hcju/proofwikiqa huggingface dataset:
According to Necessary Condition for Integral Functional to have Extremum for given function/Dependent on N Functions: Euler's Equations for functional $J_1$ are:  :$\ds F_{\mathbf y} - \frac \d {\d x} F_{\mathbf y'} = 0$ Equivalently, for $J_2$ we have {{begin-eqn}} {{eqn | l = \paren {F_{\mathbf y} + \Psi_{\mathbf y} } - \map {\frac \d {\d x} } {F_{\mathbf y'} + \Psi_{\mathbf y'} }       | r = \paren {F_{\mathbf y} - \frac \d {\d x} F_{\mathbf y'} } + \paren {\Psi_{\mathbf y} - \frac \d {\d x} \Psi_{\mathbf y'} } }} {{eqn | r = 0       | c = condition for the existence of extremum }} {{end-eqn}} Furthermore: {{begin-eqn}} {{eqn | l = \Psi_{\mathbf y}       | r = \frac {\partial^2 \Phi} {\partial \mathbf y \partial x} + \frac \partial {\partial \mathbf y} \sum_{j \mathop = 1}^n \frac {\partial \Phi} {\partial y_j} y_j' }} {{eqn | r = \frac {\partial^2 \Phi} {\partial \mathbf y \partial x} + \sum_{j \mathop = 1}^n \frac {\partial^2 \Phi} {\partial \mathbf y \partial y_j} y_j' }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \Psi_{\mathbf y'}       | r = \frac {\partial^2 \Phi} {\partial \mathbf y' \partial x} + \frac \partial {\partial \mathbf y'} \sum_{j \mathop = 1}^n \frac {\partial \Phi} {\partial y_j} y_j' }} {{eqn | r = 0 + \sum_{j \mathop = 1}^n \paren {\frac {\partial^2 \Phi} {\partial \mathbf y' \partial y_j} y_j' + \frac {\partial \Phi} {\partial y_j} \frac {\partial y_j'} {\partial \mathbf y'} } }} {{eqn | r = \frac {\partial \Phi} {\partial \mathbf y}       | c = as $\frac {\partial y_i'} {\partial y_j'} = \delta_{i j} $, where $\delta_{i j}$ is the Kronecker Delta }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \frac {\d \Psi_{\mathbf y'} } {\d x}       | r = \frac {\partial \Psi_{\mathbf y'} } {\partial x} + \sum_{j \mathop = 1}^n \frac {\partial \Psi_{\mathbf y'} } {\partial y_j} \frac {\d y_j} {\d x} + \sum_{j \mathop = 1}^n \frac {\partial \Psi_{\mathbf y'} } {\partial y_j'} \frac {\d y_j'} {\d x} }} {{eqn | r = \frac {\partial^2 \Phi} {\partial x \partial \mathbf y} + \sum_{j \mathop = 1}^n \frac {\partial^2 \Phi} {\partial \mathbf y \partial y_j} y_j' }} {{end-eqn}} Since $\Phi$ is twice differentiable, by Schwarz-Clairaut Theorem partial derivatives commute and: :$\Psi_{\mathbf y} - \dfrac \d {\d x} \Psi_{\mathbf y'} = 0$ Therefore, $J_1$ and $J_2$ have same Euler's Equations. {{qed}}
{{begin-eqn}} {{eqn | l = T_a - T_b       | r = \dfrac {a^2 + a} 2 - \dfrac {b^2 + b} 2       | c = Closed Form for Triangular Numbers }} {{end-eqn}} Let $a = 3b + 1$ {{begin-eqn}} {{eqn | l = T_{3 b + 1} - T_b       | r = \dfrac {\paren {3 b + 1}^2 + 3 b + 1} 2 - \dfrac {b^2 + b} 2       | c = }} {{eqn | r = \dfrac {\paren {3 b + 1}^2 + 3 b + 1 - b^2 + b} 2       | c =  }} {{eqn | r = \dfrac {9 b^2 + 6 b + 1 + 3 b + 1 - b^2 - b} 2       | c =  }} {{eqn | r = \dfrac {8 b^2 + 8 b + 2} 2       | c =  }} {{eqn | r = 4 b^2 + 4 b + 1       | c =  }} {{eqn | r = \paren {2 b + 1}^2       | c =  }} {{end-eqn}} The square of every odd number can be made through this method. {{begin-eqn}} {{eqn | l = T_{3 b + 1}       | r = \dfrac {\paren {3 b + 1} ^ 2 + 3 b + 1} 2       | c = }} {{eqn | r = \dfrac {9 b^2 + 9 b + 2} 2       | c =  }} {{eqn | r = 9 \paren {\dfrac {b^2 + b} 2} + 1       | c =  }} {{eqn | ll= \leadsto       | l = T_{3 b + 1} - 9 T_b       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \gcd \set {T_{3 b + 1}, T_b}       | r = 1       | c = Bézout's Lemma }} {{end-eqn}} Both triangular numbers are coprime. {{qed}}
By Intersection is Subset: :$A \cap B \subseteq A$ and $A \cap B \subseteq B$ By definition: :$A \cap B$ is lower bound for $\left\{ {A, B}\right\}$ We will prove that :$\forall C \in X: C$ is lower bound for $\left\{ {A, B}\right\} \implies C \subseteq A \cap B$ Let $C \in X$ such that :$C$ is lower bound for $\left\{ {A, B}\right\}$. By definition of lower bound: :$C \subseteq A$ and $C \subseteq B$ Thus by Intersection is Largest Subset: :$C \subseteq A \cap B$ {{qed|lemma}} By definition of infimum: :$\inf \left\{ {A, B}\right\} = A \cap B$ Thus by definition of meet: :$A \wedge B = A \cap B$ {{qed}}
=== Sufficient Condition === Let $A$ be meet irreducible in $P$. By Top in Ordered Set of Topology: :$A \ne S$ By Relative Complement of Empty Set and Relative Complement of Relative Complement: :$\complement_S\left({A}\right) \ne \varnothing$ Thus by definition: :$\complement_S\left({A}\right)$ is non-empty. Thus by definition: :$\complement_S\left({A}\right)$ is closed. Let $B, C$ be closed subsets of $S$ such that :$\complement_S\left({A}\right) = B \cup C$ By De Morgan's Laws (Set Theory)/Relative Complement/Complement of Union and Relative Complement of Relative Complement: :$A = \complement_S\left({B}\right) \cap \complement_S\left({C}\right)$ By definition of closed set: :$\complement_S\left({B}\right) \in \tau$ and $\complement_S\left({C}\right) \in \tau$ By definition of topological space: :$\complement_S\left({B}\right) \cap \complement_S\left({C}\right) \in \tau$ By Meet in Inclusion Ordered Set: :$\complement_S\left({B}\right) \wedge \complement_S\left({C}\right) = A$ By definition of meet irreducible: :$A = \complement_S\left({B}\right)$ or $A = \complement_S\left({C}\right)$ Thus by Relative Complement of Relative Complement: :$\complement_S\left({A}\right) = B$ or $\complement_S\left({A}\right) = C$ {{qed|lemma}} === Necessary Condition === This follows from Complement of Element is Irreducible implies Element is Meet Irreducible. {{qed}}
{{begin-eqn}} {{eqn | l = {T_n}^2 - {T_{n - 1} }^2       | r = \paren {\frac {n \paren {n + 1} } 2}^2 - \paren {\frac {\paren {n - 1} n} 2}^2       | c = Closed Form for Triangular Numbers }} {{eqn | r = \frac {\paren {n^2 + n}^2 - \paren {n^2 - n}^2} 4       | c =  }} {{eqn | r = \frac {\paren {n^4 + 2 n^3 + n^2} - \paren {n^4 - 2 n^3 + n^2} } 4       | c =  }} {{eqn | r = \frac {4 n^3} 4       | c =  }} {{eqn | r = n^3       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = 3 \sum_{i \mathop = 1}^n i^5       | r = {T_n}^2 \paren {4 T_n - 1}       | c = Sum of Sequence of Fifth Powers }} {{eqn | r = k {T_n}^2       | c = where $k = 4 T_n - 1$ }} {{eqn | ll= \leadsto       | l = {T_n}^2       | o = \divides       | r = 3 \sum_{i \mathop = 1}^n i^5       | c = {{Defof|Divisor of Integer}} }} {{eqn | ll= \leadsto       | l = \sum_{i \mathop = 1}^n i^3       | o = \divides       | r = 3 \sum_{i \mathop = 1}^n i^5       | c = Square of Triangular Number equals Sum of Sequence of Cubes }} {{end-eqn}} {{qed}}
<onlyinclude> The proof proceeds by induction. For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle \sum_{i \mathop = 1}^n i^5 = \dfrac { {T_n}^2 \paren {4 T_n - 1} } 3$ === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = \dfrac { {T_1}^2 \paren {4 T_1 - 1} } 3       | r = \dfrac {\paren {\frac {1 \paren {1 + 1} } 2}^2 \paren {4 \frac {1 \paren {1 + 1} } 2 - 1} } 3       | c = Closed Form for Triangular Numbers }} {{eqn | r = \dfrac {1^2 \times \paren {4 \times 1 - 1} } 3       | c =  }} {{eqn | r = 1       | c =  }} {{eqn | r = \sum_{i \mathop = 1}^1 i^5       | c =  }} {{end-eqn}} 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 $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \sum_{i \mathop = 1}^k i^5 = \dfrac { {T_k}^2 \paren {4 T_k - 1} } 3$ from which it is to be shown that: :$\displaystyle \sum_{i \mathop = 1}^{k + 1} i^5 = \dfrac { {T_{k + 1} }^2 \paren {4 T_{k + 1} - 1} } 3$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^{k + 1} i^5       | r = \sum_{i \mathop = 1}^k i^5 + \paren {k + 1}^5       | c =  }} {{eqn | r = \dfrac { {T_k}^2 \paren {4 T_k - 1} } 3 + \paren {k + 1}^5       | c = Induction Hypothesis }} {{eqn | r = \dfrac {\paren {\frac {k \paren {k + 1} } 2}^2 \paren {4 \frac {k \paren {k + 1} } 2 - 1} } 3 + \paren {k + 1}^5       | c = Closed Form for Triangular Numbers }} {{eqn | r = \dfrac {\paren {k \paren {k + 1} }^2 \paren {2 k \paren {k + 1} - 1} } {12} + \paren {k + 1}^5       | c =  }} {{eqn | r = \paren {k + 1}^2 \dfrac {\paren {2 k^3 \paren {k + 1} - k^2} + 12 \paren {k + 1}^3} {12}        | c =  }} {{eqn | r = \paren {k + 1}^2 \dfrac {2 k^4 + 2 k^3 - k^2 + 12 k^3 + 36 k^2 + 36 k + 12} {12}        | c =  }} {{eqn | r = \paren {k + 1}^2 \dfrac {2 k^4 + 14 k^3 + 35 k^2 + 36 k + 12} {12}        | c =  }} {{eqn | r = \paren {k + 1}^2 \paren {k + 2}^2 \dfrac {2 k^2 + 6 k + 3} {12}        | c = extracting $\paren {k + 2}^2$ as a factor }} {{eqn | r = \paren {\frac {\paren {k + 1} \paren {k + 2} } 2}^2 \dfrac {4 \frac {\paren {k + 1} \paren {k + 2} } 2 - 1} 3        | c =  }} {{eqn | r = \dfrac { {T_{k + 1} }^2 \paren {4 T_{k + 1} - 1} } 3        | c = Closed Form for Triangular Numbers }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{> 0}: \displaystyle \sum_{i \mathop = 1}^n i^5 = \dfrac { {T_n}^2 \paren {4 T_n - 1} } 3$ {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{k \mathop = n^2 + n - 1}^{n^2 + 2 n} T_n       | r = \sum_{k \mathop = 1}^{n^2 + 2 n} T_n - \sum_{k \mathop = 1}^{n^2 + n - 2} T_n }} {{eqn | r = H_{n^2 + 2 n} - H_{n^2 + n - 2}       | c = {{Defof|Tetrahedral Number}} }} {{eqn | r = \frac {\paren {n^2 + 2 n} \paren {n^2 + 2 n + 1} \paren {n^2 + 2 n + 2} } 6 - \frac {\paren {n^2 + n - 2} \paren {n^2 + n - 1} \paren {n^2 + n} } 6       | c = Closed Form for Tetrahedral Numbers }} {{eqn | r = \frac {n \paren {n + 1} \paren {n + 2} } 6 \paren {\paren {n + 1} \paren {n^2 + 2 n + 2} - \paren {n - 1} \paren {n^2 + n - 1} } }} {{eqn | r = \frac {n \paren {n + 1} \paren {n + 2} } 6 \paren {3 n^2 + 6 n + 1} }} {{eqn | r = \frac {n \paren {n + 1} \paren {n + 2} } 6 \paren {\paren {n + 3} \paren {n^2 + 3 n + 1} - \paren {n + 1} \paren {n^2 + 2 n + 2} } }} {{eqn | r = \frac {\paren {n^2 + 3 n} \paren {n^2 + 3 n + 1} \paren {n^2 + 3 n + 2} } 6 - \frac {\paren {n^2 + 2 n} \paren {n^2 + 2 n + 1} \paren {n^2 + 2 n + 2} } 6 }} {{eqn | r = H_{n^2 + 3 n} - H_{n^2 + 2 n}       | c = Closed Form for Tetrahedral Numbers }} {{eqn | r = \sum_{k \mathop = 1}^{n^2 + 3 n} T_n - \sum_{k \mathop = 1}^{n^2 + 2 n} T_n       | c = {{Defof|Tetrahedral Number}} }} {{eqn | r = \sum_{k \mathop = n^2 + 2 n + 1}^{n^2 + 3 n} T_n }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = T_{n^2 - n - 1} + T_{n^2 + n - 1}       | r = \frac {\paren {n^2 - n - 1} \paren {n^2 - n} } 2 + \frac {\paren {n^2 + n - 1} \paren {n^2 + n} } 2       | c = Closed Form for Triangular Numbers }} {{eqn | r = \frac {n \paren {n^2 - n - 1} \paren {n - 1} } 2 + \frac {n \paren {n^2 + n - 1} \paren {n + 1} } 2       | c = factoring }} {{eqn | r = \frac {n \paren {n^3 - n^2 - n - n^2 + n + 1} } 2 + \frac {n \paren {n^3 + n^2 - n + n^2 + n - 1} } 2       | c = multiplying out the numerators }} {{eqn | r = \frac {n \paren {\paren {n^3 - 2 n^2 + 1} + \paren {n^3 + 2 n^2 - 1} } } 2       | c = simplifying }} {{eqn | r = n^4       | c = simplifying }} {{end-eqn}} {{qed}}
By Conditions for Integral Functionals to have same Euler's Equations, functionals: :$\ds \int_a^b F_1 \rd x$  and: :$\ds \int_a^b F_2 \rd x = \int_a^b \paren {F_1 \pm \frac {\d \Phi} {\d x} } \rd x$  have same Euler's equations. Express the first one in canonical variables $\paren {x, \sequence{y_i}_{1 \mathop \le i \mathop \le n}, \sequence{p_i}_{1 \mathop \le i \mathop \le n}, H}$ and the second one in $\paren {x, \sequence{Y_i}_{1 \mathop \le i \mathop \le n}, \sequence {P_i}_{1 \mathop \le i \mathop \le n}, H^*}$: :$\ds \int_a^b F_1 \rd x = \int_a^b \paren {\sum_{i \mathop = 1}^n p_i y_i' - H} \rd x$ :$\ds \int_a^b F_2 \rd x = \int_a^b \paren {\sum_{i \mathop = 1}^n P_i Y_i' - H^*} \rd x$ However: :$\ds \int_a^b \paren {F_2 - F_1} \rd x = \int_a^b \pm \frac {\d \Phi} {\d x} \rd x$ Inserting new expressions for $F_1$ and $F_2$ yields: :$\ds \int_a^b \paren {\sum_{i \mathop = 1}^n P_i Y_i'-H^* - \sum_{i \mathop = 1}^n p_i y_i' + H} \rd x = \int_a^b \pm \frac {\d \Phi} {\d x}\rd x$ This is satisfied, if integrands are equal. Transform the coordinates to $\tuple {x, \sequence {y_i}_{1 \mathop \le i \mathop \le n}, \sequence {Y_i}_{1 \mathop \le i \mathop \le n} }$ and write out the full derivative of $\Phi$: :$\ds \sum_{i \mathop = 1}^n P_i Y_i' - H^* - \sum_{i \mathop = 1}^n p_i y_i' + H = \pm \frac {\partial \Phi} {\partial x} \pm \sum_{i \mathop = 1}^n \frac {\partial \Phi} {\partial y_i} y_i' \pm \sum_{i \mathop = 1}^n \frac {\partial \Phi} {\partial Y_i} Y_i'$ Collect terms multiplied by the same the coordinates together: :$\ds \sum_{i \mathop = 1}^n Y_i' \paren {\pm \frac {\partial \Phi} {\partial Y_i} - P_i} + \sum_{i \mathop = 1}^n y_i' \paren {\pm \frac {\partial \Phi} {\partial y_i} + p_i} + \paren {\pm \frac {\partial \Phi} {\partial x} - H + H^*} = 0$ This has to hold for arbitrary values of independent coordinates $\tuple {x, \sequence {y_i}_{1 \mathop \le i \mathop \le n}, \sequence {Y_i}_{1 \mathop \le i \mathop \le n} }$. Hence: :$p_i = \mp \dfrac {\partial \Phi} {\d y_i}, \quad P_i = \pm \dfrac {\partial \Phi} {\d Y_i}, \quad H^* = H \mp \dfrac {\partial \Phi} {\partial x}$ {{qed}} {{finish|Eventually cover all four generating functions, most probably in different pages}}
{{begin-eqn}} {{eqn | l = T_n + T_{n - 1} T_{n + 1}       | r = \frac {n \paren {n + 1} } 2 + \paren {\frac {\paren {n - 1} n} 2 \frac {\paren {n + 1} \paren {n + 2} } 2}       | c = Closed Form for Triangular Numbers }} {{eqn | r = \frac {2 n^2 + 2 n + \paren {n^2 - n} \paren {n^2 + 3 n + 2} } 4       | c =  }} {{eqn | r = \frac {2 n^2 + 2 n + n^4 - n^3 + 3 n^3 - 3 n^2 + 2 n^2 - 2 n} 4       | c =  }} {{eqn | r = \frac {n^4 + 2 n^3 + n^2} 4       | c =  }} {{eqn | r = \paren {\frac {n \paren {n + 1} } 2}^2       | c =  }} {{eqn | r = {T_n}^2       | c = Closed Form for Triangular Numbers }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = 2 T_n T_{n - 1}       | r = 2 \paren {\frac {n \paren {n + 1} } 2} \paren {\frac {\paren {n - 1} n} 2}       | c = Closed Form for Triangular Numbers }} {{eqn | r = \frac {\paren {n^2 + n} \paren {n^2 - n} } 2       | c =  }} {{eqn | r = \frac {n^4 - n^2} 2       | c = Difference of Two Squares }} {{eqn | r = \frac {n^2 \paren {n^2 - 1} } 2       | c =  }} {{eqn | r = T_{n^2 - 1}       | c = Closed Form for Triangular Numbers }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{k \mathop \ge 1} \dfrac 1 {T_k}       | r = \sum_{k \mathop \ge 1} \dfrac 2 {k \paren {k + 1} }       | c = Closed Form for Triangular Numbers }} {{eqn | r = 2 \sum_{k \mathop \ge 1} \dfrac 1 {k \paren {k + 1} }       | c =  }} {{eqn | r = 2 \times 1       | c = Corollary to Sum of Sequence of Products of Consecutive Reciprocals }} {{end-eqn}} {{qed}}
By definition of topological space: :$S \in \tau$ By definition of subset: :$\forall A \in \tau: A \subseteq S$ Hence $P$ is bounded above. Thus by definition of the greatest element: :$\top_P = S$ {{qed}}
{{begin-eqn}} {{eqn | l = T_{4 n \paren {n + 1} }       | r = \frac {\paren {4 n \paren {n + 1} } \paren {4 n \paren {n + 1} + 1} } 2       | c = Closed Form for Triangular Numbers }} {{eqn | ll= \leadsto       | l = T_{4 n \paren {n + 1} }       | r = \frac {n \paren {n + 1} } 2 \times 4 \paren {4 n \paren {n + 1} + 1}       | c = Extract $T_n$ }} {{eqn | ll= \leadsto       | l = T_{4 n \paren {n + 1} }       | r = T_n \times \paren {16 n^2 + 16 n + 4}       | c = Simplify }} {{eqn | ll= \leadsto       | l = T_{4 n \paren {n + 1} }       | r = T_n \times \paren {4 n + 2}^2       | c = Factorise }} {{end-eqn}} The product of two squares is also square. Let $T_n$ be square. Therefore $T_{4 n \paren {n + 1} }$ is also square. {{qed}}
Suppose $T_n = x^3$ for some $x \in \Z$. Then by Odd Square is Eight Triangles Plus One: :$\exists y \in \Z: 8 T_n + 1 = \paren {2 x}^3 + 1 = y^2$ By Cube which is One Less than a Square: :$2 x = 2$, $y = 3$ giving the unique solution: :$T_n = 1^3 = 1$ {{qed}}
{{rename|{{begin-eqn}} {{eqn | l = 31 \times 16^n       | r = \paren {16 + 3 \times 4 + 3 \times 1} \times 4^{2 n} }} {{eqn | r = 4^{2 n + 2} + 3 \times 4^{2 n + 1} + 3 \times 4^{2 n} }} {{end-eqn}} A sum of 7 powers of 4.</br>whilst the link above links to biquadrates (fourth powers), in contrast to title of this page. In that case:}} Observe that for an even number $2 k$: :$\paren {2 k}^4 = 16 k^4 \equiv 0 \pmod {16}$ For an odd number $2 k + 1$: :$\paren {2 k + 1}^4 = 16 k^4 + 32 k^3 + 24 k^2 + 8 k + 1 \equiv 8 k \paren {k + 1} + 1 \pmod {16} \equiv 1 \pmod {16}$ since $k \paren {k + 1}$ is even. It is obvious that for $n = 0$, $31$ requires $16$ powers of $4$ to express: :$31 = 2^4 + 15 \times 1^4$ {{AimForCont}} for some $n > 0$, $31 \times 16^n$ requires less than $16$ powers of $4$ to express. Let $m$ be the smallest of those $n$. Suppose $x$ of the summands are odd, where $x < 16$. By the above, we must have $31 \times 16^n \equiv x \pmod {16}$. Since $31 \times 16^m$ is divisible by $16$, $x = 0$. Hence each summand is even. Dividing each summand by $2$ gives a representation of $31 \times 16^{m - 1}$ as a sum of less than $16$ powers of $4$. This contradicts the minimality condition on $m$. Hence the result by proof by contradiction. {{qed}} Category:Hilbert-Waring Theorem Category:16 8y90otfd5rmancdvvyfnezjl92og8p3
Apart from $2$, all primes are odd. From Sigma Function Odd iff Argument is Square or Twice Square, for $\map \sigma n$ to be odd it needs to be of the form $m^2$ or $2 m^2$. Suppose $n$ has two coprime divisors $p$ and $q$, each to power $k_p$ and $k_q$ respectively. Then $\map \sigma n$ will have $\map \sigma {p^{k_p} }$ and $\map \sigma {q^{k_q} }$ as divisors. Hence $\map \sigma n$ will not be prime. So for $\map \sigma n$ to be prime, $n$ can have only one prime factor. This gives possible values for $n$ as: :powers of $2$, either odd or even or: :even powers of a prime number. These can be investigated in turn, using Sigma Function of Power of Prime: :$\map \sigma {p^k} = \dfrac {p^{k + 1} - 1} {p - 1}$ Note that as $\map \sigma {2^k} = \dfrac {2^{k + 1} - 1} {2 - 1} = 2^{k + 1} - 1$ it is necessary for powers of $2$ merely to report the appropriate Mersenne prime. Hence when $k + 1$ is not prime, $\map \sigma {2^k}$ will not be prime and there is no need to test it. Thus we test all $n$ such that: :$n = p^{2 k}$ for prime $p$ :$n = 2^k$ where $k + 1$ is prime and so: {{begin-eqn}} {{eqn | l = \map \sigma 2       | r = 2^2 - 1       | c =  }} {{eqn | r = 3       | c = which is a Mersenne prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map \sigma 4       | r = \map \sigma {2^2} }} {{eqn | r = 2^3 - 1 }} {{eqn | r = 7       | c = which is a Mersenne prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map \sigma 9       | r = \map \sigma {3^2} }} {{eqn | r = \dfrac {3^3 - 1} {3 - 1} }} {{eqn | r = \dfrac {26} 2 }} {{eqn | r = 13       | c = which is prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map \sigma {16}       | r = \map \sigma {2^4} }} {{eqn | r = \dfrac {2^5 - 1} {2 - 1} }} {{eqn | r = 31       | c = which is prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map \sigma {25}       | r = \map \sigma {5^2} }} {{eqn | r = \dfrac {5^3 - 1} {5 - 1} }} {{eqn | r = \dfrac {124} 4 }} {{eqn | r = 31       | c = which is prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map \sigma {49}       | r = \map \sigma {7^2} }} {{eqn | r = \dfrac {7^3 - 1} {7 - 1} }} {{eqn | r = \dfrac {342} 6 }} {{eqn | r = 57 = 3 \times 19       | c = which is not prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map \sigma {64}       | r = \map \sigma {2^6} }} {{eqn | r = \dfrac {2^7 - 1} {2 - 1} }} {{eqn | r = 127       | c = which is a Mersenne prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map \sigma {121}       | r = \map \sigma {11^2} }} {{eqn | r = \dfrac {11^3 - 1} {11 - 1} }} {{eqn | r = \dfrac {1330} {10} }} {{eqn | r = 133 = 7 \times 19       | c = which is not prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map \sigma {169}       | r = \map \sigma {13^2} }} {{eqn | r = \dfrac {13^3 - 1} {11 - 1} }} {{eqn | r = \dfrac {2196} {12} }} {{eqn | r = 183 = 3 \times 61       | c = which is not prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map \sigma {289}       | r = \map \sigma {17^2} }} {{eqn | r = \dfrac {17^3 - 1} {17 - 1} }} {{eqn | r = \dfrac {4912} {16} }} {{eqn | r = 307       | c = which is prime }} {{end-eqn}} Hence the sequence as given. {{qed}}
=== Sufficient Condition === Assume that :$A$ is a dense element in $P$. By Bottom in Ordered Set of Topology: :$\bot_P = \O$ We will prove that :for every open subset $U$ of $S$: $U \ne \O \implies U \cap A \ne \O$ Let $U$ be an open subset of $S$ such that :$U \ne \O$ By definition of open set: :$U \in \tau$ By definition of topological space: :$U \cap A \in \tau$ By Meet in Inclusion Ordered Set: :$U \wedge A = U \cap A$ Thus by definition of dense element: :$U \cap A \ne \O$ {{qed|lemma}} By Characterization of Closure by Open Sets: :$A^- = S$ where $A^-$ denotes the topological closure of $A$. Hence $A$ is everywhere dense. {{qed|lemma}} === Necessary Condition === Assume that :$A$ is everywhere dense. Let $B \in \tau$ such that :$B \ne \bot_P$ By definition of non-empty set: :$\exists x: x \in B$ By definition: :$B$ is an open set. By definition of everywhere dense: :$A^- = S$ By Characterization of Closure by Open Sets: :$B \cap A \ne \O$ By definition of topological space: :$B \cap A \in \tau$ Thus by Meet in Inclusion Ordered Set: :$B \wedge A \ne \bot_P$ {{qed}}
Let $S$ be a square whose area equals its perimeter. Let $A$ be the area of $S$. Let $P$ be the perimeter of $S$. Let $b$ be the length of one side of $S$. From Area of Square: :$A = b^2$ From Perimeter of Rectangle: :$P = 2 b + 2 b = 4 b$ Setting $A = P$ :$b^2 = 4 b$ and so: :$b = 4$ and so: :$A = 16 = P$ {{qed}}
Let $M_4$ denote an order $4$ magic square By Sum of Terms of Magic Square, the total of all the entries in $M_4$ is given by: :$T_4 = \dfrac {4^2 \left({4^2 + 1}\right)} 2 = \dfrac {16 \times 17} 2 = 136$ As there are $4$ rows of $M_4$, the magic constant of $M_4$ is given by: :$S_4 = \dfrac {136} 4 = 34$ {{qed}}
Let $M_n$ denote the magic square of order $n$. By Magic Constant of Magic Square, the magic constant of $M_n$ is given by: :$S_n = \dfrac {n \left({n^2 + 1}\right)} 2$ Setting $n = 4$: :$S_4 = \dfrac {4 \times 17} 2 = 34$ {{qed}}
Recall Moessner's order $4$ magic square: {{:Definition:Moessner's Order 4 Magic Square}} {{begin-eqn}} {{eqn | l = 12^3 + 3^3 + 14^3 + 5^3       | r = 1728 + 27 + 2744 + 125       | c =  }} {{eqn | r = 4624       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 9^3 + 2^3 + 15^3 + 8^3       | r = 729 + 8 + 3375 + 512       | c =  }} {{eqn | r = 4624       | c =  }} {{end-eqn}} {{qed}}
Recall Moessner's order $4$ magic square: {{:Definition:Moessner's Order 4 Magic Square}} === Rows $1$ and $4$ === {{begin-eqn}} {{eqn | l = 12^2 + 13^2 + 1^2 + 8^2       | r = 144 + 169 + 1 + 64       | c =  }} {{eqn | r = 378       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 9^2 + 16^2 + 4^2 + 5^2       | r = 81 + 256 + 16 + 25       | c =  }} {{eqn | r = 378       | c =  }} {{end-eqn}} {{qed|lemma}} === Rows $2$ and $3$ === {{begin-eqn}} {{eqn | l = 6^2 + 3^2 + 15^2 + 10^2       | r = 36 + 9 + 225 + 100       | c =  }} {{eqn | r = 370       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 7^2 + 2^2 + 14^2 + 11^2       | r = 49 + 4 + 196 + 121       | c =  }} {{eqn | r = 370       | c =  }} {{end-eqn}} {{qed|lemma}} === Columns $1$ and $4$ === {{begin-eqn}} {{eqn | l = 12^2 + 6^2 + 7^2 + 9^2       | r = 144 + 36 + 49 + 81       | c =  }} {{eqn | r = 310       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 8^2 + 10^2 + 11^2 + 5^2       | r = 64 + 100 + 121 + 25       | c =  }} {{eqn | r = 310       | c =  }} {{end-eqn}} {{qed|lemma}} === Columns $2$ and $3$ === {{begin-eqn}} {{eqn | l = 13^2 + 3^2 + 2^2 + 16^2       | r = 169 + 9 + 4 + 256       | c =  }} {{eqn | r = 438       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 1^2 + 15^2 + 14^2 + 4^2       | r = 1 + 225 + 196 + 16       | c =  }} {{eqn | r = 438       | c =  }} {{end-eqn}} {{qed}}