n^2}$
where $m, n \in \Z$ such that:
:$m, n \in \Z_{>0}$ are (strictly) positive integers
:$m \perp n$, that is, $m$ and $n$ are coprime
:$m$ and $n$ are of opposite parity
:$m > n$.
=== Necessary Condition ===
Let the generator $G$ for $T$ be of the form:
:$G = \set {n, n + 1}$
Recall that from Solutions of Pythagorean Equation, $T$ is of the form:
:$\tuple {2 m n, m^2 - n^2, m^2 + n^2}$
where $m, n \in \Z$ such that:
:$m, n \in \Z_{>0}$ are (strictly) positive integers
:$m \perp n$, that is, $m$ and $n$ are coprime
:$m$ and $n$ are of opposite parity
:$m > n$.
We are given that $n \in \Z_{>0}$ such that $m = n + 1$.
We have that Consecutive Integers are Coprime.
They are also of opposite parity.
Therefore the Pythagorean triple generated by $n$ and $n + 1$ is primitive.
So:
Thus it is seen that $P$ is a Pythagorean triangle where:
:one legs is of length $2 n \paren {n + 1}$
:the hypotenuse is of length $2 n \paren {n + 1} + 1$
Thus $P$ is a Pythagorean triangle whose hypotenuse of $P$ is $1$ greater than one of its legs.
=== Sufficient Condition ===
Let the hypotenuse of $P$ be $1$ greater than one of its legs.
Let the legs of $P$ be $a$ and $b$.
Let the hypotenuse of $P$ be $h$ such that $h = b + 1$.
By Consecutive Integers are Coprime, $h$ is coprime to $b$.
Then by Elements of Primitive Pythagorean Triple are Pairwise Coprime, $P$ is primitive.
Because $h$ and $b$ are of opposite parity, it follows that $b$ is even.
Thus by Solutions of Pythagorean Equation:
:$a = m^2 - n^2$
:$b = 2 m n$
:$h = m^2 + n^2$
for some $m, n \in \Z_{>0}$ where $m > n$.
Thus:
Hence the result.
== Sequence ==
== Sources ==
* : $13$
* : $13$
Category:Pythagorean Triangles
Pythagorean Triangle whose Hypotenuse and Leg differ by 1/Sequence

== Sequence ==
The sequence of Pythagorean triangles whose hypotenuse and leg differ by 1 can be tabulated as follows:
:$\begin{array} {r r | r r | r r r}
m & n & m^2 & n^2 & 2 m n & m^2 - n^2 & m^2 + n^2 \\
\hline
2 & 1 & 4 & 1 & 4 & 3 & 5 \\
3 & 2 & 9 & 4 & 12 & 5 & 13 \\
4 & 3 & 16 & 9 & 24 & 7 & 25 \\
5 & 4 & 25 & 16 & 40 & 9 & 41 \\
6 & 5 & 36 & 25 & 60 & 11 & 61 \\
7 & 6 & 49 & 36 & 84 & 13 & 85 \\
\hline
\end{array}$
== Sources ==
* : $13$
Category:Pythagorean Triples
Category:Pythagorean Triangles
Pythagorean Triangle with Sides in Arithmetic Sequence

== Theorem ==
The $3-4-5$ triangle is the only Pythagorean triangle such that:
:the lengths of whose sides are in arithmetic sequence
and:
:the lengths of whose sides form a primitive Pythagorean triple.
== Proof ==
Let $a, b, c$ be the lengths of the sides of a Pythagorean triangle such that $a < b < c$.
Let $a, b, c$ be in arithmetic sequence:
:$b - a = c - b$
Let $a, b, c$ form a primitive Pythagorean triple:
:$a \perp b$
By definition of primitive Pythagorean triple, $a, b, c$ are in the form:
:$2 m n, m^2 - n^2, m^2 + n^2$
We have that $m^2 + n^2$ is always the hypotenuse.
There are two cases:
:$(1): \quad 2 m n > m^2 - n^2$, as in, for example, $3, 4, 5$, where $m = 2, n = 1$.
:$(2): \quad 2 m n < m^2 - n^2$, as in, for example, $8-15-17$, where $m = 4, n = 1$.
First, let $2 m n > m^2 - n^2$:
:$a = m^2 - n^2$
:$b = 2 m n$
:$c = m^2 + n^2$
Then:
From Solutions of Pythagorean Equation: Primitive, $m$ and $n$ must be coprime.
Hence $n = 1$ and $m = 2$ are the only $m$ and $n$ which fulfil the requirements.
This leads to the $3-4-5$ triangle.
Now let $2 m n < m^2 - n^2$:
:$a = 2 m n$
:$b = m^2 - n^2$
:$c = m^2 + n^2$
Then:
In order for $a, b, c$ to form a primitive Pythagorean triple, then $m$ and $n$ must be of opposite parity.
If $m$ is even, then $m^2 - 2 m n$ is even.
But then $3 n^2$ is even, which makes $n$ even.
Otherwise, if $m$ is odd, then $m^2 - 2 m n$ is odd.
But then $3 n^2$ is odd, which makes $n$ odd.
So when $2 m n < m^2 - n^2$, $a, b, c$ cannot be both in arithmetic sequence and be a primitive Pythagorean triple.
Hence follows the result.
== Sources ==
* : $5$
* : $13$
* : $5$
* : $13$
Category:Pythagorean Triangles
Pythagorean Triangle with Sides in Arithmetic Sequence/Mistake

== Source Work ==
:
:The Dictionary
:: $5$
:
:The Dictionary
:: $5$
== Mistake ==
:The $3-4-5$ triangle is the only Pythagorean triangle whose sides are in arithmetical sequence.
It needs to be pointed out that, as has defined Pythagorean triangle, $6-8-10$ is also a Pythagorean triangle, and $6, 8, 10$ is an arithmetical sequence.
The correct statement needs to include the condition that the sides form a primitive Pythagorean triple.
== Sources ==
* : $5$
* : $5$
Category:Mistakes/Curious and Interesting Numbers
Pythagorean Triangles whose Area equal their Perimeter

== Theorem ==
There exist exactly $2$ Pythagorean triples which define a Pythagorean triangle whose area equals its perimeter:
:$(1): \quad \tuple {6, 8, 10}$, leading to an area and perimeter of $24$
:$(2): \quad \tuple {5, 12, 13}$, leading to an area and perimeter of $30$.
== Proof ==
From Area of Right Triangle, the area $\AA$ is:
:$\AA = \dfrac {a b} 2$
where $a$ and $b$ are the legs.
$(1): \quad$ The area of the $\tuple {6, 8, 10}$ triangle is $\dfrac {6 \times 8} 2 = 24$.
Its perimeter equals $6 + 8 + 10 = 24$.
$(2): \quad$ The area of the $\tuple {5, 12, 13}$ triangle is $\dfrac {5 \times 12} 2 = 30$.
Its perimeter equals $5 + 12 + 13 = 30$.
It remains to prove that these are the only ones.
Let $a, b, c$ be the lengths of the sides of a Pythagorean triangle $T$.
Thus $a, b, c$ form a Pythagorean triple.
By definition of Pythagorean triple, $a, b, c$ are in the form:
:$2 m n, m^2 - n^2, m^2 + n^2$
We have that $m^2 + n^2$ is always the hypotenuse.
Thus the area of $T$ is given by:
:$\AA = m n \paren {m^2 - n^2}$
The perimeter of $T$ is given by:
:$\PP = m^2 - n^2 + 2 m n + m^2 + n^2 = 2 m^2 + 2 m n$
We need to find all $m$ and $n$ such that $\PP = \AA$. 
Thus:
As $m$ and $n$ are both (strictly) positive integers, it follows immediately that either:
:$n = 1$
:$m - n = 2$
and so:
:$m = 3, n = 1$
leading to the triangle:
: $a = 6, b = 8, c = 10$
or:
:$n = 2$
:$m - n = 1$
and so:
:$m = 3, n = 2$
leading to the triangle:
: $a = 12, b = 5, c = 13$
and the result follows.
== Sources ==
* : $30$
* : Bachet: $110$
* : $30$
Category:Pythagorean Triangles
Category:24
Category:30
Pythagorean Triangles whose Areas are Repdigit Numbers

== Theorem ==
The following Pythagorean triangles have areas consisting of repdigit numbers:
=== $3-4-5$ Triangle ===
It has area $6$.
=== $693-1924-2045$ Triangle ===
It has area $666 \, 666$.
== Proof ==
From Pythagorean Triangle whose Area is Half Perimeter, the area of the $3-4-5$ triangle is $6$, which is trivially repdigit.
The next Pythagorean triangles in area are:
:the $6-8-10$ triangle, which has area $\dfrac {6 \times 8} 2 = 24$
:the $5-2-13$ triangle, which has area $\dfrac {5 \times 12} 2 = 30$
So there are no more Pythagorean triangles whose areas consist of a single digit.
We have that the $693-1924-2045$ triangle is Pythagorean.
Then its area $A$ is given by:
== Sources ==
* : $13$
* : $13$
Category:Pythagorean Triangles
Category:693
Category:1924
Category:2045
Category:666,666
Pythagorean Triple/Examples

== Examples of Pythagorean Triples ==
=== Example: $\tuple {3, 4, 5}$ ===
=== Example: $\tuple {6, 8, 10}$ ===
=== Example: $\tuple {9, 12, 15}$ ===
=== Example: $\tuple {5, 12, 13}$ ===
=== Example: $\tuple {8, 15, 17}$ ===
Category:Examples of Pythagorean Triples
Pythagorean Triple/Examples/3, 4, 5

== Example of Pythagorean Triple ==
The triple $\tuple {3, 4, 5}$ forms a Pythagorean triple which is also a primitive Pythagorean triple.
== Proof ==
Hence $\tuple {3, 4, 5}$ is a Pythagorean triple by definition.
We also have that $3 \perp 4$, demonstrating that $\tuple {3, 4, 5}$ is a primitive Pythagorean triple.
== Sources ==
* 
Category:Examples of Pythagorean Triples
Pythagorean Triple/Examples/5, 12, 13

== Example of Pythagorean Triple ==
The triple $\tuple {5, 12, 13}$ forms a Pythagorean triple which is also a primitive Pythagorean triple.
== Proof ==
Hence $\tuple {5, 12, 13}$ is a Pythagorean triple by definition.
We also have that $5 \perp 12$, demonstrating that $\tuple {5, 12, 13}$ is a primitive Pythagorean triple.
== Sources ==
* 
Category:Examples of Pythagorean Triples
Pythagorean Triple/Examples/6, 8, 10

== Example of Pythagorean Triple ==
The triple $\tuple {6, 8, 10}$ forms a Pythagorean triple which is not a primitive Pythagorean triple.
== Proof ==
Hence $\tuple {6, 8, 10}$ is a Pythagorean triple by definition.
We also have that $\gcd \set {6, 8} = 2$, demonstrating that $\tuple {6, 8, 10}$ is not a primitive Pythagorean triple.
== Sources ==
* 
Category:Examples of Pythagorean Triples
Pythagorean Triple/Examples/8, 15, 17

== Example of Pythagorean Triple ==
The triple $\tuple {8, 15, 17}$ forms a Pythagorean triple which is also a primitive Pythagorean triple.
== Proof ==
Hence $\tuple {8, 15, 17}$ is a Pythagorean triple by definition.
We also have that $8 \perp 15$, demonstrating that $\tuple {8, 15, 17}$ is a primitive Pythagorean triple.
== Sources ==
* 
Category:Examples of Pythagorean Triples
Pythagorean Triple/Examples/9, 12, 15

== Example of Pythagorean Triple ==
The triple $\tuple {9, 12, 15}$ forms a Pythagorean triple which is not a primitive Pythagorean triple.
== Proof ==
Hence $\tuple {9, 12, 15}$ is a Pythagorean triple by definition.
We also have that $\gcd \set {9, 12} = 3$, demonstrating that $\tuple {9, 12, 15}$ is not a primitive Pythagorean triple.
== Sources ==
* 
Category:Examples of Pythagorean Triples
Pépin's Test

== Theorem ==
Let $F_n = 2^{2^n} + 1$ be a Fermat number.
Then $F_n$ is prime :
:$3^{\paren {F_n - 1} / 2} \equiv -1 \pmod {F_n}$
== Proof ==
=== Sufficient Condition ===
Let this congruence hold:
:$3^{\paren {F_n - 1} / 2} \equiv -1 \pmod {F_n}$
Then:
:$3^{F_n - 1} \equiv 1 \pmod {F_n}$
Thus the multiplicative order of $3$ modulo $F_n$ is a divisor of $F_n - 1 = 2^{2^n}$.
This is a power of $2$.
On the other hand, the multiplicative order of $3$ modulo $F_n$ is not a divisor of $\dfrac {F_n - 1} 2$.
Therefore it must be equal to $F_n - 1$.
In particular, there are at least $F_n - 1$ numbers below $F_n$ which are coprime to $F_n<$.
This can happen only if $F_n$ is prime.
=== Necessary Condition ===
Let $F_n$ be prime.
By Euler's Criterion for Quadratic Residue:
:$3^{\paren {F_n - 1} / 2} \equiv \paren {\dfrac 3 {F_n} } \pmod {F_n}$
$\paren {\dfrac 3 {F_n} }$ denotes the Legendre symbol.
By repeated squaring:
:$2^{2^n} \equiv 1 \pmod 3$
Thus:
:$F_n \equiv 2 \pmod 3$
and so:
:$\paren {\dfrac {F_n} 3} = -1$
As $F_n \equiv 1 \pmod 4$, it follows from the Law of Quadratic Reciprocity that:
:$\paren {\dfrac 3 {F_n} } = -1$
== Sources ==
* 
* : $257$
* : $257$
Category:Fermat Numbers
Pépin's Test/Mistake

== Source Work ==
:
:The Dictionary
::$257$
== Mistake ==
:Such tests are easily performed today on computers using this criterion, which is similar to Lucas's test for the primality of Mersenne numbers: $F_n$ is prime it divides $3^{1/x \paren {F_n - 1} } + 1$.
== Correction ==
That last expression should read $3^{\frac 1 2 \paren {F_n - 1} } + 1$, or more neatly: $3^{\paren {F_n - 1} / 2} + 1$.
This misprint was not present in the $1986$ edition of .
== Sources ==
* : $257$
Category:Mistakes/Curious and Interesting Numbers
Pólya-Vinogradov Inequality

== Theorem ==
Let $p$ be a positive odd prime.
Then:
:$\forall m, n \in \N: \ds \size {\sum_{k \mathop = m}^{m + n} \paren {\frac k p} } < \sqrt p \, \ln p$
where $\paren {\dfrac k p}$ is the Legendre symbol.
== Proof ==
Start with the following manipulations:
The expression:
:$\ds \sum_{k \mathop = 0}^{p - 1} \paren {\dfrac k p} e^{-2 \pi i a t / p}$
is a Gauss sum, and has magnitude $\sqrt p$.
Hence:
Here $\norm x$ denotes the difference between $x$ and the closest integer to $x$, that is:
:$\ds \norm x = \inf_{z \mathop \in \Z} \set {\size {x - z} }$
Since $p$ is odd, we have:
Now:
:$\ln \dfrac {2 x + 1} {2 x - 1} > \dfrac 1 x$
for $x > 1$.
To prove this, it suffices to show that the function $f: \hointr 1 \infty \to \R$ given by:
:$\map f x = x \ln \dfrac {2 x + 1} {2 x - 1}$
is decreasing and approaches $1$ as $x \to \infty$.
To prove the latter statement, substitute $v = 1/x$ and take the limit as $v \to 0$ using L'Hospital's Rule.
To prove the former statement, it will suffice to show that $f'$ is less than zero on the interval $\hointr 1 \infty$.
Now we have that:
:$\map {f} x = \dfrac {-4} {4 x^2 - 1} \paren {1 - \dfrac {4 x^2 + 1} {4 x^2 - 1} } > 0$
for $x > 1$.
So $f'$ is increasing on $\hointr 1 \infty$.
But $\map {f'} x \to 0$ as $x \to \infty$.
So $f'$ is less than zero for $x > 1$.
With this in hand, we have:
== Sources ==
* 
Category:Legendre Symbol
Quadratic Equation/Examples

== Examples of Quadratic Equation ==
=== Example: $x^2 + 1 = 0$ ===
=== Example: $x^2 + 4 = 0$ ===
=== Example: $z^2 - \paren {3 + i} z + 4 + 3 i = 0$ ===
=== Example: $z^2 + \paren {2 i - 3} z + 5 - i = 0$ ===
=== Example: $z^2 - 4 z + 5 = 0$ ===
=== Example: $5 z^2 + 2 z + 10 = 0$ ===
=== Example: $z^4 + z^2 + 1 = 0$ ===
=== Example: $z^2 + \paren {i - 2} z + \paren {3 - i} = 0$ ===
=== Example: $z^4 - 2 z^2 + 4 = 0$ ===
=== Example: $z^6 + z^3 + 1 = 0$ ===
=== Example: $3 x^2 + 24 x + 9 = 0$ ===
Category:Examples of Quadratic Equations
Quadratic Equation/Examples/3 x^2 + 24 x + 9 = 0

== Example of Quadratic Equation ==
The quadratic equation:
:$3 x^2 + 24 x + 9 = 0$
has the roots:
:$x = -4 \pm \sqrt {13}$
== Proof ==
The result follows.
== Sources ==
* 
* 
Category:Examples of Quadratic Equations
Quadratic Equation/Examples/5z^2 + 2z + 10 = 0

== Example of Quadratic Equation ==
The quadratic equation in $\C$:
:$5 z^2 + 2 z + 10 = 0$
has the solutions:
:$z = \dfrac {-1 \pm 7 i} 5$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polynomial Equations: $100 \ \text{(a)}$
Category:Examples of Quadratic Equations
Quadratic Equation/Examples/x^2 + 1 = 0

== Example of Quadratic Equation ==
The quadratic equation:
:$x^2 + 1 = 0$
has no root in the set of real numbers $\R$:
:$x = \pm i$
where $i = \sqrt {-1}$ is the imaginary unit.
== Proof ==
From the Quadratic Formula:
== Sources ==
* : $\S 1.1$. Number Systems: $(1.1)$
Category:Examples of Quadratic Equations
Category:Complex Numbers
Quadratic Equation/Examples/x^2 + 4 = 0

== Example of Quadratic Equation ==
The quadratic equation:
:$x^2 + 4 = 0$
has the wholly imaginary roots:
:$x = \pm 2 i$
where $i = \sqrt {-1}$ is the imaginary unit.
== Proof ==
From the Quadratic Formula:
== Sources ==
* 
* 
Category:Examples of Quadratic Equations
Category:Complex Numbers
Quadratic Equation/Examples/z^2 + (2i-3)z + 5-i = 0

== Example of Quadratic Equation ==
The quadratic equation in $\C$:
:$z^2 + \paren {2 i - 3} z + 5 - i = 0$
has the solutions:
:$z = \begin {cases} 2 - 3 i \\ 1 + i \end {cases}$
== Proof ==
From the Quadratic Formula:
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Polynomial Equations: $32$
Category:Examples of Quadratic Equations
Quadratic Equation/Examples/z^2 + (i-2)z + (3-i) = 0

== Example of Quadratic Equation ==
The quadratic equation in $\C$:
:$z^2 + \paren {i - 2} z + \paren {3 - i} = 0$
has the solutions:
:$z = \begin {cases} 1 + i \\ 1 - 2 i \end {cases}$
== Proof ==
From the Quadratic Formula:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polynomial Equations: $100 \ \text{(b)}$
Category:Examples of Quadratic Equations
Quadratic Equation/Examples/z^2 - (3+i)z + 4+3i = 0

== Example of Quadratic Equation ==
The quadratic equation in $\C$:
:$z^2 - \paren {3 + i} z + 4 + 3 i = 0$
has the solutions:
:$z = \begin{cases} 1 + 2 i \\ 2 - i \end{cases}$
== Proof ==
From the Quadratic Formula:
== Sources ==
* : $\S 1$. Algebraic Theory of Complex Numbers: Exercise $7$
Category:Examples of Quadratic Equations
Quadratic Equation/Examples/z^2 - 4z + 5 = 0

== Example of Quadratic Equation ==
Let $b, c \in \R$ be real.
Let the quadratic equation:
:$z^2 + b z + c = 0$
have the root:
:$z = 2 + i$
Then the other root is:
:$z = 2 - i$
and it follows that:
== Proof ==
We are given that one of the roots of $z^2 + b z + c$ is $2 + i$.
It follows from Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs that the other root is $2 - i$.
From Viète's Formulas, it follows that:
Hence the result.
== Sources ==
* : $\S 1$. Algebraic Theory of Complex Numbers: Exercise $8$
Category:Examples of Quadratic Equations
Quadratic Equation/Examples/z^4 + z^2 + 1 = 0

== Example of Quadratic Equation ==
The quartic equation:
:$z^4 + z^2 + 1 = 0$
has the solutions:
:$z = \dfrac {\pm 1 \pm i \sqrt 3} 2$
== Proof ==
Although this is a quartic in $z$, it can be solved as a quadratic in $z^2$.
$-\dfrac 1 2 \pm i \dfrac {\sqrt 3} 2$ are recognised as the complex cube roots of unity, and so:
Taking each in turn:
and:
== Illustration ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polynomial Equations: $102$
Category:Examples of Quadratic Equations
Quadratic Equation/Examples/z^4 + z^2 + 1 = 0/Illustration

== Illustration of Equation $z^4 + z^2 + 1 = 0$ ==
The roots of the quartic equation:
:$z^4 + z^2 + 1 = 0$
can be depicted as follows:
:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polynomial Equations: $102 \ \text {(b)}$
Category:Examples of Quartic Equations
Quadratic Equation/Examples/z^4 - 2z^2 + 4 = 0

== Example of Quadratic Equation ==
The quartic equation:
:$z^4 - 2 z^2 + 4 = 0$
has the solutions:
:$z = \begin {cases} \dfrac {\sqrt 6} 2 + i \dfrac {\sqrt 2} 2 \\
\dfrac {\sqrt 6} 2 - i \dfrac {\sqrt 2} 2 \\
-\dfrac {\sqrt 6} 2 + i \dfrac {\sqrt 2} 2 \\
-\dfrac {\sqrt 6} 2 - i \dfrac {\sqrt 2} 2 \end {cases}$
== Proof ==
Although this is a quartic in $z$, it can be solved as a quadratic in $z^2$.
From the Quadratic Formula:
It remains to solve for $z$.
Let $z = x + i y$.
We have:
Thus we have:
:$z = \begin{cases} \dfrac {\sqrt 6} 2 + i \dfrac {\sqrt 2} 2 \\
\dfrac {\sqrt 6} 2 - i \dfrac {\sqrt 2} 2 \\
-\dfrac {\sqrt 6} 2 + i \dfrac {\sqrt 2} 2 \\
-\dfrac {\sqrt 6} 2 - i \dfrac {\sqrt 2} 2 \end{cases}$
== Sources ==
* : $\S 1$. Algebraic Theory of Complex Numbers: Exercise $10$
Category:Examples of Quadratic Equations
Quadratic Equation/Examples/z^6 + z^3 + 1 = 0

== Example of Quadratic Equation ==
The sextic equation:
:$z^6 + z^3 + 1 = 0$
has the solutions:
:$z = \begin{cases} \cos \dfrac {2 \pi} 9 \pm i \sin \dfrac {2 \pi} 9 \\ \cos \dfrac {4 \pi} 9 \pm i \sin \dfrac {4 \pi} 9 \\ \cos \dfrac {8 \pi} 9 \pm i \sin \dfrac {8 \pi} 9 \end{cases}$
== Proof 1 ==
== Proof 2 ==
Category:Examples of Quadratic Equations
Category:Quadratic Equation/Examples/z^6 + z^3 + 1 = 0
Quadratic Equation/Examples/z^6 + z^3 + 1 = 0/Proof 1

== Example of Quadratic Equation ==
== Proof ==
Although this is a sextic in $z$, it can be solved as a quadratic in $z^3$.
From the Quadratic Formula:
$-\dfrac 1 2 \pm i \dfrac {\sqrt 3} 2$ are recognised as the complex cube roots of unity, and so:
It remains to solve for $z$.
By Roots of Complex Number:
:$z = \set {w \alpha^k: k \in \set {1, 2, \ldots, n - 1} }$
where:
: $w$ is a cube root of $e^{\pm 2 i \pi / 3}$
: $\alpha$ is one of the complex cube roots of unity.
Taking $z^3 = e^{2 i \pi / 3}$:
The roots obtained by taking $z^3 = e^{-2 i \pi / 3}$ can be calculated as above:
We note that:
and:
Category:Quadratic Equation/Examples/z^6 + z^3 + 1 = 0
Quadratic Equation/Examples/z^6 + z^3 + 1 = 0/Proof 2

== Example of Quadratic Equation ==
== Proof ==
From Complex Algebra Examples $z^6 + z^3 + 1$:
:$z^6 + z^3 + 1 = \paren {z^2 - 2 z \cos \dfrac {2 \pi} 9 + 1} \paren {z^2 - 2 z \cos \dfrac {4 \pi} 9 + 1} \paren {z^2 - 2 z \cos \dfrac {8 \pi} 9 + 1}$
Then from Complex Roots of Unity occur in Conjugate Pairs:
:$\paren {z^2 - 2 z \cos \dfrac {2 k \pi} 9 + 1} = \paren {z - \paren {\cos \dfrac {2 k \pi} 9 + i \sin \dfrac {2 k \pi} 9} } \paren {z - \paren {\cos \dfrac {2 k \pi} 9 - i \sin \dfrac {2 k \pi} 9} }$
for each $k$.
The result follows.
Category:Quadratic Equation/Examples/z^6 + z^3 + 1 = 0