hcju/proofwikiqa huggingface dataset:
{{refactor|Extract individual proofs into their own pages and transclude them}} === Weierstrass Form equivalent to Euler Form === First it is shown that the Weierstrass form is equivalent to the Euler form. {{begin-eqn}} {{eqn | l = \frac 1 {\map \Gamma z}       | r = z e^{\gamma z} \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac z n} e^{-z/n} }       | c = Weierstrass Form of $\Gamma$ Function }} {{eqn | r = z \paren {\lim_{m \mathop \to \infty} \exp \paren {\paren {1 + \frac 1 2 + \cdots + \frac 1 m - \ln \paren m} z} } \paren {\lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac z n} e^{-z/n} } }       | c = {{Defof|Euler-Mascheroni Constant}} }} {{end-eqn}} Combining the limits: {{begin-eqn}}      {{eqn | l = \frac 1 {\map \Gamma z}       | r = z \lim_{m \mathop \to \infty} \paren {\exp \paren {\paren {1 + \frac 1 2 + \cdots + \frac 1 m - \ln \paren m} z} \prod_{n \mathop = 1}^m \paren {\paren {1 + \frac z n} e^{-z/n} } }       | c =  }} {{eqn | r = z \lim_{m \mathop \to \infty} \paren {\exp \paren {\paren {1 + \frac 1 2 + \cdots + \frac 1 m - \ln \paren m} z} \exp \paren {\frac {-z} 1 + \frac {-z} 2 + \cdots + \frac {-z} m} \prod_{n \mathop = 1}^m \paren {1 + \frac z n} }       | c =  }} {{eqn | r = z \lim_{m \mathop \to \infty} \paren {\exp \paren {\paren {1 - 1 + \frac 1 2 - \frac 1 2 + \cdots + \frac 1 m - \frac 1 m - \ln \paren m} z} \prod_{n \mathop = 1}^m \paren {1 + \frac z n} }       | c = Exponential of Sum }} {{eqn | r = z \lim_{m \mathop \to \infty} \paren {m^{-z} \prod_{n \mathop = 1}^m \paren {1 + \frac z n} }       | c =  }} {{end-eqn}} But: :$(1): \quad m = \dfrac {m!} {\paren {m - 1}!} = \dfrac 2 1 \cdot \dfrac 3 2 \cdots \dfrac {x + 1} x \cdots \dfrac m {m - 1}$ Each term in $(1)$ is just $\dfrac {x + 1} x = 1 + \dfrac 1 x$, so: :$\displaystyle m = \prod_{n \mathop = 1}^{m - 1} \paren {1 + \frac 1 n}$ Thus the expression for $\dfrac 1 {\map \Gamma z}$ becomes: {{begin-eqn}}      {{eqn | o =        | r = z \lim_{m \mathop \to \infty} \paren {\prod_{n \mathop = 1}^{m - 1} \paren {1 + \frac 1 n}^{-z} \prod_{n \mathop = 1}^m \paren {1 + \frac z n} }       | c =  }} {{eqn | r = z \lim_{m \mathop \to \infty} \paren {\paren {1 + \frac 1 m}^z \prod_{n \mathop = 1}^m \paren {1 + \frac 1 n}^{-z} \paren {1 + \frac z n} }       | c =  }} {{eqn | r = z \lim_{m \mathop \to \infty} \paren {1 + \frac 1 m}^z \lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^m \paren {1 + \frac 1 n}^{-z} \paren {1 + \frac z n}       | c = Product Rule for Complex Sequences }} {{eqn | r = z \prod_{n \mathop = 1}^\infty \paren {1 + \frac 1 n}^{-z} \paren {1 + \frac z n} }} {{end-eqn}} Hence: :$\displaystyle \map \Gamma z = \frac 1 z \prod_{n \mathop = 1}^\infty \paren {1 + \frac 1 n}^z \paren {1 + \frac z n}^{-1}$ which is the Euler form of the Gamma function. {{qed|lemma}} === Integral Form equivalent to Euler Form === This is proved in the page: :Integral Form of Gamma Function equivalent to Euler Form {{qed}}
Let $\dfrac a b$ be a rational number in canonical form such that $b < q_{n + 1}$. Suppose it is not true that $a = p_n$ and $b = q_n$, in which case the equality certainly holds. Consider the system of equations: {{begin-eqn}} {{eqn | l = a       | r = r p_n + s p_{n - 1}       | c =  }} {{eqn | l = b       | r = r q_n + s q_{n - 1}       | c =  }} {{end-eqn}} Multiplying the first by $b_n$, and the second by $a_n$, then subtracting, we get: :$a q_n - b p_n = s \left({p_{n + 1} q_n - p_n q_{n + 1} }\right)$ After applying Difference between Adjacent Convergents of Simple Continued Fraction we get: {{begin-eqn}} {{eqn | l = s       | r = \left({-1}\right)^{n + 1} \left({a q_n - b p_n}\right)       | c =  }} {{eqn | l = r       | r = \left({-1}\right)^{n + 1} \left({b p_{n + 1} - a q_{n + 1} }\right)       | c = by a similar process }} {{end-eqn}} So $r$ and $s$ are integers. Neither of them is $0$ because: : if $r = 0$ then $a q_{n + 1} = b p_{n + 1}$, and Euclid's Lemma means $q_{n + 1} \mathrel \backslash b$ as $p_{n + 1} \perp q_{n + 1}$, which contradicts $0 < b < q_{n + 1}$ : if $s = 0$ we have $\dfrac a b = \dfrac {p_n} {q_n}$ and this we have already excluded as a special case. {{questionable|the below needs to link to a stronger result}} From Even Convergent of Simple Continued Fraction is Strictly Smaller than Odd Convergent, the convergents are alternately greater than and less than $x$. Hence since $0 < b = r q_n + s q_{n + 1} < q_{n + 1}$, the integers $r$ and $s$ must have opposite sign. It follows that $r \left({q_n x - p_n}\right)$ and $s \left({q_{n + 1} x - p_{n + 1} }\right)$ have the same sign. This is necessary for the Triangle Inequality to hold. So: {{begin-eqn}} {{eqn | l = \left\vert{b x - a}\right\vert       | r = \left\vert{\left({r q_n + s q_{n + 1} }\right) x - \left({r p_n + s p_{n+1} }\right)}\right\vert       | c =  }} {{eqn | r = \left\vert{r \left({q_n x - p_n}\right) + s \left({q_{n + 1} x - p_{n + 1} }\right)}\right\vert       | c =  }} {{eqn | r = \left\vert{r}\right\vert \left\vert{q_n x - p_n}\right\vert + \left\vert{s}\right\vert \left\vert{q_{n + 1} x - p_{n + 1} }\right\vert       | c =  }} {{eqn | o = >       | r = \left\vert{r}\right\vert \left\vert{q_n x - p_n}\right\vert       | c =  }} {{eqn | o = \ge       | r = \left\vert{q_n x - p_n}\right\vert       | c =  }} {{end-eqn}} as we wanted to prove. {{qed}} Category:Continued Fractions h5aaptp4jnt0q17c0xa28zafgrp1mbp
Suppose to the contrary, that $\left\vert{x - \dfrac a b}\right\vert < \dfrac 1 {2 b^2}$ but that $\dfrac a b$ is ''not'' one of the convergents $\dfrac {p_n} {q_n}$ of $x$. Let $r$ be the unique integer for which $q_r \le b \le q_{r+1}$. Then: {{begin-eqn}} {{eqn | l=\left\vert{q_r x - p_r}\right\vert       | o=\le       | r=\left\vert{b x - a}\right\vert       | c=Convergents are Best Approximations }} {{eqn | r=b \left\vert{x - \frac a b}\right\vert       | c= }} {{eqn | o=<       | r=b \times \frac 1 {2 b^2}       | c= }} {{eqn | r=\frac 1 {2b}       | c= }} {{end-eqn}} Therefore: : $\displaystyle q_r \left\vert{x - \frac {p_r} {q_r}}\right\vert < \frac 1 {2b}$ and so: : $\displaystyle \left\vert{x - \frac {p_r} {q_r}}\right\vert < \frac 1 {2 q_r b}$ Hence: {{begin-eqn}} {{eqn | l=\left\vert{\frac a b - \frac {p_r} {q_r} }\right\vert       | o=\le       | r=\left\vert{x - \frac {p_r} {q_r} }\right\vert + \left\vert{x - \frac a b}\right\vert       | c=Triangle Inequality }} {{eqn | n=1       | o=<       | r=\frac 1 {2 q_r b} + \frac 1 {2b^2}       | c= }} {{end-eqn}} Now note that $q_r a - p_r b$ is a integer, and also non-zero otherwise $\dfrac a b = \dfrac {p_r} {q_r}$ and we supposed (at the top of this proof) that it's not. But we have: {{begin-eqn}} {{eqn | l=\left\vert{\frac a b - \frac {p_r} {q_r} }\right\vert       | r=\left\vert{\frac {q_r a - p_r b} {q_r b} }\right\vert       | c= }} {{eqn | n=2       | o=\ge       | r=\frac 1 {q_r b}       | c= }} {{end-eqn}} So, combining results $(1)$ and $(2)$, we get: :$\displaystyle \frac 1 {q_r b} < \frac 1 {2 q_r b} + \frac 1 {2b^2}$ This simplifies to $q_r > b$, which contradicts our initial assumptions. {{qed}} Category:Continued Fractions pgo2asb3upq47p9w33i6qiqfc28i9me
Let $x = r + s \sqrt n$. From Solution to Quadratic Equation, the solutions of $a x^2 + b x + c$ are: :$x = \dfrac {-b \pm \sqrt {b^2 - 4 a c}} {2 a}$ given the appropriate condition on the discriminant. So if $x = r + s \sqrt n$ is a solution, then so is $x = r - s \sqrt n$. Hence we have: {{begin-eqn}} {{eqn | l = \left({x - r + s \sqrt n}\right) \left({x - r - s \sqrt n}\right)       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \left({x-r}\right)^2 - \left({s \sqrt n}\right)^2       | r = 0       | c = Difference of Two Squares }} {{eqn | ll= \leadsto       | l = x^2 - 2 r x + r^2 - s^2 n       | r = 0       | c =  }} {{end-eqn}} As $r$ and $s$ are rational and $n$ is an integer, it follows that $-2 r$ and $r^2 - s^2 n$ are also rational from Rational Numbers form Field. Hence the result. {{qed}} Category:Quadratic Equations Category:Quadratic Irrationals 1ap9dfp1fp6nlk45w543fu8dppl4pr4
We use: :Expansion of Associated Reduced Quadratic Irrational to establish a series of reduced quadratic irrationals associated to $n$  and then use :Finitely Many Reduced Associated Quadratic Irrationals to assert this series must repeat (hence be periodic) due to the finite number of such irrationals. Let: :$a_0 = \floor {\sqrt n}$ :$\sqrt n = a_0 + \dfrac 1 {\alpha_0}$ For all $i \in \N: i > 0$, let: :$a_{i + 1} = \floor {\alpha_i}$ :$\alpha_i = a_{i + 1} + \dfrac 1 {\alpha_{i + 1} }$ From here, we will prove: :$\alpha_0$ is a reduced quadratic irrational associated to $n$ :If all $\alpha_i: 0 \le i \le k$ are all reduced quadratic irrationals associated to $n$, then so is $\alpha_{k + 1}$. Since $\dfrac 1 {\alpha_0}$ is the fractional part of the irrational $\sqrt n$, we have: :$0 < \dfrac 1 {\alpha_0} < 1 \implies \alpha_0 > 1$ By simple algebra, we have: :$\alpha_0 = \dfrac {a_0 + \sqrt n} {n - a_0^2}$ :$\tilde {\alpha_0} = \dfrac {a_0 - \sqrt n} {n - a_0^2}$ where $\tilde {\alpha_0}$ is the conjugate of $\alpha_0$. Since $a_0$ is the floor of $\sqrt n$, we know that: :$a_0 - \sqrt n < 0 \implies \tilde {\alpha_0} < 0$ Since $n \in \Z \implies \sqrt n > 1$ and $\sqrt n > a_0$, we have: {{begin-eqn}} {{eqn | l = 1       | o = <       | r = \sqrt n + a_0       | c =  }} {{eqn | ll= \leadsto       | l = \sqrt n - a_0       | o = <       | r = n - a_0^2       | c =  }} {{eqn | ll= \leadsto       | l = a_0 - \sqrt n       | o = >       | r = -\paren {n - a_0^2}       | c =  }} {{eqn | ll= \leadsto       | l = \tilde {\alpha_0}       | o = >       | r = -1       | c =  }} {{end-eqn}} Thus $\alpha_0$ is a reduced quadratic irrational. Since $P = a_0$ and $Q = n - a_0^2 = n - P^2$, $Q$ clearly divides $n - P^2$. Thus $\alpha_0$ is associated to $n$ as well. We have that each $\alpha_i$ is a reduced quadratic irrational. So, following the recurrence defined, each $a_i \ge 1$. Also, by Expansion of Associated Reduced Quadratic Irrational, each $\alpha_{i + 1}$ is reduced and associated to $n$ since $\alpha_0$ is. By Finitely Many Reduced Associated Quadratic Irrationals, we only have finitely many choices for these. Hence there must be some smallest $k$ for which $\alpha_k = \alpha_0$. Since $\alpha_{i + 1}$ is determined completely by $\alpha_i$ we will then have: :$\alpha_{k + j} = \alpha_j$ for all $j > 0$. Hence the $\alpha_i$ are periodic. Similarly, as the $a_i$ for $i > 0$ are determined completely by $\alpha_{i - 1}$, the $a_i$ must be periodic as well. This forces the continued fraction expansion: :$\sqrt n = a_0 + \cfrac 1 {a_1 + \cfrac 1 {a_2 + \ddots} }$ to be periodic. {{wtd|This gives us the fact that it repeats. We still need explain the pattern.}}
Let $a^2 - n b^2 = 1$. Then we have: :$\left({a - b \sqrt n}\right) \left({a + b \sqrt n}\right) = 1$. So: :$a - b \sqrt n = \dfrac 1 {a + b \sqrt n} > 0$ and so $a > b \sqrt n$. Therefore: {{begin-eqn}} {{eqn | l = \left\vert{\sqrt n - \frac a b}\right\vert       | r = \frac {a - b \sqrt n} b       | c =  }} {{eqn | r = \frac 1 {b \left({a + b \sqrt n}\right)}       | c =  }} {{eqn | o = <       | r = \frac 1 {b \left({b \sqrt n + b \sqrt n}\right)}       | c =  }} {{eqn | r = \frac 1 {2 b^2 \sqrt n}       | c =  }} {{eqn | o = <       | r = \frac 1 {2 b^2}       | c =  }} {{end-eqn}} The result follows from Condition for Rational to be a Convergent. {{qed}} Category:Pell's Equation Category:Continued Fractions pegxezxos9u7ukr2oax3nzot2rhq3nv
From Smaller Elements of Pythagorean Triple not both Odd, $x$ and $y$ are not both odd. {{AimForCont}} $x$ and $y$ are both even. Then by definition they have $2$ as a common divisor. But from Elements of Primitive Pythagorean Triple are Pairwise Coprime this cannot be the case. So by Proof by Contradiction, $x$ and $y$ cannot both be even. It follows that $x$ and $y$ are of opposite parity. {{qed}} Category:Pythagorean Triples lutvr5hektr7ps1ssx5cnyixsv1qvfr
$\gcd \set {h + 1, h} = \gcd \set {h, 1} = \gcd \set {1, 0} = 1$ from the Euclidean Algorithm. {{qed}}
Let $k \in \Z: k \divides h$. Also assume $k \divides \paren {h + 1}$. Thus: :$\exists a, b \in \N: a k = h, b k = \paren {h + 1}$ Then: :$\paren {h + 1} - h = b k - a k$ and so: :$1 = \paren {b - a} k$ Since the integers form an integral domain, $\paren {b - a} \in \Z$. Thus either $k = 1$ and $b - a = 1$, or $k = -1$ and $b - a = -1$. Therefore, only $1$ and $-1$ can be factors of both $h$ and $\paren {h + 1}$. {{qed}}
We have that $x \perp y$ by definition. Suppose there is a prime divisor $p$ of both $x$ and $z$. That is: :$\exists p \in \mathbb P: p \divides x, p \divides z$ Then from Prime Divides Power: :$p \divides x^2, p \divides z^2$ Then from Common Divisor Divides Integer Combination: :$p \divides \paren {z^2 - x^2} = y^2$ So from Prime Divides Power again: :$p \divides y$ and: :$x \not \perp y$ This contradicts our assertion that $\tuple {x, y, z}$ is a primitive Pythagorean triple. Hence $x \perp z$. The same argument shows that $y \perp z$. {{qed}} Category:Pythagorean Triples Category:Coprime Integers 8qsgq5gevl1ehybvgql79khf41s4nji
Let $m = k + 1, n = k$ where $k \in \Z: k \ge 1$. From Consecutive Integers are Coprime, $m \perp n$. Then we have: : $m, n \in \Z$ are positive integers : $m \perp n$, i.e. $m$ and $n$ are coprime : $m$ and $n$ are of opposite parity : $m > n$. From Solutions of Pythagorean Equation, these conditions are necessary and sufficient for $\left({2 m n, m^2 - n^2, m^2 + n^2}\right)$ to be a primitive Pythagorean triple. Substituting for $m$ and $n$, we get: :$\left({2 k^2 + 2 k, 2 k + 1, 2 k^2 + 2 k + 1}\right)$ is a primitive Pythagorean triple. So we see that for all $k \ge 1$, $2 k + 1$ is an element of a primitive Pythagorean triple. So every odd integer from $3$ upwards is an element of some primitive Pythagorean triple. That is, any integer $x$ such that $x \equiv 1$ or $x \equiv 3 \pmod 4$. Now, consider $m = 2 k, n = 1$ where $k \in \Z: k \ge 1$. These also fit the criteria for $\left({2 m n, m^2 - n^2, m^2 + n^2}\right)$ to be a primitive Pythagorean triple. Substituting for $m$ and $n$, we get: :$\left({4 k, 4 k^2 - 1, 4 k^2 + 1}\right)$ is a primitive Pythagorean triple. This means that for all $k \ge 1$, $4 k$ is an element of a primitive Pythagorean triple. So every multiple of $4$ is an element of some primitive Pythagorean triple. That is, any integer $x$ such that $x \equiv 0 \pmod 4$. Now, consider the general primitive Pythagorean triple $\left({2 m n, m^2 - n^2, m^2 + n^2}\right)$. As $m$ and $n$ are of opposite parity, they can be expressed as $2r$ and $2s + 1$. So $2 m n = 2 \left({2r}\right)\left({2s + 1}\right) = 4 r \left({2s + 1}\right)$. So $2 m n$ is divisible by $4$. As the only even elements of a primitive Pythagorean triple are of this form $2 m n$ from Parity of Smaller Elements of Primitive Pythagorean Triple, there can be no such elements $x$ of the form $x = 2 \pmod 4$. Hence the result. {{qed}} Category:Pythagorean Triples kvgurvy8akz6meisuv0ulp2qrntn1dd
Let $x \in \Z$ be an odd square. Then $x = n^2$ where $n$ is also odd. Thus $n$ can be expressed as $2 k + 1$ for some $k \in \Z$. Hence: {{begin-eqn}} {{eqn | l = x       | r = n^2       | c =  }} {{eqn | r = \paren {2 k + 1}^2       | c =  }} {{eqn | r = 4 k^2 + 4 k + 1       | c =  }} {{eqn | r = 4 k \paren {k + 1} + 1       | c =  }} {{end-eqn}} But $k$ and $k + 1$ are of opposite parity and can therefore be expressed as $2 r$ and $2 s + 1$ (either way round). {{begin-eqn}} {{eqn | l = x       | r = 4 k \paren {k + 1} + 1       | c =  }} {{eqn | r = 4 \paren {2 r} \paren {2 s + 1} + 1       | c =  }} {{eqn | r = 8 r \paren {2 s + 1} + 1       | c =  }} {{end-eqn}} Hence the result. {{qed}}
Let $T_{n - 1}$ and $T_n$ be two consecutive triangular numbers. From Closed Form for Triangular Numbers, we have: :$T_{n - 1} = \dfrac {\paren {n - 1} n} 2$ :$T_n = \dfrac {n \paren {n + 1} } 2$ So: {{begin-eqn}}      {{eqn | l = T_{n - 1} + T_n       | r = \frac {\paren {n - 1} n} 2 + \frac {n \paren {n + 1} } 2       | c =  }} {{eqn | r = \frac {\paren {n - 1 + n + 1} n} 2       | c =  }} {{eqn | r = \frac {2 n^2} 2       | c =  }} {{eqn | r = n^2       | c =  }} {{end-eqn}} {{qed}}
Let $n$ be triangular. Then: :$\exists k \in \Z: n = \dfrac {k \paren {k + 1} } 2$ So: {{begin-eqn}} {{eqn | l = 9 n + 1       | r = 9 \frac {k \paren {k + 1} } 2 + 1       | c =  }} {{eqn | r = \frac {9 k^2 + 9 k + 2} 2       | c =  }} {{eqn | r = \frac {\paren {3 k + 1} \paren {3 k + 2} } 2       | c =  }} {{end-eqn}} which is triangular. {{qed}}
Follows directly from the identity: {{begin-eqn}} {{eqn | l = 8 \frac {k \paren {k + 1} } 2 + 1       | r = 4 k^2 + 4 k + 1       | c =  }} {{eqn | r = \paren {2 k + 1}^2       | c =  }} {{end-eqn}} as follows: Let $m$ be triangular. Then from Closed Form for Triangular Numbers: :$\exists k \in \Z: m = \dfrac {k \paren {k + 1} } 2$ From the above identity: :$8 m + 1 = \paren {2 k + 1}^2$ which is an odd square. Let $n$ be an odd square. Then $n = r^2$ where $r$ is odd. Let $r = 2 k + 1$, so that $n = \paren {2 k + 1}^2$. From the above identity: :$n = 8 \dfrac {k \paren {k + 1} } 2 + 1 = 8 m + 1$ where $m$ is triangular. {{qed}}
Let $n = T_r$. Then from Closed Form for Triangular Numbers: : $n = \dfrac {r \left({r + 1}\right)} 2$ It suffices from Euclid's Lemma to investigate the nature of $r \left({r + 1}\right)$ modulo $3$. There are three cases to consider: : $r \equiv 0 \pmod 3$ : $r \equiv 1 \pmod 3$ : $r \equiv 2 \pmod 3$ Let $r \equiv 0 \pmod 3$. Then: :$r \left({r + 1}\right) \equiv 0 \pmod 3$ and so $T_r \equiv 0 \pmod 3$. {{qed|lemma}} Let $r \equiv 2 \pmod 3$. Then: :$r + 1 \equiv 3 \equiv 0 \pmod 3$ So: :$r \left({r + 1}\right) \equiv 0 \pmod 3$ and so $T_r \equiv 0 \pmod 3$. {{qed|lemma}} Let $r \equiv 1 \pmod 3$. Then $\exists k \in \Z: r = 3 k + 1$. So $r \left({r + 1}\right) = \left({3 k + 1}\right) \left({3 k + 2}\right)$ {{begin-eqn}} {{eqn | l = r \left({r + 1}\right)       | r = \left({3 k + 1}\right) \left({3 k + 2}\right)       | c =  }} {{eqn | r = 9 k^2 + 9 k + 2       | c =  }} {{eqn | r = 9 k \left({k + 1}\right) + 2       | c =  }} {{end-eqn}} So: : $T_r = 9 \dfrac {k \left({k + 1}\right)} 2 + 1$ Thus $T_r \equiv 1 \pmod 9$. {{qed}} Category:Triangular Numbers d8p7lmplwo6ux72hwl31arfnp41va2z
By definition of the $n$th $k$-gonal number: $\map P {k, n} = \begin{cases} 0 & : n = 0 \\ \map P {k, n - 1} + \paren {k - 2} \paren {n - 1} + 1 & : n > 0 \end{cases}$ Then: :$\paren {\paren {k - 2} \paren {j - 1} + 1}$ is an arithmetic sequence. Its initial term $a$ is $1$, and its common difference $d$ is $k - 2$. Hence: {{begin-eqn}} {{eqn | l = \map P {k, n}       | r = \sum_{j \mathop = 1}^n \paren {\paren {k - 2} \paren {j - 1} + 1}       | c = Sum of Arithmetic Sequence }} {{eqn | r = \frac {n \paren {2 + \paren {n - 1} \paren {k - 2} } } 2       | c =  }} {{eqn | r = \frac n 2 \paren {\paren {k - 2} n - \paren {k - 2} + 2}       | c =  }} {{eqn | r = \frac n 2 \paren {\paren {k - 2} n - k + 4}       | c =  }} {{end-eqn}} as required. {{qed}}
From Closed Form for Polygonal Numbers: :$\map P {k, n} = \dfrac n 2 \paren {\paren {k - 2} n - k + 4}$ Thus: {{begin-eqn}} {{eqn | l = \map P {k + 1, n} - \map P {k, n}       | r = \dfrac n 2 \paren {\paren {\paren {k + 1} - 2} n - \paren {k + 1} + 4} - \dfrac n 2 \paren {\paren {k - 2} n - k + 4}       | c =  }} {{eqn | r = \dfrac n 2 \paren {\paren {k - 1} n - k + 3} - \dfrac n 2 \paren {\paren {k - 2} n - k + 4}       | c =  }} {{eqn | r = \dfrac n 2 \paren {k n - n - k + 3} - \dfrac n 2 \paren {k n - 2 n - k + 4}       | c =  }} {{eqn | r = \dfrac n 2 \paren {k n - n - k + 3 - k n + 2 n + k - 4}       | c =  }} {{eqn | r = \dfrac n 2 \paren {- n + 3 + 2 n - 4}       | c =  }} {{eqn | r = \frac {n \paren {n - 1} } 2       | c =  }} {{eqn | r = T_{n - 1}       | c =  }} {{end-eqn}} {{qed}}
Suppose that $P \left({n_\alpha}\right)$ holds. Then from the descent step, $\exists n_\beta \in \N_{n_\alpha}: P \left({n_\beta}\right)$. The descent step then tell us we can deduce a smaller positive solution, $n_\gamma$, such that $P \left({n_\gamma}\right)$ is true and $n_\gamma \in \N_{n_\beta}$. And again, the descent step tells us we can deduce a still smaller positive solution, $n_\delta$, such that $P \left({n_\delta}\right)$ is true and $n_\delta \in \N_{n_\gamma}$. Now, consider the unending sequence: $n_\alpha > n_\beta > n_\gamma > n_\delta > \cdots > 0$. The set $S = \left\{{n_\alpha, n_\beta, n_\gamma, n_\delta, \ldots}\right\}$ is not bounded below, as for any $\forall x \in S: \exists y \in S: y < x$. By the Well-Ordering Principle, any non-empty subset of $\N$ must have a least element. As $S$ is not bounded below, it has no least element, so must be empty. {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \left({a c + b d}\right)^2 + \left({a d - b c}\right)^2       | c =  }} {{eqn | r = \left({\left({a c}\right)^2 + 2 \left({a c}\right) \left({b d}\right) + \left({b d}\right)^2}\right) + \left({\left({a d}\right)^2 - 2 \left({a b}\right) \left({c d}\right) + \left({b c}\right)^2}\right)       | c = Square of Sum, Square of Difference }} {{eqn | r = a^2 c^2 + 2 a b c d + b^2 d^2 + a^2 d^2 - 2 a b c d + b^2 c^2       | c = multiplying out }} {{eqn | r = a^2 c^2 + a^2 d^2 + b^2 c^2 + b^2 d^2       | c = simplifying }} {{eqn | r = \left({a^2 + b^2}\right) \left({c^2 + d^2}\right)       | c =  }} {{end-eqn}} {{qed}}
From the more general version of Brahmagupta-Fibonacci Identity: :$\left({a^2 + n b^2}\right) \left({c^2 + n d^2}\right) = \left({a c + n b d}\right)^2 + n \left({a d - b c}\right)^2$ The result follows by setting $n = 1$. {{qed}}
Lagrange's Identity gives: :$\displaystyle \paren {\sum_{k \mathop = 1}^n {a_k}^2} \paren {\sum_{k \mathop = 1}^n {b_k}^2} - \paren {\sum_{k \mathop = 1}^n a_k b_k}^2 = \sum_{i \mathop = 1}^{n - 1} \sum_{j \mathop = i + 1}^n \paren {a_i b_j - a_j b_i}^2$ Setting $n = 2$: {{begin-eqn}} {{eqn | l = \paren {\sum_{k \mathop = 1}^2 {a_k}^2} \paren {\sum_{k \mathop = 1}^2 {b_k}^2} - \paren {\sum_{k \mathop = 1}^2 a_k b_k}^2       | r = \sum_{i \mathop = 1}^1 \sum_{j \mathop = i + 1}^2 \paren {a_i b_j - a_j b_i}^2       | c =  }} {{eqn | ll= \leadsto       | l = \paren { {a_1}^2 + {a_2}^2} \paren { {b_1}^2 + {b_2}^2} - \paren {a_1 b_1 + a_2 b_2}^2       | r = \paren {a_1 b_2 - a_2 b_1}^2       | c = expanding out summations }} {{eqn | ll= \leadsto       | l = \paren {a^2 + b^2} \paren {c^2 + d^2} - \paren {a c + b d}^2       | r = \paren {a d - b c}^2       | c = renaming $a_1 \to a, a_2 \to b, b_1 \to c, b_2 \to d$ }} {{eqn | ll= \leadsto       | l = \paren {a^2 + b^2} \paren {c^2 + d^2}       | r = \paren {a c + b d}^2 + \paren {a d - b c}^2       | c =  }} {{end-eqn}} {{qed}}
Let $z_1 = a + b i, z_2 = c + d i$ be complex numbers. Let $\left\lvert{z}\right\rvert$ denote the complex modulus of a given complex number $z \in \C$. By definition of complex multiplication: :$(1): \quad z_1 z_2 = \left({a c - b d}\right) + \left({a d + b c}\right) i$ Then we have: {{begin-eqn}} {{eqn | l = \left\lvert{z_1 z_2}\right\rvert        | r = \left\lvert{z_1}\right\rvert \cdot \left\lvert{z_2}\right\rvert       | c = Complex Modulus of Product of Complex Numbers }} {{eqn | ll= \leadsto       | l = \left\lvert{\left({a + b i}\right) \left({c + d i}\right)}\right\rvert^2       | r = \left\lvert{a + b i}\right\rvert^2 \cdot \left\lvert{c + d i}\right\rvert^2       | c =  }} {{eqn | ll= \leadsto       | l = \left({a c - b d}\right)^2 + \left({a d + b c}\right)^2       | r = \left({a^2 + b^2}\right) \left({c^2 + d^2}\right)       | c = {{Defof|Complex Modulus}}, and from $(1)$ }} {{end-eqn}} {{qed}}