hcju/proofwikiqa huggingface dataset:
{{AimForCont}} that: :$\displaystyle \exists x_0,y_0 \in D : \map \alpha {x_0,y_0} > 0$ $\alpha$ is continuous in $D$.  Therefore, there exists a closed ball $B^-_{\epsilon}$ defined by: :$\map {B^-_{\epsilon}} {x_0, y_0} := \set {\tuple{x,y} \in D : \paren {x - x_0}^2 + \paren {y - y_0}^2 \le \epsilon^2}$ such that: :$\forall x,y \in B^-_{\epsilon} : \map \alpha {x,y} > 0$ Choose $\map h {x, y}$ in the following way: :$\map h {x,y} = \begin {cases} 0, \forall x, y \notin B^-_{\epsilon} \\ \sqbrk {\epsilon^2 - \paren {x - x_0}^2 - \paren {y - y_0}^2}^3, \forall x,y \in B^-_{\epsilon} \end{cases}$ Such a choice for $\map h {x,y}$ satisfies the conditions of the lemma. But then both $\alpha$ and $h$ are positive inside this ball. Hence, the integral is positive. This contradicts assumptions of the lemma. Hence: :$\displaystyle \forall x, y \in D : \map \alpha {x,y} = 0$ {{qed}}
An $n$-gon (either stellated or otherwise) is constructed by connecting all $n$ points in some order. From Number of Permutations, there are ${}^n P_n = n!$ ways of ordering $n$ points. However, for each $n$-gon formed in this way, we have: :$n$ ways of choosing the first vertex :$2$ different sides from that vertex to use as the first one (starting off either clockwise or anticlockwise). and all of these end up describing the same $n$-gon. Thus there are $\dfrac n 2 n! = \dfrac {\paren {n - 1}!} 2$ different $n$-gons. {{qed}}
Let $P$ be a regular stellated $n$-gons. Let the $n$ vertices of $P$ be $p_1, p_2, \dotsc, p_n$. These will be arranged on the circumference of a circle $C$, dividing $C$ into $n$ arcs of equal length. Once we have chosen the first side of $P$, the others are all the same length and are completely determined by that first side. {{WLOG}}, the first vertex of $P$ is chosen to be $p_1$ We can choose that first side as follows: :$p_1 p_2, p_1 n_3, \ldots, p_1 p_{n - 1}, p_1 p_n$ But we have that: :$\size {p_1 p_2} = \size {p_1 p_n}$ :$\size {p_1 p_3} = \size {p_1 p_{n - 1} }$ and so on, down to: :$\size {p_1 p_{\paren {n - 1} / 2} }= \size {p_1 p_{\paren {n + 1} / 2} }$ where $\size {p_a p_b}$ denotes the length of the line $p_1 p_b$. So for the $n - 1$ lines that are chosen for the first side of $P$, each is paired with another of the same length. Hence there are $\dfrac {n - 1} 2$ ways of choosing the first side of $P$. Thus there are $\dfrac {n - 1} 2$ distinct regular stellated $n$-gons. {{qed}}
From Wilson's Theorem, $p$ divides $\paren {p - 1}! + 1$. Let $q$ be a prime number less than $p$. Then $q$ is a divisor of $\paren {p - 1}!$ and so does not divide $\paren {p - 1}! + 1$. {{qed}}
We prove the theorem by establishing a bijection between the set of partitions of $2 n$ with that of $2 n + 1$, under the constraint where each partition is an integer power of $2$. We have that $2^k$ is even for all $k > 0$. Also, we have that $2 n + 1$ is odd for all $n$. So, for each partition of $2 n + 1$ into integer powers of $2$, there is at least one part of size $1$. Removing this part gives a partition of $2 n$ into integer powers of $2$. The inverse of this is adding a part of size $1$ to a partition of $2 n$. This will give a partition of $2 n + 1$ into integer powers of $2$. Hence both removing and adding are well-defined mappings. Thus they are bijections. Therefore: :$\map b {2 n} = \map b {2 n + 1}$ {{qed}}
Proof by Counterexample: We have that: :$11 - 5 = 6 = 3 \times 2$ and so by definition of congruence modulo $2$: :$10 \equiv 4 \pmod 2$ But $11 \ne 5$. {{qed}}
By definition of the $p$-adic norm: :$\norm p_p = \frac 1 p < 1$ By definition of the absolute value: :$\size p = p > 1$ By definition of open unit ball equivalence, $\norm {\,\cdot\,}_p$ and $\size {\,\cdot\,}$ are not equivalent norms. By Equivalence of Definitions of Equivalent Division Ring Norms and the definition of topologically equivalent norms then the topology induced by $\norm {\,\cdot\,}_p$ does not equal the topology induced by $\size{\,\cdot\,}$. {{qed}}
It is noted that: :$\sup \set {\size n: n \in \Z} = +\infty$ By a corollary of Characterisation of Non-Archimedean Division Ring Norms then $\size {\,\cdot\,}$ is Archimedean. By P-adic Norm is Non-Archimedean Norm then $\norm {\,\cdot\,}_p$ is non-Archimedean. By Equivalent Norms are both Non-Archimedean or both Archimedean, $\norm {\,\cdot\,}_p$ and $\size {\,\cdot\,}$ are not equivalent norms. {{qed}}
Consider $p_1/p_2 \in \Q$.  With $\norm {\,\cdot\,}_{p_1}$: {{begin-eqn}} {{eqn| l = \norm{p_1/p_2}_{p_1}      | r = \norm{p_1}_{p_1} \norm{1/p_2}_{p_1}      | c = Norm axiom (M2) (Multiplicativity) }} {{eqn| r = \norm{p_1}_{p_1} \times 1      | c = $p_1$ does not divide $p_2$ }} {{eqn| r = 1/{p_1} }} {{eqn| r = 1      | o = \lt }} {{end-eqn}} On the other hand, with $\norm {\,\cdot\,}_{p_2}$: {{begin-eqn}} {{eqn| l = \norm{p_1/p_2}_{p_2}      | r = \norm{p_1}_{p_2} \norm{1/p_2}_{p_2}      | c = Norm axiom (M2) (Multiplicativity) }} {{eqn| r = 1 \times \norm{1/p_2}_{p_2}      | c = $p_2$ does not divide $p_1$ }} {{eqn| r = p_2 }} {{eqn| r = 1      | o = \gt }} {{end-eqn}} By open unit ball equivalence, $\norm {\,\cdot\,}_{p_1}$ and $\norm {\,\cdot\,}_{p_2}$ are not equivalent norms. By Equivalence of Definitions of Equivalent Division Ring Norms and topological equivalence then the topology induced by $\norm {\,\cdot\,}_{p_1}$ does not equal the topology induced by $\norm {\,\cdot\,}_{p_2}$. {{qed}}
We have that: :$x \equiv y \pmod m$ From Congruence of Powers: :$x^k \equiv y^k \pmod m$ From Modulo Multiplication is Well-Defined: :$\forall k \in \set {0, 2, \ldots, r}: a_k x^k \equiv a_k y^k \pmod m$ The result follows from Modulo Addition is Well-Defined. {{qed}}
Let: :$t_0 = 0, t_1 = 1, \dots, t_{m - 1} = m - 1$ Let $n \in \Z$. Then from the Division Theorem there exist unique integers $q$ and $u$ such that: :$n = m q + u$ such that $0 \le u < m$. That is: :$n \equiv u \pmod m$ and $u$ is one of $t_0, t_1, \ldots, t_{m - 1}$. Also, since $\size {t_i - t_j} < m$, no two elements of $\set {t_0, t_1, \ldots, t_{m - 1} }$ are congruent. Thus $\set {t_0, t_1, \ldots, t_{m - 1} }$ is a complete residue system modulo $m$. So each $r_i$ is congruent to exactly one element of $\set {t_0, t_1, \ldots, t_{m - 1} }$. So: :$s \le m$ And since $\set {r_1, r_2, \dotsb, r_s}$ forms a complete residue system modulo $m$, every element of $\set {t_0, t_1, \ldots, t_{m - 1} }$ is congruent to exactly one element of $S$. So: :$m \le s$ Hence the result. {{qed}}
Let $n \in \Z$. From the Division Theorem there exist unique integers $q$ and $u$ such that: :$n = m q + u$ such that $0 \le u < m$. That is: :$n \equiv u \pmod m$ and $u$ is one of $0, 1, \ldots, m - 1$. Also, since $\forall i, j \in \N_m: \size {i - j} < m$, no two elements of $\N_m$ are congruent. Thus $\N_m = \set {0, 1, 2, \ldots, m - 1}$ is a complete residue system modulo $m$. {{qed}}
Proof by induction on $s$: === Basis for the Induction === For $s = 0$, we apply the definition of a complete residue system modulo $p$: :$\forall n \in \Z: \exists a_i \in S: n \equiv a_i \pmod p$ This is our base case. === Induction Hypothesis === This is our induction hypothesis: :For some $k \in \Z_{\ge 0}$, for all integers $n \in \Z$, there exists a congruence of the form: ::$n \equiv \displaystyle \sum_{j \mathop = 0}^k b_j p^j \pmod {p^{k + 1} }$ It is to be demonstrated that: :For all integers $n \in \Z$, there exists a congruence of the form: ::$n \equiv \displaystyle \sum_{j \mathop = 0}^{k + 1} b_j p^j \pmod {p^{k + 2} }$ === Induction Step === This is our induction step: From $n \equiv \displaystyle \sum_{j \mathop = 0}^k b_j p^j \pmod {p^{k + 1} }$ we have: :$\exists q \in \Z: n = \displaystyle \sum_{j \mathop = 0}^k b_j p^j + q p^{k + 1}$ From the definition of a complete residue system modulo $p$: :$\exists a_i \in S: q \equiv a_i \pmod p$ This gives: :$\exists r \in \Z: q = a_i + r p$ Substituting this to the original equation we have: {{begin-eqn}} {{eqn | l = n       | r = \sum_{j \mathop = 0}^k b_j p^j + \paren {a_i + r p} p^{k + 1} }} {{eqn | r = a_i p^{k + 1} + \sum_{j \mathop = 0}^k b_j p^j + r p^{k + 2} }} {{eqn | o = \equiv        | r = a_i p^{k + 1} + \sum_{j \mathop = 0}^k b_j p^j       | rr = \pmod {p^{k + 2} } }} {{end-eqn}} This shows that $n$ can be expressed as the form: :$n \equiv \displaystyle \sum_{j \mathop = 0}^{k + 1} b_j p^j \pmod {p^{k + 2} }$ By the Principle of Mathematical Induction, the theorem is true for any $s$. Note that in the proof above, we did not use the fact that $p$ is a prime number. {{qed}}
Let $P = \set {p < n: p \text { prime}, p \nmid n}$. Let $Q = \set {0 < q < n: q \perp n}$, where $q \perp n$ denotes that $q$ and $n$ are coprime. Let $p \in P$. From Prime not Divisor implies Coprime, $p$ is coprime to $n$. That is: :$p \in Q$ So, by definition of subset: :$P \subseteq Q$ From Integer is Coprime to 1: :$1 \in Q$ But as One is not Prime: :$1 \notin P$ Thus $P \subsetneq Q$ and so: :$\card P < \card Q$ By definition of $\map w n$: :$\card P = \map w n$ and by definition of Euler $\phi$ function: :$\card Q = \map \phi n$ Hence the result. {{qed}}
The above equation is equivalent to the property that all numbers greater than $1$ that are coprime to it but less are prime. For an integer to have this property: If it is greater than $p^2$ for some prime $p$, then it must be divisible by $p$. If not, it will be coprime to $p^2$, a composite number. Let $p_n$ denote the $n$th prime. Suppose $N$ has this property. By the argument above, if $p_{n + 1}^2 \ge N > p_n^2$, we must have $p_1 p_2 \cdots p_n \divides N$. By Absolute Value of Integer is not less than Divisors, we have $p_1 p_2 \cdots p_n \le N$. Bertrand-Chebyshev Theorem asserts that there is a prime between $p_n$ and $2 p_n$. Thus we have $2 p_n > p_{n + 1}$. Hence for $n \ge 5$: {{begin-eqn}} {{eqn | l = N       | o = \ge       | r = p_1 p_2 \cdots p_n }} {{eqn | r = 2 \times 3 \times 5 p_4 \cdots p_n }} {{eqn | o = >       | r = 8 p_{n - 1} p_n }} {{eqn | o = >       | r = 4 p_n^2       | c = Bertrand-Chebyshev Theorem }} {{eqn | o = >       | r = p_{n + 1}^2       | c = Bertrand-Chebyshev Theorem }} {{eqn | o = \ge       | r = N       | c = From assumption }} {{end-eqn}} This is a contradiction.  Hence we must have $N \le p_5^2 = 121$. From the argument above we also have: :$2 \divides N$ for $4 < N \le 9$ :$2, 3 \divides N$ for $9 < N \le 25$ :$2, 3, 5 \divides N$ for $25 < N \le 49$ :$2, 3, 5, 7 \divides N$ for $49 < N \le 121$ So we end up with the list $N = 1, 2, 3, 4, 6, 8, 12, 18, 24, 30$. This list is verified in Integers such that all Coprime and Less are Prime. {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition: :After $n$ modified perfect faro shuffles, $C$ will be in position $a_n$ where $a_n \equiv 2^n x \pmod  {2 r + 1}$ === Basis for the Induction === $\map P 1$ is the case: :After $1$ modified perfect faro shuffle, $C$ will be in position $a_1$ where $a_1 \equiv 2 x \pmod  {2 r + 1}$ which is the definition of a modified perfect faro shuffle. Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :After $k$ modified perfect faro shuffles, $C$ will be in position $a_k$ where $a_k \equiv 2^k x \pmod  {2 r + 1}$ from which it is to be shown that: :After $k + 1$ modified perfect faro shuffles, $C$ will be in position $a_{k + 1}$ where $a_{k + 1} \equiv 2^{k + 1} x \pmod  {2 r + 1}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = a_{k + 1}       | o = \equiv       | r = 2 a_k       | rr= \pmod {2 r + 1}       | c = Basis for the Induction }} {{eqn | o = \equiv       | r = 2 \paren {2^k x}       | rr= \pmod {2 r + 1}       | c = Induction Hypothesis }} {{eqn | o = \equiv       | r = 2^{k + 1} x       | rr= \pmod {2 r + 1}       | c = Congruence of Product }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{>0}$: after $n$ modified perfect faro shuffles, $C$ will be in position $a_n$ where $a_n \equiv 2^n x \pmod  {2 r + 1}$ {{qed}}
From Position of Card after n Modified Perfect Faro Shuffles, after $n$ shuffles a card in position $x$ will be in position $2^n x \pmod {m + 1}$. So for all $2 m$ cards in $D$, we need to find $n$ such that: :$2^n x \equiv x \pmod {2 m + 1}$ Because $2 m + 1$ is odd, we have: :$\gcd \set {2, 2 m + 1}$ and so from Cancellability of Congruences: :$2^n \equiv 1 \pmod {2 m + 1}$ {{qed}}
From Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order, the cards of $D$ will return to their original order after $n$ such shuffles, where: :$2^n \equiv 1 \pmod {63}$ We have that: :$63 = 2^6 - 1$ and so: :$2^6 \equiv 1 \pmod {63}$ Hence the result. {{qed}}
From Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order, the cards of $D$ will return to their original order after $n$ such shuffles, where: :$2^n \equiv 1 \pmod 9$ Inspecting $2^n$ for $n$ from $1$: {{begin-eqn}} {{eqn | l = 2^1       | o = \equiv       | r = 2       | rr= \pmod 9       | c =  }} {{eqn | l = 2^2       | o = \equiv       | r = 4       | rr= \pmod 9       | c =  }} {{eqn | l = 2^3       | o = \equiv       | r = 8       | rr= \pmod 9       | c =  }} {{eqn | l = 2^4       | o = \equiv       | r = 7       | rr= \pmod 9       | c =  }} {{eqn | l = 2^5       | o = \equiv       | r = 7       | rr= \pmod 9       | c =  }} {{eqn | l = 2^6       | o = \equiv       | r = 1       | rr= \pmod 9       | c =  }} {{end-eqn}} Hence the result. {{qed}}
From Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order, the cards of $D$ will return to their original order after $n$ such shuffles, where: :$2^n \equiv 1 \pmod {13}$ From Fermat's Little Theorem: :$2^{12} \equiv 1 \pmod {13}$ so we know that $n$ is at most $12$. But $n$ may be smaller, so it is worth checking the values: Inspecting $2^n$ for $n$ from $1$: {{begin-eqn}} {{eqn | l = 2^1       | o = \equiv       | r = 2       | rr= \pmod {13}       | c =  }} {{eqn | l = 2^2       | o = \equiv       | r = 4       | rr= \pmod {13}       | c =  }} {{eqn | l = 2^3       | o = \equiv       | r = 8       | rr= \pmod {13}       | c =  }} {{eqn | l = 2^4       | o = \equiv       | r = 3       | rr= \pmod {13}       | c =  }} {{eqn | l = 2^6       | o = \equiv       | r = 6       | rr= \pmod {13}       | c =  }} {{eqn | l = 2^6       | o = \equiv       | r = 12       | rr= \pmod {13}       | c =  }} {{eqn | l = 2^7       | o = \equiv       | r = 11       | rr= \pmod {13}       | c =  }} {{eqn | l = 2^8       | o = \equiv       | r = 9       | rr= \pmod {13}       | c =  }} {{eqn | l = 2^9       | o = \equiv       | r = 5       | rr= \pmod {13}       | c =  }} {{eqn | l = 2^{10}       | o = \equiv       | r = 10       | rr= \pmod {13}       | c =  }} {{eqn | l = 2^{11}       | o = \equiv       | r = 7       | rr= \pmod {13}       | c =  }} {{eqn | l = 2^{12}       | o = \equiv       | r = 1       | rr= \pmod {13}       | c =  }} {{end-eqn}} Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = x^4 - y^4       | r = \paren {x^2}^2 - \paren {y^2}^2       | c =  }} {{eqn | r = \paren {x^2 - y^2} \paren {x^2 + y^2}       | c = Difference of Two Squares }} {{eqn | r = \paren {x - y} \paren {x + y} \paren {x^2 + y^2}       | c = Difference of Two Squares }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = a^{2 n + 1} + b^{2 n + 1}       | r = a^{2 n + 1} - \paren {-\paren {b^{2 n + 1} } }       | c =  }} {{eqn | r = a^{2 n + 1} - \paren {-b}^{2 n + 1}       | c = as $n$ is odd }} {{eqn | r = \paren {a - \paren {-b} } \sum_{j \mathop = 0}^{2 n} a^{2 n - j} \paren {-b}^j       | c = Difference of Two Powers }} {{eqn | r = \paren {a + b} \sum_{j \mathop = 0}^{2 n} \paren {-1}^j a^{2 n - j} b^j       | c = simplifying }} {{end-eqn}} {{qed}}
From Difference of Two Powers: :$\displaystyle a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n-1} a^{n - j - 1} b^j$ Let $x = a$ and $y = -b$. Then: {{begin-eqn}} {{eqn | l = x^3 + y^3       | r = x^3 - \paren {-y^3}       | c =  }} {{eqn | r = x^3 - \paren {-y}^3       | c =  }} {{eqn | r = \paren {x - \paren {-y} } \sum_{j \mathop = 0}^2 x^{n - j - 1} \paren {-y}^j       | c =  }} {{eqn | r = \paren {x + y} \paren {x^2 + x \paren {-y} + \paren {-y}^2}       | c =  }} {{eqn | r = \paren {x + y} \paren {x^2 - x y + y^2}       | c =  }} {{end-eqn}} {{qed}}
From Sum of Two Odd Powers: :$a^{2 n + 1} + b^{2 n + 1} = \paren {a + b} \paren {a^{2 n} - a^{2 n - 1} b + a^{2 n - 2} b^2 - \dotsb + a b^{2 n - 1} + b^{2 n} }$ We have that $3 = 2 \times 1 + 1$. Hence setting $n = 1$ gives the result. {{qed}}
From Difference of Two Powers: :$\displaystyle a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$ The result follows directly by setting $n = 5$. {{qed}}
From Sum of Two Odd Powers: :$a^{2 n + 1} + b^{2 n + 1} = \paren {a + b} \paren {a^{2 n} - a^{2 n - 1} b + a^{2 n - 2} b^2 - \dotsb + a b^{2 n - 1} + b^{2 n} }$ We have that $5 = 2 \times 2 + 1$. The result follows directly by setting $n = 2$. {{qed}}
{{begin-eqn}} {{eqn | l = x^6 - y^6       | r = \paren {x^3}^2 - \paren {y^3}^2       | c =  }} {{eqn | r = \paren {x^3 - y^3} \paren {x^3 + y^3}       | c =  }} {{eqn | r = \paren {x - y} \paren {x^2 + x y + y^2} \paren {x^3 + y^3}       | c = Difference of Two Cubes }} {{eqn | r = \paren {x - y} \paren {x^2 + x y + y^2} \paren {x + y} \paren {x^2 - x y + y^2}       | c = Sum of Two Cubes }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = x^6 - y^6       | r = \paren {x - y} \paren {x + y} \paren {x^2 + 2 x y + 2 y^2} \paren {x^2 - 2 x y + 2 y^2}       | c = Difference of Two Sixth Powers }} {{eqn | r = \paren {x^2 - y^2} \paren {x^2 + 2 x y + 2 y^2} \paren {x^2 - 2 x y + 2 y^2}       | c = Difference of Two Squares }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = x^6 - y^6       | r = \paren {x^2}^3 - \paren {y^2}^3       | c =  }} {{eqn | r = \paren {x^2 - y^2} \paren {\paren {x^2}^2 + \paren {x^2} \paren {y^2} + \paren {y^2}^2}       | c = Difference of Two Cubes }} {{eqn | r = \paren {x^2 - y^2} \paren {x^4 + x^2 y^2 + y^4}       | c = simplifying }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \paren {x^2 - y^2} \paren {x^4 + x^2 y^2 + y^4}       | r = \paren {x^2 - y^2} \paren {x^2 + 2 x y + 2 y^2} \paren {x^2 - 2 x y + 2 y^2}       | c = as both equal $x^6 - y^6$ }} {{eqn | ll= \leadsto       | l = \paren {x^4 + x^2 y^2 + y^4}       | r = \paren {x^2 + 2 x y + 2 y^2} \paren {x^2 - 2 x y + 2 y^2}       | c = cancelling $\paren {x^2 - y^2}$ from both sides }} {{end-eqn}} {{qed}}
A direct application of Difference of Two Powers: {{begin-eqn}} {{eqn | l = a^n - b^n       | r = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j       | c =  }} {{eqn | r = \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \dotsb + a b^{n - 2} + b^{n - 1} }       | c =  }} {{end-eqn}} and setting $n \to 2 n + 1$. {{qed}}
{{begin-eqn}} {{eqn | l = a^{2 n} - b^{2 n}       | r = \paren {a^n}^2 - \paren {b^n}^2       | c =  }} {{eqn | r = \paren {a^n - b^n} \paren {a^n + b^n}       | c =  }} {{eqn | r = \paren {a - b} \paren {\sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j} \paren {a^n + b^n}       | c = Difference of Two Powers }} {{eqn | r = \paren {a - b} \paren {\sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j} \paren {a + b} \paren {\sum_{j \mathop = 0}^{n - 1} \paren {-1}^j a^{n - j - 1} b^j}       | c = Sum of Two Odd Powers }} {{end-eqn}} whence the result. {{qed}}
{{begin-eqn}} {{eqn | l = a^{2 n} - b^{2 n}       | r = \paren {a^2}^n - \paren {b^2}^n       | c =  }} {{eqn | r = \paren {a^2 - b^2} \paren {\sum_{j \mathop = 0}^{n - 1} \paren {a^2} ^{n - j - 1} \paren {b^2}^j}       | c = Difference of Two Powers }} {{eqn | r = \paren {a - b} \paren {a + b} \sum_{j \mathop = 0}^{n - 1} a^{2 \paren {n - j - 1} } b^{2 j}       | c = Difference of Two Squares }} {{end-eqn}} whence the result. {{qed}} Category:Difference of Two Powers 5mm6ext2i6hvk5jyif69hx9pvofs2x2
From Factorisation of $z^n - a$: :$x^{2 n + 1} - y^{2 n + 1} = \displaystyle \prod_{k \mathop = 0}^{2 n} \paren {x - \alpha^k y}$ where $\alpha$ is a primitive complex $2 n + 1$th roots of unity, for example: {{begin-eqn}} {{eqn | l = \alpha       | r = e^{2 i \pi / \paren {2 n + 1} }       | c =  }} {{eqn | r = \cos \dfrac {2 \pi} {2 n + 1} + i \sin \dfrac {2 \pi} {2 n + 1}       | c =  }} {{end-eqn}} From Complex Roots of Unity occur in Conjugate Pairs: :$U_{2 n + 1} = \set {1, \tuple {\alpha, \alpha^{2 n} }, \tuple {\alpha^2, \alpha^{2 n - 1} }, \ldots, \tuple {\alpha^k, \alpha^{2 n - k + 1} }, \ldots, \tuple {\alpha^n, \alpha^{n + 1} } }$ where $U_{2 n + 1}$ denotes the complex $2 n + 1$th roots of unity: :$U_{2 n + 1} = \set {z \in \C: z^{2 n + 1} = 1}$ The case $k = 0$ is taken care of by setting $\alpha^0 = 1$, from whence we have the factor $x - y$. Taking the product of each of the remaining factors of $x^{2 n + 1} - y^{2 n + 1}$ in pairs: {{begin-eqn}} {{eqn | l = \paren {x - \alpha^k y} \paren {x - \alpha^{2 n - k + 1} y}       | r = \paren {x - \alpha^k y} \paren {x - \overline {\alpha^k} y}       | c = Complex Roots of Unity occur in Conjugate Pairs }} {{eqn | r = x^2 - x \paren {\alpha^k + \overline {\alpha^k} } y + \alpha^k y \, \overline {\alpha^k} y       | c =  }} {{eqn | r = x^2 - x y \paren {\alpha^k + \overline {\alpha^k} } + \cmod {\alpha^k}^2 y^2       | c = Modulus in Terms of Conjugate }} {{eqn | r = x^2 - x y \paren {\alpha^k + \overline {\alpha^k} } + y^2       | c = Modulus of Complex Root of Unity equals 1 }} {{eqn | r = x^2 - x y \paren {\cos \dfrac {2 k \pi} {2 n + 1} + i \sin \dfrac {2 k \pi} {2 n + 1} + \cos \dfrac {2 k \pi} {2 n + 1} - i \sin \dfrac {2 k \pi} {2 n + 1} } + y^2       | c = Definition of $\alpha$ }} {{eqn | r = x^2 - 2 x y \cos \dfrac {2 k \pi} {2 n + 1} + y^2       | c = simplification }} {{end-eqn}} Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = x^{2 n + 1} + y^{2 n + 1}       | r = x^{2 n + 1} - \paren {-\paren {y^{2 n + 1} } }       | c =  }} {{eqn | r = x^{2 n + 1} - \paren {-y}^{2 n + 1}       | c =  }} {{eqn | r = \paren {x - \paren {-y} } \displaystyle \prod_{k \mathop = 1}^n \paren {x^2 - 2 x \paren {-y} \cos \dfrac {2 \pi k} {2 n + 1} + \paren {-y}^2}       | c = Factors of Difference of Two Odd Powers }} {{eqn | r = \paren {x + y} \displaystyle \prod_{k \mathop = 1}^n \paren {x^2 + 2 x y \cos \dfrac {2 \pi k} {2 n + 1} + y^2}       | c = simplification }} {{end-eqn}} {{qed}}