hcju/proofwikiqa huggingface dataset:
By definition of abundance: :$A \left({n}\right) = \sigma \left({n}\right) - 2 n$ By definition of $\sigma$ function: :$\sigma \left({n}\right)$ is the sum of all the divisors of $n$. Thus $\sigma \left({n}\right) - n$ is the sum of the aliquot parts of $n$. Hence the result. {{qed}} Category:Almost Perfect Numbers fmfbrdd4kbcsp5zqor0pyd135ppfznu
By definition of abundance: :$A \left({n}\right) = \sigma \left({n}\right) - 2 n$ By definition of $\sigma$ function: :$\sigma \left({n}\right)$ is the sum of all the divisors of $n$. Thus $\sigma \left({n}\right) - n$ is the sum of the aliquot parts of $n$. Hence the result. {{qed}} Category:Quasiperfect Numbers 1oae6dj8sczu3khj0qwqnbg8xfx0th5
By definition of quasiperfect number: :$\map \sigma n = 2 n + 1$ where $\map \sigma n$ denotes the $\sigma$ function of $n$. That is, $\map \sigma n$ is odd. Then from Sigma Function Odd iff Argument is Square or Twice Square: $n$ is either square or twice a square. Suppose $n = 2^k m^2$ is a quasiperfect number, where $m$ is odd and $k \in \Z_{\ge 0}$. Then: {{begin-eqn}} {{eqn | l = \map \sigma n       | r = \map \sigma {2^k} \map \sigma {m^2}       | c = Sigma Function is Multiplicative }} {{eqn | r = \paren {2^{k + 1} - 1} \map \sigma {m^2}       | c = Sigma Function of Power of Prime }} {{eqn | r = 2 n + 1       | c = {{Defof|Quasiperfect Number}} }} {{eqn | r = 2^{k + 1} m^2 + 1 }} {{end-eqn}} Hence we have: :$\paren {2^{k + 1} - 1} \divides \paren {2^{k + 1} m^2 + 1}$ Since: :$\paren {2^{k + 1} - 1} \divides \paren {2^{k + 1} m^2 - m^2}$ we have: :$\paren {2^{k + 1} - 1} \divides \paren {1 + m^2}$ {{AimForCont}} $k > 0$. Write: :$m^2 \equiv -1 \pmod {2^{k + 1} - 1}$ But by First Supplement to Law of Quadratic Reciprocity (extended to Jacobi symbols): :$\paren {\dfrac {-1} {2^{k + 1} - 1} } = \paren {-1}^{\frac {2^{k + 1} - 1 - 1} 2} = \paren {-1}^{2^k - 1} = -1$ {{WIP|Maybe it is worth it to extend quadratic residue/reciprocity results to Jacobi symbols.}} Hence $-1$ is not a quadratic residue modulo $2^{k + 1} - 1$, contradicting the above. Therefore we must have $k = 0$. In this case, $n = m^2$, an odd square. {{qed}}
By Empty Set is Element of Topology: :$\varnothing \in \tau$ By Empty Set is Subset of All Sets: :$\forall A \in \tau: \varnothing \subseteq A$ Hence $P$ is bounded below. Thus by definition of the smallest element: :$\bot_P = \varnothing$ {{qed}}
We have that: {{begin-eqn}} {{eqn | l = 17^3       | r = 4913       | c =  }} {{eqn | l = 17       | r = 4 + 9 + 1 + 3       | c =  }} {{end-eqn}} From Sequence of Dudeney Numbers, the complete set of positive integers with this property are: :$0, 1, 8, 17, 18, 26, 27$ Of these, only $17$ is prime. {{qed}}
We have trivially that: {{begin-eqn}} {{eqn | l = 0^3       | r = 0       | c =  }} {{eqn | l = 1^3       | r = 1       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = 8^3       | r = 512       | c =  }} {{eqn | l = 8       | r = 5 + 1 + 2       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 17^3       | r = 4913       | c =  }} {{eqn | l = 17       | r = 4 + 9 + 1 + 3       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 18^3       | r = 5832       | c =  }} {{eqn | l = 18       | r = 5 + 8 + 3 + 2       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 26^3       | r = 17576       | c =  }} {{eqn | l = 26       | r = 1 + 7 + 5 + 7 + 6       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 27^3       | r = 19683       | c =  }} {{eqn | l = 27       | r = 1 + 9 + 6 + 8 + 3       | c =  }} {{end-eqn}} A quick empirical test shows that when $n = 46$, it is already too large to be the sum of the digits of its cube. For $46 < n \le 54$, $n^3 \le 54^3 < 200 \, 000$. Hence the sum of the digits of $n^3$ is less than: :$1 + 5 \times 9 = 46 < n$ For $54 < n < 100$, $n^3 < 10^6$. Hence the sum of the digits of $n^3$ is less than: :$6 \times 9 = 54 < n$ For $n \ge 100$, let $n$ be a $d$-digit number, where $d \ge 3$. Then $10^{d - 1} \le n < 10^d$ and $n^3 < 10^{3 d}$. Hence the sum of the digits of $n^3$ is less than: {{begin-eqn}} {{eqn | l = 3 d \times 9       | r = 27 d }} {{eqn | o = \le       | r = 27 d + 63 d - 189 }} {{eqn | o = <       | r = 90 d - 170 }} {{eqn | r = 10 \paren {1 + 9 \paren {d - 2} } }} {{eqn | o = \le       | r = 10 \times \paren {1 + 9}^{d - 2}       | c = Bernoulli's Inequality }} {{eqn | o = \le       | r = n }} {{end-eqn}} so no numbers greater than $46$ can have this property. {{qed}}
Consider the increment $\Delta S$: :$\Delta S = \map S {x + \Delta x, \mathbf y + \Delta \mathbf y} - \map S {x, \mathbf y}$ Note that the change of function $\mathbf y$ denoted by $\Delta \mathbf y$ is dependent on the manner $\Delta x$ is chosen through the definition of geodetic distance. For sufficiently smooth $S$, $\size {\Delta \mathbf y} \to 0$ as $\size {\Delta \mathbf x} \to 0$. {{explain|sufficiently}} By definition of differential, $\Delta S$ can be written as: :$\map {\Delta S} {x, \mathbf y; \Delta x, \Delta \mathbf y} = \map {\d S} {x, \mathbf y; \Delta x, \Delta \mathbf y} + \epsilon \Delta x + \boldsymbol \epsilon \cdot \Delta \boldsymbol y$ where $\epsilon \to 0$ as $\Delta x \to 0$, and $\size {\mathbf y} \to 0$ as $\size {\Delta \mathbf x} \to 0$. By definition of the geodetic distance, :$\Delta S = J \sqbrk {\gamma^*} - J \sqbrk \gamma$ where $\gamma$ and $\gamma^*$ are extremal curves, connecting the fixed initial point with points $\tuple {x, \mathbf y}$ and $\tuple {x + \Delta x, \mathbf y + \mathbf h}$ respectively. By definition of increment of functional: :$J \sqbrk {\gamma^*} - J \sqbrk \gamma = \Delta J \sqbrk {\gamma; \Delta \gamma}$ where $\Delta \gamma = \gamma^* - \gamma$. A differentiable $J$ can be expressed as: :$\Delta J \sqbrk {\gamma; \Delta \gamma} = \delta J \sqbrk {\gamma; \Delta \gamma} + \epsilon_\gamma \cdot \size {\Delta \gamma}$ where $\epsilon_\gamma \to 0$ as $\size {\Delta \gamma} \to 0$, and $\size {\Delta \gamma} \to 0$ as $\size {\Delta \mathbf x} \to 0$ for sufficiently smooth $S$. To summarise: :$\Delta \map S {x, \mathbf y; \Delta x, \Delta \mathbf y} = \Delta J \sqbrk {\gamma; \Delta \gamma}$ {{explain|change in $x$ should be reflected on both sides, but literature omits this}} Both sides contain terms linear in $\size {\Delta x}$, $\size {\Delta \mathbf y}$, $\size {\Delta \gamma}$ as well terms of higher order. Higher order terms on both sides are the same. {{explain|why}} Hence, the principal parts match: :$\d S = \delta J$ The variation of extremal $J$ is expressible as :$\displaystyle \delta J = \sum_{i \mathop = 1}^n p_i \Delta y_i - H \Delta x$ while the differential of $S$ is  :$\displaystyle \d S = \frac {\partial S} {\partial x} \Delta x + \sum_{i \mathop = 1}^n \frac {\partial S} {\partial y_i} \Delta y_i$ Equivalently: :$\displaystyle \paren {\frac {\partial S} {\partial x} + H} \Delta x + \sum_{i \mathop = 1}^n \paren {\frac {\partial S} {\partial y_i} - p_i} \Delta y_i = 0$ $\Delta x$ and $\Delta y_i$ are independent variables. The equation holds only if all the coefficients in front of $\Delta x$ and $\Delta y_i$ vanish simultaneously: :$\dfrac {\partial S} {\partial x} = -H, \quad \dfrac {\partial S} {\partial y_i} = p_i$ Since $H = \map H {x, \mathbf y, \mathbf p}$, using the second relation to replace $\mathbf p$ together with the first one proves the formula. {{finish|the proof makes sense, but very technical aspects are missing}} {{NamedforDef|William Rowan Hamilton|name2 = Carl Gustav Jacob Jacobi|cat = Hamilton|cat2 = Jacobi}}
Consider the total derivative of $\dfrac {\partial S} {\partial \alpha_i}$ {{WRT|Differentiation}} $x$: {{begin-eqn}} {{eqn | l = \frac \d {\d x} \frac {\partial S} {\partial \alpha_i}       | r = \frac {\partial^2 S} {\partial x \partial\alpha_i} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial \alpha_i} \frac {\d y_j} {\d x} + \sum_{j \mathop = 1}^m + \frac {\partial^2 S} {\partial \alpha_j \partial \alpha_i} \frac {\d \alpha_j} {\d x}       | c = Total derivative of $S$ {{WRT|Differentiation}} $x$ }} {{eqn | r = \frac {\partial^2 S} {\partial x \partial \alpha_i} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial \alpha_i} \frac {\d y_j} {\d x}       | c = $ \alpha_i$ is parameter, independent of $x$ }} {{eqn | r = -\frac {\partial H} {\partial \alpha_i} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial \alpha_i} \frac {\d y_j} {\d x}       | c = $S$ satisfies Hamilton-Jacobi equation }} {{eqn | r = -\frac {\partial x} {\partial \alpha_i} \frac {\partial H} {\partial x} - \sum_{j \mathop = 1}^n \frac {\partial y_j} {\partial \alpha_i} \frac {\partial H} {\partial y_j} - \sum_{j \mathop = 1}^n \frac {\partial p_j} {\partial \alpha_i} \frac {\partial H} {\partial p_j} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial \alpha_i} \frac {\d y_j} {\d x}       | c = Partial derivative of multivariate composite function $\map H {x, \mathbf y, \mathbf p}$ }} {{eqn | r = -\sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial \alpha_i} \frac {\partial H} {\partial p_j} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial \alpha_i} \frac {\partial y_j} {\partial x}           | c = $x$, $y_j$ independent of $\alpha_i$; $S$ satisfies Hamilton-Jacobi equation, thus $p_j = \dfrac {\partial S} {\partial y_j}$ }} {{eqn | r = \sum_{j \mathop = 1}^n \map {\frac {\partial^2 S} {\partial y_j \partial \alpha_i} } {\frac {\d y_j} {\d x} - \frac {\partial H} {\partial p_j} }     }} {{end-eqn}} If Euler's equations are satisfied, the {{RHS}} vanishes. Hence :$\dfrac \d {\d x} \dfrac {\partial S} {\partial \alpha_i} = 0$ or: :$\dfrac {\partial S} {\partial \alpha_i} = C_i$ where $C_i$ is a constant. {{qed}}
Consider canonical Euler's equations: :$\dfrac {\d y_i} {\d x} = \dfrac {\partial H} {\partial p_i}$ :$\dfrac {\d p_i} {\d x} = -\dfrac {\partial H} {\partial y_i}$ Apply a canonical transformation: :$\tuple {x, \mathbf y, \mathbf p, H} \to \tuple {x, \boldsymbol \alpha, \boldsymbol \beta, H^*}$ where $\Phi = S$. By Conditions for Transformation to be Canonical: :$p_i = \dfrac {\partial S} {\partial y_i}$ :$\beta_i = \dfrac {\partial S} {\partial \alpha_i}$ :$H^* = H + \dfrac {\partial S} {\partial x}$ Because $S$ is a solution to the Hamilton-Jacobi equation: :$H^* = 0$ In these new coordinates canonical Euler's equations are: :$\dfrac {\d\alpha_i} {\d x} = \dfrac {\partial H^*} {\partial \beta_i}$ :$\dfrac {\d \beta_i} {\d x} = -\dfrac {\partial H^*} {\partial \alpha_i}$ By $H^* = 0$: :$\dfrac {\d \alpha_i} {\d x} = 0$ :$\dfrac {\d\beta_i} {\d x} = 0$ which imply that $ \alpha_i$ and $\beta_i$ are constant along each extremal. {{explain|the meaning of "extremal"}} $\beta_i$ constancy provides with $n$ first integrals: :$\dfrac {\partial S} {\partial \alpha_i} = \beta_i$ Because $S = \map S {x, \mathbf y, \boldsymbol \alpha}$, the aforementioned set of first integrals is also a system of equations for functions $y_i$. Thus, functions $y_i$ can be found. Functions $p_i$ are found by the results of Conditions for Transformation to be Canonical: :$p_i = \dfrac {\partial} {\partial y_i} \map S {x, \mathbf y, \boldsymbol \alpha}$ Then: :$\map {y_i} {x, \boldsymbol \alpha, \boldsymbol \beta}$ :$\map {p_i} {x, \boldsymbol \alpha, \boldsymbol \beta}$ are solutions to canonical Euler's equations. {{qed}}
=== Sufficient Condition === Assume that :$\exists x \in I: x$ is upper bound for $I$ We will prove that :$I \subseteq x^\preceq$ Let $y \in I$. By definition of upper bound: :$y \preceq x$ Thus by definition of lower closure of element: :$y \in x^\preceq$ {{qed|lemma}} We will prove that :$x^\preceq \subseteq I$ Let $y \in x^\preceq$. By definition of lower closure of element: :$y \preceq x$ Thus by definition of lower set: :$y \in I$ {{qed|lemma}} Thus by definition of set equality: :$\exists x \in S: I = x^\preceq$ {{qed|lemma}} === Necessary Condition === Assume that :$\exists x \in S: I = x^\preceq$ By definition of reflexivity: :$x \preceq x$ Thus by definition of lower closure of element: :$x \in I$ Let $y \in I$. Thus by definition of lower closure of element: :$y \preceq x$ {{qed}}
First note that $3$ is not so expressible: {{begin-eqn}} {{eqn | l = 3 - 2 \times 1^2       | r = 1       | c = which is not prime }} {{end-eqn}} and so $3$ cannot be written in such a form. Then we have: {{begin-eqn}} {{eqn | l = 5       | r = 2 \times 1^2 + 3       | c =  }} {{eqn | l = 7       | r = 2 \times 1^2 + 5       | c =  }} {{eqn | l = 9       | r = 2 \times 1^2 + 7       | c =  }} {{eqn | l = 11       | r = 2 \times 2^2 + 3       | c =  }} {{eqn | l = 13       | r = 2 \times 1^2 + 11       | c =  }} {{eqn | r = 2 \times 2^2 + 5       | c =  }} {{eqn | l = 15       | r = 2 \times 1^2 + 13       | c =  }} {{eqn | r = 2 \times 2^2 + 7       | c =  }} {{end-eqn}} However, we have: {{begin-eqn}} {{eqn | l = 17 - 2 \times 1^2       | r = 15       | c = which is not prime }} {{eqn | l = 17 - 2 \times 2^2       | r = 9       | c = which is not prime }} {{end-eqn}} {{qed}}
By Ideals are Continuous Lattice Subframe of Power Set: :$P$ is continuous lattice subframe of $\left({\mathcal P\left({S}\right), \subseteq'}\right)$ where  :$\mathcal P\left({S}\right)$ denotes the power set of $S$, :$\mathord\subseteq' = \mathord\subseteq \cap \left({\mathcal P\left({S}\right) \times \mathcal P\left({S}\right)}\right)$ === Sufficient Condition === Assume that :$x$ is compact element. By Compact Element iff Existence of Finite Subset that Element equals Intersection and Includes Subset: :$\exists F \in \mathit{Fin}\left({S}\right): x = \bigcap \left\{ {I \in \mathit{Ids}\left({L}\right): F \subseteq I}\right\} \land F \subseteq x$ where $\mathit{Fin}\left({S}\right)$ denotes the set of all finite subsets of $S$. We will prove that :$\exists y \in x: y$ is upper bound for $x$. Define $y = \sup_L F$. By Directed in Join Semilattice with Finite Suprema: :$F \ne \varnothing \implies y \in x$ By Supremum of Empty Set is Smallest Element: :$F = \varnothing \implies y = \bot_L$ where $\bot_L$ is the smallest element in $L$. By Bottom in Ideal: :$F = \varnothing \implies y \in x$ Thus $y \in x$ Let $z \in x$. We will prove that :$F \subseteq y^\preceq$ Let $u \in F$. By definitions of supremum and upper bound: :$u \preceq y$ Thus by definition of lower closure of element: :$u \in y^\preceq$ {{qed|lemma}} By Lower Closure of Element is Ideal: :$y^\preceq$ is ideal in $L$. Then :$y^\preceq \in \left\{ {I \in \mathit{Ids}\left({L}\right): F \subseteq I}\right\}$ By definition of intersection: :$z \in y^\preceq$ Thus by definition of lower closure of element: :$z \preceq y$ {{qed|lemma}} Hence $x$ is principal ideal. {{qed|lemma}} === Necessary Condition === Assume that $x$ is principal ideal. By definition of principal ideal: :$\exists y \in x: y$ is upper bound for $x$. We will prove that :$\exists F \in \mathit{Fin}\left({S}\right): F \subseteq x \land x = \bigcap \left\{ {I \in \mathit{Ids}\left({L}\right): F \subseteq I}\right\}$ Define $F = \left\{ {y}\right\}$ By Singleton is Finite: :$F$ is finite. Thus by definition of $\mathit{Fin}$: :$F \in \mathit{Fin}\left({S}\right)$ Thus by definitions of subset and singleton: :$F \subseteq x$ We will prove that :$\forall z: z \in x \iff z \in \bigcap \left\{ {I \in \mathit{Ids}\left({L}\right): F \subseteq I}\right\}$ Thus by definition of set equality: :$x = \bigcap \left\{ {I \in \mathit{Ids}\left({L}\right): F \subseteq I}\right\}$ {{qed|lemma}} Thus by Compact Element iff Existence of Finite Subset that Element equals Intersection and Includes Subset: :$x$ is compact element. {{qed}}
=== Case $1$: $n$ is even === There is some integer $k > 2$ such that one of the following holds: :$n = 6 k = 2 + 3 + \paren {6 k - 5}$ :$n = 6 k + 2 = 4 + 3 + \paren {6 k - 7}$ :$n = 6 k + 4 = 2 + 3 + \paren {6 k - 5}$ All terms are greater than $1$. The first two terms are powers of $2$ and $3$, so they are coprime. The last term is not divisible by $2$ or $3$, so the three numbers are pairwise coprime. {{qed|lemma}} === Case $2$: $n$ is odd === There is some integer $k \ge 1$ such that one of the following holds: :$n = 12 k + 1 = 3 + \paren {6 k - 7} + \paren {6 k + 5}$ and $k \ge 2$ :$n = 12 k + 3 = 9 + \paren {6 k - 5} + \paren {6 k - 1}$ and $k \ge 2$ :$n = 12 k + 5 = 3 + \paren {6 k - 5} + \paren {6 k + 7}$ and $k \ge 2$ :$n = 12 k + 7 = 3 + \paren {6 k - 1} + \paren {6 k + 5}$ :$n = 12 k + 9 = 3 + \paren {6 k - 1} + \paren {6 k + 7}$ :$n = 12 k + 11 = 3 + \paren {6 k + 1} + \paren {6 k + 7}$ All terms are greater than $1$. The first term is a power of $3$. The last two terms are not divisible by $3$, so they are coprime with $3$. {{AimForCont}} the last two terms are not coprime. Then they must have a common divisor greater than $1$. This common divisor must also divide their difference. However their difference is only divisible by $2$ and $3$, and the terms themselves are not. Hence they cannot have a common divisor greater than $1$. So they must be coprime. Hence the result. {{qed}}
This is of course rubbish, because $p^p - 1$ and $q^q - 1$ have the obvious common factor $2$. Investigation ongoing, as it appears the author of the work this was plundered from must have been thinking about the Feit-Thompson Conjecture and got it badly wrong.
By Power of Base minus 1 is Repdigit Base minus 1, $n$ is a repdigit number consisting of $k$ occurrences of $b - 1$. Let $a = b - 1$. Thus $n$ can be expressed in base $b$ as: :$n = {\overbrace {\left[{aaa \cdots a}\right]}^k}_b$ We have that: {{begin-eqn}} {{eqn | l = n + n       | r = 2 n       | c =  }} {{eqn | r = \left[{1aa \cdots ac}\right]_b       | c = Multiple of Repdigit Base minus 1: $k - 1$ occurrences of $a$, and $c = b - 2$ }} {{end-eqn}} where there are: :$k - 1$ occurrences of $a$ :$c = b - 2$. Then: {{begin-eqn}} {{eqn | l = \left({b - 1}\right) \times n       | r = a n       | c =  }} {{eqn | r = \left[{caa \cdots a1}\right]_b       | c = Multiple of Repdigit Base minus 1 }} {{end-eqn}} where there are: :$k - 1$ occurrences of $a$ :$c = b - 2$. Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \dfrac {b^k - 1} {b - 1}       | r = \sum_{j \mathop = 0}^{k - 1} b^j       | c = Sum of Geometric Sequence }} {{eqn | ll= \leadsto       | l = n = b^k - 1       | r = \paren {b - 1} \sum_{j \mathop = 0}^{k - 1} b^j       | c =  }} {{eqn | r = \sum_{j \mathop = 0}^{k - 1} \paren {b - 1} b^j       | c =  }} {{end-eqn}} Thus, by the definition of base $b$ representation, $n$ consists of $k$ occurrences of the digit $b - 1$. {{qed}} Category:Repdigit Numbers Category:Number Bases hlz3zhxqaz2mksqgr4ujeit2i9k9wdu
Let $n$ be equal to twice the sum of its digits. Suppose $n$ has $1$ digit. Then $n$ equals the sum of its digits. Thus $n$ does not have only $1$ digit. Suppose $n$ has $3$ digits. Then $n > 99$. But the highest number that can be formed by the sum of $3$ digits is $9 + 9 + 9 = 27$. Similarly for if $n$ has more than $3$ digits. Hence, if there exists such a number, it must have exactly $2$ digits. Let the $2$ digits of $n$ be $x$ and $y$. Thus: {{begin-eqn}} {{eqn | l = n       | r = 10 x + y       | c =  }} {{eqn | r = 2 \left({x + y}\right)       | c =  }} {{eqn | ll= \leadsto       | l = 8 x       | r = y       | c =  }} {{end-eqn}} Since $x$ and $y$ are both single digits, it follows that $x = 1$ and $y = 8$. Hence the result. {{qed}}
Performing the calculation using long division: <pre>    0.05263157894736842105...   ------------------------ 19)1.00000000000000000000000      95    152     76       --    ---     --       50    180     40       38    171     38       --    ---     --       120     90     20       114     76     19       ---     --     --         60    140     100         57    133      95         --    ---     ---          30     70    ...          19     57          --     --          110    130           95    114           --    ---           150    160           133    152            --    ---            170     80            152     76 </pre> {{qed}}
{{AimForCont}} that $\map {\operatorname {diam} } S \ne \map {\operatorname {diam} } {S^-}$. $S \subseteq S^-$ by Subset of Metric Space is Subset of its Closure, so it then follows that: :$\map {\operatorname {diam} } S < \map {\operatorname {diam} } {S^-}$ Then there exists $x, y \in S^-$ such that $\map d {x, y} > \map {\operatorname {diam} } S$. By Point in Closure of Subset of Metric Space iff Limit of Sequence there exists sequences $\sequence {x_n}$ and $\sequence {y_n}$ that converge to $x$ and $y$ respectively. Let $d_\infty: \paren {A \times A} \times \paren {A \times A} \to \R$ be the metric on $A \times A$ defined as: :$\map {d_\infty} {\tuple {x, y}, \tuple {x', y'} } = \max \set {\map d {x, x'}, \map d {y, y'} }$ From this it can be said that $\sequence {\tuple {x_n, y_n} }$ is a sequence of points in $A \times A$ that converge to $\tuple {x, y}$. Let $\epsilon = \map d {x, y} - \map {\operatorname {diam} } S$. From Distance Function of Metric Space is Continuous, there exists a $N \in \N_{> 0}$ such that: :$n \ge N \implies \size {\map d {x_n, y_n} - \map d {x, y} } < \epsilon$ If $\map d {x_n, y_n} - \map d {x, y}$ is non-negative, then: :$\map d {x_n, y_n} \ge \map d {x, y} > \map {\operatorname {diam} } S$ which is a contradiction. If $\map d {x_n, y_n} - \map d {x, y}$ is a negative number, then: {{begin-eqn}} {{eqn | l = \size {\map d {x_n, y_n} - \map d {x, y} }       | r = -\map d {x_n, y_n} + \map d {x, y} }} {{eqn | o = <       | r = \map d {x, y} - \map {\operatorname {diam} } S }} {{eqn | ll= \leadsto       | l = -\map d {x_n, y_n}       | o = <       | r = -\map {\operatorname {diam} } S }} {{eqn | ll= \leadsto       | l = \map d {x_n, y_n}       | o = >       | r = \map {\operatorname {diam} } S }} {{end-eqn}} which is again a contradiction. From Proof by Contradiction it follows that $\map {\operatorname {diam} } S = \map {\operatorname {diam} } {S^-}$. Hence the result. {{qed}}
We first prove that only order $3$ magic hexagons exist apart from the order $1$ magic hexagon, by calculating the magic constant for each order. If we ignore the center tile, the rest of the hexagon can be divided into $3$ parallelogram-like parts: For example, the above can be divided into $\set {3, 17, 18, 19, 7, 1}$, $\set {11, 6, 8, 9, 14, 15}$ and $\set {2, 4, 13, 16, 12, 10}$. Each part has $n \paren {n - 1}$ tiles. Thus the order $n$ hexagon has $3 n \paren {n - 1} + 1$ tiles. Since we require each row to have the same sum, and there are $2 n - 1$ rows: :The sum of numbers in the hexagon $= \paren {2 n - 1} \times$ magic constant By Sum of Arithmetic Sequence, the sum of numbers in the tiles is: :$\dfrac {\paren {3 n \paren {n - 1} + 1} \paren {3 n \paren {n - 1} + 2}} 2$ Hence the magic constant is: {{begin-eqn}} {{eqn | o =       | r = \frac {\paren {3 n \paren {n - 1} + 1} \paren {3 n \paren {n - 1} + 2} } {2 \paren {2 n - 1} } }} {{eqn | r = \frac {9 n^4 - 18 n^3 + 18 n^2 - 9 n + 2} {2 \paren {2 n - 1} } }} {{end-eqn}} Since the magic constant is a sum of integers, it must be an integer. Then: :$32 \paren {\dfrac {9 n^4 - 18 n^3 + 18 n^2 - 9 n + 2} {2 \paren {2 n - 1}}} = \dfrac {144 n^4 - 288 n^3 + 288 n^2 - 144 n + 32} {2 n - 1}$ is also an integer. Since: :$\dfrac {144 n^4 - 288 n^3 + 288 n^2 - 144 n + 32} {2 n - 1} = 72 n^3 - 108 n^2 + 90 n - 27 + \dfrac 5 {2 n - 1}$ We must require $\dfrac 5 {2 n - 1}$ to be an integer. This happens only if $n = 1$ or $n = 3$. Thus there are no magic hexagons that are not of order $1$ or $3$. It remains to show that the order $3$ magic hexagon is unique. {{ProofWanted}}