hcju/proofwikips huggingface dataset:
Differential of Differentiable Functional is Unique/Lemma
 == Lemma ==  Let $\phi \left[{y; h}\right]$ be a linear functional w.r.t. $h$. Let: :$\displaystyle \lim_{\left \vert{h}\right\vert \to 0} \frac {\phi \left[{y; h}\right]} {\left\vert{h}\right\vert} = 0$ Then : :$\phi \left[{y; h}\right] = 0$  == Proof == This will be a proof by contradiction.  there exists a linear functional satisfying $\phi \left[{y; h_0}\right] \ne 0$ for some $h_0 \ne 0$. Also suppose: :$\displaystyle \lim_{\left \vert{h_0}\right\vert \to 0} \frac{\phi \left[{y; h_0}\right]} {\left\vert{h_0}\right\vert} = 0$ Now, define: :$h_n = \dfrac {h_0} n$ and: :$m = \dfrac{\phi \left[{y; h_0}\right]} {\left\vert{h_0}\right\vert}$ Notice that $\left\vert{h_n}\right\vert \to 0$ as $n \to \infty$.  Hence, from the assumption of the limit it should hold that: :$\displaystyle \lim_{n \to \infty} \frac{\phi \left[{y; h_n}\right]}{\left\vert{h_n}\right\vert} = \lim_{\left\vert{h_n}\right\vert \to 0} \frac{ \left[{y; h_n}\right]} {\left\vert{h_n}\right\vert} = 0$ However, using the linearity of $\phi \left[{y; h_0}\right]$ w.r.t. $h_0$: {{begin-eqn}} {{eqn | l = \lim_{n \to \infty} \frac {\phi \left[{y; h_n}\right]} {\left\vert{h_n}\right\vert}       | r = \lim_{n \to \infty} \frac {\phi \left[{y; \frac {h_0} n}\right]} { \left\vert{\frac {h_0} n}\right\vert}       | c = Definition of $h_n$ }} {{eqn | r = \lim_{n \to \infty} \frac {n \, \phi \left[{y; h_0}\right]} {n \, \left\vert{h_0}\right\vert}       | c = extract of $n$ through linearity }} {{eqn | r = \lim_{n \to \infty} \frac {\phi \left[{y; h_0}\right]} {\left\vert{h_0}\right\vert}       | c = cancel $n$ }} {{eqn | r = \frac {\phi \left[{y; h_0}\right]} {\left\vert{h_0}\right\vert}       | c = the limit does not depend on $n$ }} {{eqn | r = m       | c = definition of $m$ }} {{end-eqn}} However, by hypothesis: : $m \ne 0$ Hence, the contradiction is achieved and the initial statement of the lemma holds.  == Sources == * {{BookReference|Calculus of Variations|1963|I.M. Gelfand|author2 = S.V. Fomin}}: $\S 1.3$: The Variation of a Functional. A Necessary Condition for an Extremum
Continued Fraction Expansion of Irrational Square Root/Examples/13
== Example of Continued Fraction Expansion of Irrational Square Root ==  The continued fraction expansion of the square root of $13$ is given by: :$\sqrt {13} = \sqbrk {3, \sequence {1, 1, 1, 1, 6} }$  {{OEIS|A010122}} === Convergents === {{:Continued Fraction Expansion of Irrational Square Root/Examples/13/Convergents}} == Proof == Let $\sqrt {13} = \sqbrk {a_0, a_1, a_2, a_3, \ldots}$ From Partial Quotients of Continued Fraction Expansion of Irrational Square Root, the partial quotients of this continued fraction expansion can be calculated as: :$a_r = \floor {\dfrac {\floor {\sqrt {13} } + P_r} {Q_r} }$ where: :$P_r = \begin {cases} 0 & : r = 0 \\ a_{r - 1} Q_{r - 1} - P_{r - 1} & : r > 0 \\ \end {cases}$ :$Q_r = \begin {cases} 1 & : r = 0 \\ \dfrac {n - {P_r}^2} {Q_{r - 1} } & : r > 0 \\ \end {cases}$ {{PartialQuotientCalculator-Start | n = 13}} {{PartialQuotientCalculator | n = 13 | r = 1 | ar-1 = 3 | Qr-1 = 1 | Pr-1 = 0}} {{PartialQuotientCalculator | n = 13 | r = 2 | ar-1 = 1 | Qr-1 = 4 | Pr-1 = 3}} {{PartialQuotientCalculator | n = 13 | r = 3 | ar-1 = 1 | Qr-1 = 3 | Pr-1 = 1}} {{PartialQuotientCalculator | n = 13 | r = 4 | ar-1 = 1 | Qr-1 = 3 | Pr-1 = 2}} {{PartialQuotientCalculator | n = 13 | r = 5 | ar-1 = 1 | Qr-1 = 4 | Pr-1 = 1}} |} and the cycle is complete: :$\sequence {1, 1, 1, 1, 6}$
Continued Fraction Expansion of Irrational Square Root/Examples/29
== Examples of Continued Fraction Expansion of Irrational Square Root ==  The continued fraction expansion of the square root of $29$ is given by: :$\sqrt {29} = \sqbrk {5, \sequence {2, 1, 1, 2, 10} }$  {{OEIS|A010128}} === Convergents === {{:Continued Fraction Expansion of Irrational Square Root/Examples/29/Convergents}} == Proof == Let $\sqrt {29} = \sqbrk {a_0, a_1, a_2, a_3, \ldots}$ From Partial Quotients of Continued Fraction Expansion of Irrational Square Root, the partial quotients of this continued fraction expansion can be calculated as: :$a_r = \floor {\dfrac {\floor {\sqrt {29} } + P_r} {Q_r} }$ where: :$P_r = \begin {cases} 0 & : r = 0 \\ a_{r - 1} Q_{r - 1} - P_{r - 1} & : r > 0 \\ \end {cases}$ :$Q_r = \begin {cases} 1 & : r = 0 \\ \dfrac {n - {P_r}^2} {Q_{r - 1} } & : r > 0 \\ \end {cases}$ {{PartialQuotientCalculator-Start | n = 29}} {{PartialQuotientCalculator | n = 29 | r = 1 | ar-1 = 5 | Qr-1 = 1 | Pr-1 = 0}} {{PartialQuotientCalculator | n = 29 | r = 2 | ar-1 = 2 | Qr-1 = 4 | Pr-1 = 5}} {{PartialQuotientCalculator | n = 29 | r = 3 | ar-1 = 1 | Qr-1 = 5 | Pr-1 = 3}} {{PartialQuotientCalculator | n = 29 | r = 4 | ar-1 = 1 | Qr-1 = 5 | Pr-1 = 2}} {{PartialQuotientCalculator | n = 29 | r = 5 | ar-1 = 2 | Qr-1 = 4 | Pr-1 = 3}} |} and the cycle is complete: :$\sequence {2, 1, 1, 2, 10}$
Continued Fraction Expansion of Irrational Square Root/Examples/13/Convergents
== Convergents to Continued Fraction Expansion of $\sqrt {13}$ ==  The sequence of convergents to the continued fraction expansion of the square root of $13$ begins: :$\dfrac 3 1, \dfrac 4 1, \dfrac 7 2, \dfrac {11} 3, \dfrac {18} 5, \dfrac {119} {33}, \dfrac {137} {38}, \dfrac {256} {71}, \dfrac {393} {109}, \dfrac {649} {180}, \ldots$  {{OEIS-Numerators|A041018}} {{OEIS-Denominators|A041019}} == Proof == Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be its continued fraction expansion. Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be its numerators and denominators. Then the $n$th convergent is $\dfrac {p_n} {q_n}$. By definition: :$p_k = \begin {cases} a_0 & : k = 0 \\ a_0 a_1 + 1 & : k = 1 \\ a_k p_{k - 1} + p_{k - 2} & : k > 1 \end {cases}$ :$q_k = \begin {cases} 1 & : k = 0 \\ a_1 & : k = 1 \\ a_k q_{k - 1} + q_{k - 2} & : k > 1 \end {cases}$ From Continued Fraction Expansion of $\sqrt {13}$: :$\sqrt {13} = \sqbrk {3, \sequence {1, 1, 1, 1, 6} }$ Thus the convergents are assembled: {{ConvergentCalculator-Start | a0 = 3 | a1 = 1}} {{ConvergentCalculator | k = 2 | ak = 1 | pk-1 = 4 | pk-2 = 3 | qk-1 = 1 | qk-2 = 1}} {{ConvergentCalculator | k = 3 | ak = 1 | pk-1 = 7 | pk-2 = 4 | qk-1 = 2 | qk-2 = 1}} {{ConvergentCalculator | k = 4 | ak = 1 | pk-1 = 11 | pk-2 = 7 | qk-1 = 3 | qk-2 = 2 }} {{ConvergentCalculator | k = 5 | ak = 6 | pk-1 = 18 | pk-2 = 11 | qk-1 = 5 | qk-2 = 3}} {{ConvergentCalculator | k = 6 | ak = 1 | pk-1 = 119 | pk-2 = 18 | qk-1 = 33 | qk-2 = 5}} {{ConvergentCalculator | k = 7 | ak = 1 | pk-1 = 137 | pk-2 = 119 | qk-1 = 38 | qk-2 = 33}} {{ConvergentCalculator | k = 8 | ak = 1 | pk-1 = 256 | pk-2 = 137 | qk-1 = 71 | qk-2 = 38}} {{ConvergentCalculator | k = 9 | ak = 1 | pk-1 = 393 | pk-2 = 256 | qk-1 = 109 | qk-2 = 71}} |}
Continued Fraction Expansion of Irrational Square Root/Examples/29/Convergents
== Convergents to Continued Fraction Expansion of $\sqrt {29}$ ==  The sequence of convergents to the continued fraction expansion of the square root of $29$ begins: :$\dfrac 5 1, \dfrac {11} 2, \dfrac {16} 3, \dfrac {27} 5, \dfrac {70} {13}, \dfrac {727} {135}, \dfrac {1524} {283}, \dfrac {2251} {418}, \dfrac {3775} {701}, \dfrac {9801} {1820}, \ldots$  {{OEIS-Numerators|A041046}} {{OEIS-Denominators|A041047}} == Proof == Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be its continued fraction expansion. Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be its numerators and denominators. Then the $n$th convergent is $p_n / q_n$. By definition: :$p_k = \begin {cases} a_0 & : k = 0 \\ a_0 a_1 + 1 & : k = 1 \\ a_k p_{k - 1} + p_{k - 2} & : k > 1\end {cases}$ :$q_k = \begin {cases} 1 & : k = 0 \\ a_1 & : k = 1 \\ a_k q_{k - 1} + q_{k - 2} & : k > 1\end {cases}$ From Continued Fraction Expansion of $\sqrt {29}$: :$\sqrt {29} = \sqbrk {5, \sequence {2, 1, 1, 2, 10} }$ Thus the convergents are assembled: {{ConvergentCalculator-Start | a0 = 5 | a1 = 2}} {{ConvergentCalculator | k = 2 | ak =  1 | pk-1 =   11 | pk-2 =    5 | qk-1 =   2 | qk-2 =   1}} {{ConvergentCalculator | k = 3 | ak =  1 | pk-1 =   16 | pk-2 =   11 | qk-1 =   3 | qk-2 =   2}} {{ConvergentCalculator | k = 4 | ak =  2 | pk-1 =   27 | pk-2 =   16 | qk-1 =   5 | qk-2 =   3}} {{ConvergentCalculator | k = 5 | ak = 10 | pk-1 =   70 | pk-2 =   27 | qk-1 =  13 | qk-2 =   5}} {{ConvergentCalculator | k = 6 | ak =  2 | pk-1 =  727 | pk-2 =   70 | qk-1 = 135 | qk-2 =  13}} {{ConvergentCalculator | k = 7 | ak =  1 | pk-1 = 1524 | pk-2 =  727 | qk-1 = 283 | qk-2 = 135}} {{ConvergentCalculator | k = 8 | ak =  1 | pk-1 = 2251 | pk-2 = 1524 | qk-1 = 418 | qk-2 = 283}} {{ConvergentCalculator | k = 9 | ak =  2 | pk-1 = 3775 | pk-2 = 2251 | qk-1 = 701 | qk-2 = 418}} |}
Lifting The Exponent Lemma/Lemma
== Lemma ==  Let $x, y \in \Z$ be distinct integers. Let $p$ be an odd prime. Let: :$p \divides x - y$ and: :$p \nmid x y$. Then  :$\map {\nu_p} {x^p - y^p} = \map {\nu_p} {x - y} + 1$  where $\nu_p$ denotes $p$-adic valuation. == Proof == Let $\map {\nu_p} {x - y} = k$. Then $x=p^k m + y$ where $p \nmid m$. We have: {{begin-eqn}} {{eqn | l = x^p - y^p       | r = (p^k m + y)^p - y^p       | c =  }} {{eqn | r = \sum_{i \mathop = 0}^p \paren {\binom p i \paren {p^k m}^{p - i} y^i} - y^p       | c = Binomial Theorem }} {{eqn | r = \sum_{i \mathop = 0}^{p - 2} \paren {\binom p i \paren {p^k m}^{p - i} y^i} + \binom p {p - 1} \paren {p^k m} y^{p - 1}       | c = picking out the last two terms from the summation }} {{eqn | r = \sum_{i \mathop = 0}^{p - 2} \paren {\binom p i \paren {p^k m}^{p - i} y^i} + p^{k + 1} m y^{p - 1}       | c =  }} {{end-eqn}} Note that all terms in the above expression have a factor of $p$ to the order at least $k+1$. So, $p^{k + 1} \mid x^p - y^p$. Also note that all terms in the summation have a factor of $p$ to the order at least $k + 2$. But in the term $p^{k + 1} m y^{p - 1}$, since $p \nmid m$ and $p \nmid y$, we have: :$p^{k + 2} \nmid p^{k + 1} m y^{p - 1}$ So: :$p^{k + 2} \nmid x^p - y^p$ So by definition of $p$-adic valuation: :$\map {\nu_p} {x^p - y^p} = k + 1$  == Sources == * {{citation|date = July 1904|title = On the Integral Divisors of $a^n - b^n$|journal = Annals of Mathematics|abbr = Ann. Math.|volume = 5|startpage = 173|endpage = 180|jstor = 2007263|author = George David Birkhoff|author2 = Harry Schultz Vandiver|jstorcat = yes}}: Theorem $\text{III}$.
Solutions of Pythagorean Equation/Sequence
== Sequence ==  The sequence of solutions of the Pythagorean equation can be tabulated as follows: :$\begin{array} {r r | r r | r r r | c} m & n & m^2 & n^2 & 2 m n & m^2 - n^2 & m^2 + n^2 \\ \hline 2 & 1 & 4 & 1 & 4 & 3 & 5 & \text{Primitive} \\ \hline 3 & 1 & 9 & 1 & 6 & 8 & 10 \\ 3 & 2 & 9 & 4 & 12 & 5 & 13 & \text{Primitive} \\ \hline 4 & 1 & 16 & 1 & 8 & 15 & 17 & \text{Primitive} \\ 4 & 2 & 16 & 4 & 16 & 12 & 20 \\ 4 & 3 & 16 & 9 & 24 & 7 & 25 & \text{Primitive} \\ \hline 5 & 1 & 25 & 1 & 10 & 24 & 26 \\ 5 & 2 & 25 & 4 & 20 & 21 & 29 & \text{Primitive} \\ 5 & 3 & 25 & 9 & 30 & 16 & 34 \\ 5 & 4 & 25 & 16 & 40 & 9 & 41 & \text{Primitive} \\ \hline 6 & 1 & 36 & 1 & 12 & 35 & 37 & \text{Primitive} \\ 6 & 2 & 36 & 4 & 24 & 32 & 40 \\ 6 & 3 & 36 & 9 & 36 & 27 & 45 \\ 6 & 4 & 36 & 16 & 48 & 20 & 52 \\ 6 & 5 & 36 & 25 & 60 & 11 & 61 & \text{Primitive} \\ \hline 7 & 1 & 49 & 1 & 14 & 48 & 50 \\ 7 & 2 & 49 & 4 & 28 & 45 & 53 & \text{Primitive} \\ 7 & 3 & 49 & 9 & 42 & 40 & 58 \\ 7 & 4 & 49 & 16 & 56 & 33 & 65 & \text{Primitive} \\ 7 & 5 & 49 & 25 & 70 & 24 & 74 \\ 7 & 6 & 49 & 36 & 84 & 13 & 85 & \text{Primitive} \\ \hline \end{array}$
Pythagorean Triangle/Examples/5-12-13
== Example of Primitive Pythagorean Triangle ==  The triangle whose sides are of length $5$, $12$ and $13$ is a primitive Pythagorean triangle.  :400px == Proof == {{begin-eqn}} {{eqn | l = 5^2 + 12^2       | r = 25 + 144       | c =  }} {{eqn | r = 169       | c =  }} {{eqn | r = 13^2       | c =  }} {{end-eqn}} It follows by Pythagoras's Theorem that $5$, $12$ and $13$ form a Pythagorean triple. Note that $5$ and $12$ are coprime. Hence, by definition, $5$, $12$ and $13$ form a primitive Pythagorean triple. The result follows by definition of a primitive Pythagorean triangle.  == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Smallest Pythagorean Triangle is 3-4-5|next = Pythagorean Triangle/Examples/6-8-10}}: $13$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Smallest Pythagorean Triangle is 3-4-5|next = Pythagorean Triangle/Examples/6-8-10}}: $13$
Pythagorean Triangle/Examples/6-8-10
== Example of Pythagorean Triangle ==  The triangle whose sides are of length $6$, $8$ and $10$ is a Pythagorean triangle. This is not a primitive Pythagorean triangle.  :300px == Proof == {{begin-eqn}} {{eqn | l = 6^2 + 8^2       | r = 2^2 \times 3^2 + 2^2 \times 4^2       | c =  }} {{eqn | r = 4 \times \paren {9 + 16}       | c =  }} {{eqn | r = 4 \times 25       | c =  }} {{eqn | r = 2^2 \times 5^2       | c =  }} {{eqn | r = 10^2       | c =  }} {{end-eqn}} It follows by Pythagoras's Theorem that $6$, $8$ and $10$ form a Pythagorean triple. Note that $6$ and $8$ are not coprime as $\gcd \set {6, 8} = 2$. Hence, by definition, $6$, $8$ and $10$ do not form a primitive Pythagorean triple. The result follows by definition of a primitive Pythagorean triangle.  == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Pythagorean Triangle/Examples/5-12-13|next = Pythagorean Triangle/Examples/7-24-25}}: $13$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Pythagorean Triangle/Examples/5-12-13|next = Pythagorean Triangle/Examples/7-24-25}}: $13$
Convergence of Complex Sequence in Polar Form/Corollary
== Corollary to Convergence of Complex Sequence in Polar Form ==  Let $\left\langle{z_n}\right\rangle$ be a sequence of nonzero complex numbers. Let $z \ne 0$ be a complex number with modulus $r$ and argument $\theta$. Let $I$ be a real interval of length at most $2 \pi$ that contains $\theta$. Suppose $\theta$ is not an endpoint of $I$. Suppose each $z_n$ admits an argument $\theta_n \in I$. Let $r_n$ be the modulus of $z_n$. Then $z_n$ converges to $z$  $r_n$ converges to $r$ and $\theta_n$ converges to $\theta$.  == Proof == Suppose $r_n \to r$ and $\theta_n \to \theta$. Then by Convergence of Complex Sequence in Polar Form, $z_n \to z$. Conversely, suppose $z_n \to z$. By Convergence of Complex Sequence in Polar Form, we have: :$(1): \quad r_n \to r$ :$(2): \quad$ There exists a sequence $\left\langle{k_n}\right\rangle$ of integers such that $\theta_n + 2 k_n \pi$ converges to $\theta$. It remains to be proved that $\theta_n \to \theta$. Let $N \in \N$ such that $\left\vert{\theta_N + 2 k_N \pi - \theta}\right\vert \le \pi / 2$ for all $n \ge N$. By the Triangle Inequality for Real Numbers: :$\left\vert{2 k_n \pi - 2 k_N \pi}\right\vert \le \left\vert{\theta_n + 2 k_n \pi - \theta}\right\vert + \left\vert{\theta_N + 2 k_N \pi - \theta}\right\vert \le \pi$ for all $n \ge N$. Thus $\left\vert{k_n - k_N}\right\vert \le 1 / 2$, so $k_n = k_N$ for all $n \ge N$. By the Triangle Inequality for Real Numbers: :$\left\vert{2 \pi k_N}\right\vert \le \left\vert{\theta_n - \theta}\right\vert + \left\vert{\theta_n + 2 \pi k_N - \theta}\right\vert$ for all $n \in \N$. Because $\theta_n \in I$ and $\theta$ is not an endpoint of $I$: :$\left\vert{\theta_n - \theta}\right\vert < 2 \pi$ for all $n \in \N$. Because $\theta_n + 2 \pi k_N - \theta \to 0$: :$\left\vert{2 \pi k_N}\right\vert < 2 \pi$ Thus $k_n = 0$ for all $n \ge N$. Thus $\theta_n \to \theta$.
Pythagorean Triangle/Examples/693-1924-2045
== Example of Primitive Pythagorean Triangle ==  The triangle whose sides are of length $693$, $1924$ and $2045$ is a primitive Pythagorean triangle.  :700px == Proof == {{begin-eqn}} {{eqn | l = 693^2 + 1924^2       | r = 480 \, 249 + 3 \, 701 \, 776       | c =  }} {{eqn | r = 4 \, 182 \, 025       | c =  }} {{eqn | r = 2045^2       | c =  }} {{end-eqn}} It follows by Pythagoras's Theorem that $693$, $1924$ and $2045$  form a Pythagorean triple. Note that $693$ and $1924$ are coprime. Hence, by definition, $693$, $1924$ and $2045$ form a primitive Pythagorean triple. The result follows by definition of a primitive Pythagorean triangle.  == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Pythagorean Triangle whose Area is Half Perimeter|next = Pythagorean Triangles whose Areas are Repdigit Numbers}}: $13$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Pythagorean Triangle whose Area is Half Perimeter|next = Pythagorean Triangles whose Areas are Repdigit Numbers}}: $13$
693
{{NumberPageLink|prev = 692|next = 694}} == Number == $693$ ('''six hundred and ninety three''') is: :$3^2 \times 7 \times 11$ :One of the legs of the Pythagorean triangle $693-1924-2045$, whose area is $666 \, 666$. == Also see == * Pythagorean Triangles whose Areas are Repdigit Numbers
1924
{{NumberPageLink|prev = 1923|next = 1925}} == Number == $1924$ ('''one thousand nine hundred and twenty four''') is: :$2^2 \times 13 \times 37$ :One of the legs of the Pythagorean triangle $693-1924-2045$, whose area is $666 \, 666$. == Also see == * Pythagorean Triangles whose Areas are Repdigit Numbers
111
{{NumberPageLink|prev = 110|next = 112}} == Number == $111$ ('''one hundred and eleven''') is: :$3 \times 37$ :The magic constant of the smallest prime magic square. :The $3$rd repuint after $1$, $11$ :The $4$th of the $17$ positive integers for which the value of the Euler $\phi$ function is $72$: ::$73$, $91$, $95$, $111$, $117$, $135$, $146$, $148$, $152$, $182$, $190$, $216$, $222$, $228$, $234$, $252$, $270$ :The $6$th positive integer after $1$, $2$, $7$, $11$, $101$ whose cube is palindromic: ::$111^3 = 1 \, 367 \, 631$ :The magic constant of a magic square of order $6$, after $1$, $(5)$, $15$, $34$, $65$: ::$111 = \displaystyle \dfrac 1 6 \sum_{k \mathop = 1}^{6^2} k = \dfrac {6 \paren {6^2 + 1} } 2$ :The $7$th palindromic lucky number: ::$1$, $3$, $7$, $9$, $33$, $99$, $111$, $\ldots$ :The $8$th palindromic integer after $0$, $1$, $2$, $3$, $11$, $22$, $101$ whose square is also palindromic integer ::$111^2 = 12 \, 321$ :The $15$th Zuckerman number after $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, $11$, $12$, $15$, $24$, $36$: ::$111 = 111 \times 1 = 111 \times \paren {1 \times 1 \times 1}$ :The $24$th lucky number: ::$1$, $3$, $7$, $9$, $13$, $15$, $21$, $25$, $31$, $33$, $37$, $43$, $49$, $51$, $63$, $67$, $73$, $75$, $79$, $87$, $93$, $99$, $105$, $111$, $\ldots$ :The $35$th semiprime: ::$111 = 3 \times 37$ == Also see == * Magic Constant of Smallest Prime Magic Square * {{NumberPageLink|prev = 11|next = 1111|type = Repunit|cat = Repunits}} * {{NumberPageLink|prev = 36|next = 112|type = Zuckerman Number|cat = Zuckerman Numbers}} * {{NumberPageLink|prev = 65|next = 175|result = Magic Constant of Magic Square}} * {{NumberPageLink|prev = 95|next = 117|result = Numbers with Euler Phi Value of 72}} * {{NumberPageLink|prev = 99|next = 141|result = Sequence of Palindromic Lucky Numbers}} * {{NumberPageLink|prev = 101|next = 121|result = Square of Small-Digit Palindromic Number is Palindromic}} * {{NumberPageLink|prev = 101|next = 1001|result = Sequence of Integers whose Cube is Palindromic}} * {{NumberPageLink|prev = 105|next = 115|type = Lucky Number|cat = Lucky Numbers}} * {{NumberPageLink|prev = 106|next = 115|type = Semiprime Number|cat = Semiprimes}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Magic Constant of Smallest Prime Magic Square|next = 112/Historical Note}}: $111$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Magic Constant of Smallest Prime Magic Square|next = Sequence of Palindromic Lucky Numbers}}: $111$
1001
{{NumberPageLink|prev = 1000|next = 1002}} == Number == $1001$ ('''one thousand and one''') is: :$7 \times 11 \times 13$ :The $4$th pentagonal number after $1$, $5$, $22$ which is also palindromic: ::$1001 = \displaystyle \sum_{k \mathop = 1}^{26} \paren {3 k - 2} = \dfrac {26 \paren {3 \times 26 - 1} } 2$ :The $7$th positive integer after $1$, $2$, $7$, $11$, $101$, $111$ whose cube is palindromic: ::$1001^3 = 1 \, 003 \, 003 \, 001$ :The $11$th pentatope number after $1$, $5$, $15$, $35$, $70$, $126$, $210$, $330$, $495$, $715$: ::$1001 = \displaystyle \sum_{k \mathop = 1}^{11} \dfrac {k \paren {k + 1} \paren {k + 2} } 6 = \dfrac {11 \paren {11 + 1} \paren {11 + 2} \paren {11 + 3} } {24}$ :The $12$th palindromic integer after $0$, $1$, $2$, $3$, $11$, $22$, $101$, $111$, $121$, $202$, $212$ whose square is also palindromic integer ::$1001^2 = 1 \, 002 \, 001$ :The $26$th pentagonal number after $1$, $5$, $12$, $22$, $35$, $\ldots$, $477$, $532$, $590$, $651$, $715$, $782$, $852$, $925$: ::$1001 = \displaystyle \sum_{k \mathop = 1}^{26} \paren {3 k - 2} = \dfrac {26 \paren {3 \times 26 - 1} } 2$ :The $51$st generalized pentagonal number after $1$, $2$, $5$, $7$, $12$, $15$, $\ldots$, $610$, $651$, $672$, $715$, $737$, $782$, $805$, $852$, $876$, $925$, $950$: ::$1001 = \displaystyle \sum_{k \mathop = 1}^{26} \paren {3 k - 2} = \dfrac {26 \paren {3 \times 26 - 1} } 2$ == Also see == * {{NumberPageLink|prev = 22|next = 2882|result = Sequence of Palindromic Pentagonal Numbers}} * {{NumberPageLink|prev = 101|next = 10,001|result = Prime Factors of One More than Power of 10}} * {{NumberPageLink|prev = 111|next = 2201|result = Sequence of Integers whose Cube is Palindromic}} * {{NumberPageLink|prev = 212|next = 1111|result = Square of Small-Digit Palindromic Number is Palindromic}} * {{NumberPageLink|prev = 715|next = 1365|type = Pentatope Number|cat = Pentatope Numbers}} * {{NumberPageLink|prev = 925|next = 1080|type = Pentagonal Number|cat = Pentagonal Numbers}} * {{NumberPageLink|prev = 950|next = 1027|type = Generalized Pentagonal Number|cat = Generalized Pentagonal Numbers}} == Historical Note == {{:1001/Historical Note}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = 1000|next = Divisibility Test for 7, 11 and 13}}: $1001$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = 1000|next = Divisibility Test for 7, 11 and 13}}: $1001$
Brahmagupta-Fibonacci Identity/General
== General version of Brahmagupta-Fibonacci Identity ==  Let $a, b, c, d, n$ be numbers. :$\left({a^2 + n b^2}\right) \left({c^2 + n d^2}\right) = \left({a c + n b d}\right)^2 + n \left({a d - b c}\right)^2$  === Corollary === {{:Brahmagupta-Fibonacci Identity/General/Corollary}} === Extension === {{:Brahmagupta-Fibonacci Identity/General/Extension}} == Proof == {{begin-eqn}} {{eqn | o =        | r = \left({a c + n b d}\right)^2 + n \left({a d - b c}\right)^2       | c =  }} {{eqn | r = \left({\left({a c}\right)^2 + 2 \left({a c}\right) \left({n b d}\right) + \left({n b d}\right)^2}\right) + n \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 n a b c d + n^2 b^2 d^2 + n a^2 d^2 - 2 n a b c d + n b^2 c^2       | c = multiplying out }} {{eqn | r = a^2 c^2 + n a^2 d^2 + n b^2 c^2 + n^2 b^2 d^2       | c = simplifying }} {{eqn | r = \left({a^2 + n b^2}\right) \left({c^2 + n d^2}\right)       | c =  }} {{end-eqn}}  {{Namedfor|Brahmagupta|name2 = Leonardo Fibonacci}}