hcju/proofwikiqa huggingface dataset:
The result follows from Vector Space has Basis between Linearly Independent Set and Spanning Set. It suffices to find a linearly independent subset $L \subseteq V$ that is contained in a spanning set $S \subseteq V$. By Empty Set is Linearly Independent, $L$ can be taken to be the empty set. Or if $V$ is nonzero, by Singleton is Linearly Independent, $L$ can be taken to be any singleton of $V$. $S$ can be taken to be $V$, since $V$ trivially spans itself. Therefore, $L$ and $S$ exist and $L \subseteq S$ so $V$ has a basis $B$ with $L \subseteq B \subseteq S$. {{qed}}
Let $m\in M$. Let $[m]_{\mathcal A}$ be its coordinate vector relative to $\mathcal A$, and similary for $\mathcal B$ and $\mathcal C$. On the one hand: {{begin-eqn}} {{eqn | l = [m]_{\mathcal A}       | r = \mathbf M_{\mathcal A,\mathcal C} \cdot [m]_{\mathcal C}       | c = Change of Coordinate Vector Under Change of Basis }} {{end-eqn}} On the other hand: {{begin-eqn}} {{eqn | l = [m]_{\mathcal A}       | r = \mathbf M_{\mathcal A,\mathcal B} \cdot [m]_{\mathcal B}       | c = Change of Coordinate Vector Under Change of Basis }} {{eqn | l =        | r = \mathbf M_{\mathcal A,\mathcal B} \cdot \mathbf M_{\mathcal B,\mathcal C}  \cdot [m]_{\mathcal C}       | c = Change of Coordinate Vector Under Change of Basis }} {{end-eqn}} Thus $(\mathbf M_{\mathcal A,\mathcal C} - \mathbf M_{\mathcal A,\mathcal B} \cdot \mathbf M_{\mathcal B,\mathcal C}) \cdot [m]_{\mathcal C} = 0$ for all $m\in M$. Because $m$ is arbitrary, the result follows. {{explain|find a link for this}} {{qed}}
Let $m\in M$, and $[m]_{\mathcal A}$ be its coordinate vector with respect to $\mathcal A$. On the one hand: {{begin-eqn}} {{eqn | l = [g(f(m))]_{\mathcal C}       | r = \mathbf M_{(g\mathop\circ f), \mathcal C, \mathcal A} \cdot [m]_{\mathcal A}       | c = Change of Coordinate Vectors Under Linear Mapping applied to $g\circ f$ }} {{end-eqn}} On the other hand: {{begin-eqn}} {{eqn | l = [g(f(m))]_{\mathcal C}       | r = \mathbf M_{g, \mathcal C, \mathcal B} \cdot [f(m)]_{\mathcal B}       | c = Change of Coordinate Vectors Under Linear Mapping applied to $g$ }} {{eqn | l =        | r = \mathbf M_{g, \mathcal C, \mathcal B} \cdot \mathbf M_{f, \mathcal B, \mathcal A} \cdot [m]_{\mathcal A}       | c = Change of Coordinate Vectors Under Linear Mapping applied to $f$ }} {{end-eqn}} Thus $(\mathbf M_{(g\mathop\circ f), \mathcal C, \mathcal A} - \mathbf M_{g, \mathcal C, \mathcal B} \cdot \mathbf M_{f, \mathcal B, \mathcal A}) \cdot [m]_{\mathcal A} = 0$ for all $m\in M$. The result follows. {{explain|find a link why}} {{qed}}
Let $m\in M$, and let $[m]_{\mathcal A}$ and $[m]_{\mathcal B}$ denote its coordinate vectors relative to $\mathcal A$ and $\mathcal B$. We have: {{begin-eqn}} {{eqn | l = f(m)       | r = [m]_{\mathcal A}^\intercal \cdot \mathbf M_{f, \mathcal A} \cdot [m]_{\mathcal A}       | c =  }} {{eqn | l =        | r = (\mathbf M_{\mathcal A, \mathcal B} \cdot [m]_{\mathcal B})^\intercal \cdot \mathbf M_{f, \mathcal A} \cdot (\mathbf M_{\mathcal A, \mathcal B} \cdot [m]_{\mathcal B})       | c = Change of Coordinate Vector Under Change of Basis }} {{eqn | l =        | r = [m]_{\mathcal B}^\intercal \cdot \mathbf M_{\mathcal A, \mathcal B}^\intercal \cdot \mathbf M_{f, \mathcal A} \cdot \mathbf M_{\mathcal A, \mathcal B} \cdot [m]_{\mathcal B}       | c = Transpose of Matrix Product }} {{end-eqn}} Thus $\mathbf M_{f, \mathcal B} = \mathbf M_{\mathcal A, \mathcal B}^\intercal \mathbf M_{f, \mathcal A} \mathbf M_{\mathcal A, \mathcal B}$. {{explain|find a link for this and the first equality above, or make it an alternative definition}} {{qed}} Category:Bilinear Forms Category:Change of Basis gsen99119fz3pxac3jib09iq7jern0k
Let $\left({v, w}\right) \in V \times V$ with $b \left({v, w}\right) = 0$. Because $b$ is symmetric, $b \left({w, v}\right) = 0$. Because $\left({v, w}\right)$ was arbitrary, $b$ is reflexive. {{qed}}
Let $\left({v, w}\right) \in V \times V$ with $b \left({v, w}\right) = 0$. We have: {{begin-eqn}} {{eqn | l = 0       | r = b \left({v + w, v + w}\right)       | c = $b$ is alternating }} {{eqn | r = b \left({v, v}\right) + b \left({v, w}\right) + b \left({w, v}\right) + b \left({w, w}\right)       | c = {{Defof|Bilinear Form}} }} {{eqn | r = b \left({v, w}\right) + b \left({w, v}\right)       | c = $b$ is alternating }} {{eqn | r = b \left({w, v}\right)       | c = $b \left({v, w}\right) = 0$ }} {{end-eqn}} Because $\left({v, w}\right)$ was arbitrary, $b$ is reflexive. {{qed}}
=== 1 implies 2 === Follows from Reflexive Bilinear Form is Symmetric or Alternating {{qed}} === 2 implies 1 === Follows from: :Symmetric Bilinear Form is Reflexive :Alternating Bilinear Form is Reflexive {{qed}} Category:Bilinear Forms gcjjc8czzhhl5k20imdbg71du2p651d
Let $x, y, z \in V$. Because $f$ is bilinear: :$\map f {x, \map f {x, y} z - \map f {x, z} y} = \map f {x, y} \, \map f {x, z} - \map f {x, z} \, \map f {x, y} = 0$ Because $f$ is reflexive: :$\map f {\map f {x, y} z - \map f {x, z} y, x} = 0$ Because $f$ is bilinear: :$(1): \quad \map f {x, y} \, \map f {z, x} = \map f {x, z} \, \map f {y, x}$ Letting $z = x$, we obtain: :$\map f {x, x} \, \map f {x, y} = \map f {x, x} \, \map f {y, x}$ Thus: :$(2): \quad$ if $\map f {x, x} \ne 0$, then $\map f {x, y} = \map f {y, x}$ for all $y \in V$. Suppose $f$ is not symmetric. We show that $\map f {u, u} = 0$ for all $u \in V$. Let $v, w \in V$ with $\map f {v, w} \ne \map f {w, v}$. By $(2)$, $\map f {v, v} = \map f {w, w} = 0$. Suppose $u \in V$ with $\map f {u, u} \ne 0$. By $(2)$, $\map f {u, v} = \map f {v, u}$ and $\map f {u, w} = \map f {w, u}$. By $(1)$, we have: :$\begin{cases} \map f {v, w} \, \map f {u, v} = \map f {v, u} \, \map f {w, v} \\ \map f {w, v} \, \map f {u, w} = \map f {w, u} \, \map f {v, w} \end{cases}$. Because $\map f {v, w} \ne \map f {w, v}$: :$\begin{cases} \map f {u, v} = \map f {v, u} = 0 \\ \map f {u, w} = \map f {w, u} = 0 \end{cases}$. Because $f$ is bilinear, $\map f {u + v, w} = \map f {v, w} \ne \map f {v, w} = \map f {u + v, w}$. By $(2)$, $\map f {u + v, u + v} = 0$. Because $f$ is bilinear and $\map f {u, v} = \map f {v, u} = \map f {v, v} = 0$, $\map f {u, u} = 0$. This contradicts the assumption $\map f {u, u} \ne 0$. Thus $f$ is alternating. {{qed}}
From Reciprocal of 61: :$\dfrac 1 {61} = 0 \cdotp \dot 01639 \, 34426 \, 22950 \, 81967 \, 21311 \, 47540 \, 98360 \, 65573 \, 77049 \, 18032 \, 78688 \, 5245 \dot 9$ From Reciprocal of 131: :$\dfrac 1 {131} = 0 \cdotp \dot 00763 \, 35877 \, 86259 \, 54198 \, 47328 \, 24427 \, 48091 \, 60305 \, 34351 \, 14503 \, 81679 \, 38931 \, 29770 \, 99236 \, 64122 \, 13740 \, 45801 \, 52671 \, 75572 \, 51908 \, 39694 \, 65648 \, 85496 \, 18320 \, 61068 \, 7022 \dot 9$
{{begin-eqn}} {{eqn | l = \sigma \left({61}\right)       | r = 62       | c = Sigma Function of Prime Number: $61$ is prime }} {{eqn | l = \sigma \left({62}\right)       | r = 96       | c = {{SigmaLink|62}} }} {{eqn | l = \sigma \left({63}\right)       | r = 104       | c = {{SigmaLink|63}} }} {{eqn | l = \sigma \left({64}\right)       | r = 127       | c = {{SigmaLink|64}} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \sigma \left({73}\right)       | r = 74       | c = Sigma Function of Prime Number: $73$ is prime }} {{eqn | l = \sigma \left({74}\right)       | r = 114       | c = {{SigmaLink|74}} }} {{eqn | l = \sigma \left({75}\right)       | r = 124       | c = {{SigmaLink|75}} }} {{eqn | l = \sigma \left({76}\right)       | r = 140       | c = {{SigmaLink|76}} }} {{end-eqn}} {{qed}}
We have that: :$n \times 9^n < 10^\left({n - 1}\right)$ when $n > 60$. So an $n$-digit integer, for $n > 60$, is always greater than the sum of the $n$th powers of its digits. {{qed}}
Because $S$ is a generating set, there is a surjective homomorphism :$\phi : R^{\paren S} \to M$ where $R^{\paren S}$ is the free $R$-module on $S$. Because $B$ is a basis, there is an isomorphism :$\psi : R^{\paren B} \to M$ Thus $f = \psi^{-1} \circ \phi: R^{\paren S} \to R^{\paren B}$ is a surjective module homomorphism. By Krull's Theorem, there exists a maximal ideal $M \subset R$. By Maximal Ideal iff Quotient Ring is Field, $R / M$ is a field. Let $k = R / M$. Let $\pi: R \to k$ denote the quotient mapping. By some universal properties, there exists a $k$-module homomorphism $\bar f: k^{\paren S} \to k^{\paren B}$ such that: :$\pi^B \circ f = \bar f \circ \pi^S$ Where $\pi^B$ denotes direct sum of module homomorphisms. Because $\pi^B \circ f$ is surjective, so is $\bar f$. Let: :$C$ be the canonical basis of $k^{\paren S}$ :$D$ be the canonical basis of $k^{\paren S}$ Because $C$ is a generator of $k^{\paren S}$ and $\bar f$ is surjective, $\map {\bar f} C$ is a generator of $k^{\paren B}$. We have: {{begin-eqn}} {{eqn | l = \size B       | r = \size D       | c = Cardinality of Canonical Basis of Free Module on Set }} {{eqn | o = \le       | r = \size {\map {\bar f} C}       | c = Basis of Vector Space Injects into Generator }} {{eqn | o = \le       | r = \size C       | c = Cardinality of Image of Mapping not greater than Cardinality of Domain }} {{eqn | r = \size S       | c = Cardinality of Canonical Basis of Free Module on Set }} {{end-eqn}} {{qed}} {{explain|find those universal properties}}
Since $f$ is differentiable on $I$, f is differentiable on $\left[{x_i \,.\,.\, x_{i+1}}\right]$ for $i = 0, \ldots, n-1$. Thus a fortiori, $f$ is also continuous on the closed interval $\left[{x_i \,.\,.\, x_{i+1}}\right]$ and differentiable on the open interval $\left({x_i \,.\,.\, x_{i+1}}\right)$. For $i = 0, \ldots, n$ we have $f \left({x_i}\right) = 0$. Hence the conditions of Rolle's Theorem are fulfilled on each interval which yields the result.
From Continued Fraction Expansion of $\sqrt {61}$: :$\sqrt {61} = \left[{7, \left\langle{1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14}\right\rangle}\right]$ The cycle is of length is $11$. By the solution of Pell's Equation, the only solutions of $x^2 - 61 y^2 = 1$ are: :${p_{11 r} }^2 - 61 {q_{11 r} }^2 = \left({-1}\right)^{11 r}$ for $r = 1, 2, 3, \ldots$ When $r = 1$ this gives: :${p_{11}}^2 - 61 {q_{11}}^2 = -1$ which is not the solution required. When $r = 2$ this gives: :${p_{22} }^2 - 61 {q_{22} }^2 = 1$ From Convergents of Continued Fraction Expansion of $\sqrt {61}$: :$p_{22} = 1 \, 766 \, 319 \, 049$ :$q_{22} = 226 \, 153 \, 980$ although on that page the numbering goes from $p_0$ to $p_{21}$, and $q_0$ to $q_{21}$. {{qed}}
Aiming for contradiction, suppose that there is a sequence like that. From the definition of a sequence, let $f$ be the the mapping that is defined by $\left\langle{x_n}\right\rangle$. Let $\omega$ denote the minimal infinite successor set. Let $g: \omega \to \N_{\gt 0}$ be defined as: :$g \left({\alpha}\right) = \alpha + 1$ Then the composition $f \circ g$ is a mapping whose domain is $\omega$ such that: :$\forall n \in \omega: \left({f \circ g}\right) \left({n^+}\right) \in \left({f \circ g}\right) \left({n}\right)$ But this contradicts No Infinitely Descending Membership Chains. Therefore by contradiction there cannot exist such a sequence. Hence the result. {{qed}} Category:Axiom of Foundation al2be08tum7r0r26t5do2vw6aes462f
Let $n \in \Z_{>0}$ be a $2$-digit positive integer. {{WLOG}}, let $n$ be expressed in decimal notation as: :$n = 10 x + y$ where: : $x > y$ :$0 \le x \le 9, 0 \le y \le 9$ The reversal of $n$ is $10 y + x$. We have: {{begin-eqn}} {{eqn | l = 10x + y - 10 y - x       | r = 9 \left({x - y}\right)       | c = where $1 \le x - y \le 9$ }} {{end-eqn}} Thus after the first iteration of Kaprekar's process, $n$ is one of: :$09$ :$18$ :$27$ :$36$ :$45$ or their reversals. Applying Kaprekar's process to $09$ gives: {{begin-eqn}} {{eqn | l = 09       | r = 90 - 09       | c = }} {{eqn | r = 81       | c = }} {{eqn | o = \to       | r = 81 - 18       | c = }} {{eqn | r = 63       | c = }} {{eqn | o = \to       | r = 63 - 36       | c = }} {{eqn | r = 27       | c = }} {{eqn | o = \to       | r = 72 - 27       | c = }} {{eqn | r = 45       | c = }} {{eqn | o = \to       | r = 45 - 54       | c = }} {{eqn | r = 09       | c = }} {{end-eqn}} All those multiples of $9$ can be seen to be in (or end up in) this loop. Hence the result. {{qed}}
From Closed Form for Centered Hexagonal Numbers: :$H_n = 3 n \paren {n - 1} + 1$ Hence: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n H_i       | r = \sum_{i \mathop = 1}^n \paren {3 i \paren {i - 1} + 1}       | c =  }} {{eqn | r = \sum_{i \mathop = 1}^n \paren {3 i^2 - 3 i + 1}       | c =  }} {{eqn | r = 3 \sum_{i \mathop = 1}^n i^2  - 3 \sum_{i \mathop = 1}^n i + \sum_{i \mathop = 1}^n 1       | c =  }} {{eqn | r = 3 \frac {n \paren {n + 1} \paren {2 n + 1} } 6 - 3 \sum_{i \mathop = 1}^n i + \sum_{i \mathop = 1}^n 1       | c = Sum of Sequence of Squares }} {{eqn | r = 3 \frac {n \paren {n + 1} \paren {2 n + 1} } 6 - 3 \frac {n \paren {n + 1} } 2 + \sum_{i \mathop = 1}^n 1       | c = Closed Form for Triangular Numbers }} {{eqn | r = \frac {n \paren {\paren {n + 1} \paren {2 n + 1} - 3 \paren {n + 1} + 2} } 2       | c = simplification }} {{eqn | r = \frac {n \paren {2 n^2 + 3 n + 1 - 3 n - 3 + 2} } 2       | c = multiplying out }} {{eqn | r = \frac {n \paren {2 n^2} } 2       | c = simplification }} {{eqn | r = n^3       | c = simplification }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = p^2       | o = \equiv       | r = 0       | rr = \pmod {p^2} }} {{eqn | l = 1       | o = \equiv       | r = 1       | rr = \pmod {p^2} }} {{eqn | ll= \leadsto       | l = p^2 + 1       | o = \equiv       | r = 0 + 1       | rr= \pmod {p^2}       | c = Modulo Addition is Well-Defined }} {{eqn | o = \equiv       | r = 1       | rr = \pmod {p^2} }} {{eqn | ll= \leadsto       | l = \paren {p^2 + 1}^{p - 1}       | o = \equiv       | r = 1^{p - 1}       | rr= \pmod {p^2}       | c = Congruence of Powers }} {{eqn | o = \equiv       | r = 1       | rr= \pmod {p^2} }} {{end-eqn}} Hence $p^2 + 1$ fulfils the conditions for the value of $n$ whose existence was required to be demonstrated. {{qed}}
From Sum of 2 Squares in 2 Distinct Ways, the smallest $2$ positive integer which are the sum of $2$ square numbers in two distinct ways are $50$ and $65$. But $50$ cannot be expressed as the sum of $2$ cube numbers: {{begin-eqn}} {{eqn | l = 50 - 1^3       | r = 49       | c = which is not cubic }} {{eqn | l = 50 - 2^3       | r = 42       | c = which is not cubic }} {{eqn | l = 50 - 3^3       | r = 23       | c = which is not cubic }} {{eqn | l = 50 - 4^3       | r = -14       | c = and we have fallen off the end }} {{end-eqn}} Hence $65$ is that smallest number. {{qed}}
Let $M_5$ denote an order $5$ magic square By Sum of Terms of Magic Square, the total of all the entries in $M_5$ is given by: :$T_5 = \dfrac {5^2 \left({5^2 + 1}\right)} 2 = \dfrac {25 \times 26} 2 = 325$ As there are $5$ rows of $M_5$, the magic constant of $M_5$ is given by: :$S_5 = \dfrac {325} 5 = 65$ {{qed}}
Throughout we use Closed Form for Triangular Numbers, which gives that the $n$th triangular number can be expressed as: :$T_n = \dfrac {11 \times 12} 2$ We have: {{begin-eqn}} {{eqn | l = 66       | r = \frac {11 \times 12} 2       | c = and so is triangular }} {{eqn | l = 105       | r = \frac {14 \times 15} 2       | c = and so is triangular }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = 66 + 105       | r = 171       | c =  }} {{eqn | r = \frac {18 \times 19} 2       | c = and so is triangular }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 105 + 105       | r = 210       | c =  }} {{eqn | r = \frac {20 \times 21} 2       | c = and so is triangular }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 66 + 105 + 105       | r = 210       | c =  }} {{eqn | r = \frac {23 \times 24} 2       | c = and so is triangular }} {{end-eqn}} {{qed}}
=== Definition 1 is equivalent to Definition 2 === This follows immediately from the definition of a strict well-ordering. {{qed|lemma}} === Definition 1 implies Definition 3 === Let $S$ be an ordinal according to Definition 1. Let $a \in S$. Then: {{begin-eqn}} {{eqn | l = S_a       | r = \set {x \in S: x \in_S a}       | c = {{Defof|Initial Segment}} }} {{eqn | r = \set {x: x \in S \land x \in a} }} {{eqn | r = S \cap a       | c = {{Defof|Set Intersection}} }} {{eqn | r = a       | c = as $a \subseteq S$ by transitivity }} {{end-eqn}} {{qed|lemma}} === Definition 3 implies Definition 1 === Let $\struct {S, \prec}$ be an ordinal according to Definition 3. Let $a \in S$. Then $a = S_a \subseteq S$ and so $S$ is transitive. Also, by the definition of set equality: {{begin-eqn}} {{eqn | l = \forall x: x \in a       | o = \iff       | r = x \in S_a }} {{eqn | ll= \leadstoandfrom       | l = \forall x: x \in a       | o = \iff       | r = \paren {x \in S \land x \prec a} }} {{end-eqn}} It has been shown that if $x, a \in S$ then: :$x \in a \iff x \prec a$ Therefore, $\operatorname \prec = \struct {S, S, R}$ where: {{begin-eqn}} {{eqn | l = R       | r = \set {\tuple {x, a} \in S \times S: x \prec a} }} {{eqn | r = \set {\tuple {x, a} \in S \times S: x \in a} }} {{eqn | r = \in_S       | c = {{Defof|Epsilon Restriction}} }} {{end-eqn}} Hence $\operatorname \prec = \Epsilon {\restriction_S}$. {{qed}}
The sequence of odd positive composite integers begins: :$9, 15, 21, 25, 27, 33, 35, 39, 45, 49, 51, 55, 57, 63, 65, \ldots$ which is more than we need for this proof. {{OEIS|A071904}} Generating the possible sums of these proceeds as follows: {{begin-eqn}} {{eqn | l = 18       | r = 9 + 9       | c =  }} {{eqn | l = 24       | r = 9 + 15       | c =  }} {{eqn | l = 30       | r = 9 + 21       | c =  }} {{eqn | r = 15 + 15       | c =  }} {{eqn | l = 34       | r = 9 + 25       | c =  }} {{eqn | l = 36       | r = 9 + 27       | c =  }} {{eqn | r = 15 + 21       | c =  }} {{eqn | l = 40       | r = 15 + 25       | c =  }} {{eqn | l = 42       | r = 9 + 33       | c =  }} {{eqn | r = 15 + 27       | c =  }} {{eqn | r = 21 + 21       | c =  }} {{eqn | l = 44       | r = 9 + 35       | c =  }} {{eqn | l = 46       | r = 21 + 25       | c =  }} {{eqn | l = 48       | r = 9 + 39       | c =  }} {{eqn | r = 15 + 33       | c =  }} {{eqn | r = 21 + 27       | c =  }} {{eqn | l = 50       | r = 15 + 35       | c =  }} {{eqn | r = 25 + 25       | c =  }} {{eqn | l = 52       | r = 25 + 27       | c =  }} {{eqn | l = 54       | r = 9 + 45       | c =  }} {{eqn | r = 15 + 39       | c =  }} {{eqn | r = 21 + 33       | c =  }} {{eqn | r = 27 + 27       | c =  }} {{eqn | l = 56       | r = 21 + 35       | c =  }} {{eqn | l = 58       | r = 9 + 49       | c =  }} {{eqn | r = 25 + 33       | c =  }} {{eqn | l = 60       | r = 9 + 51       | c =  }} {{eqn | r = 15 + 45       | c =  }} {{eqn | r = 21 + 39       | c =  }} {{eqn | r = 25 + 35       | c =  }} {{eqn | r = 27 + 33       | c =  }} {{eqn | l = 62       | r = 27 + 35       | c =  }} {{eqn | l = 64       | r = 9 + 55       | c =  }} {{eqn | r = 15 + 49       | c =  }} {{eqn | r = 25 + 39       | c =  }} {{eqn | l = 66       | r = 9 + 57       | c =  }} {{eqn | r = 15 + 51       | c =  }} {{eqn | r = 21 + 45       | c =  }} {{eqn | r = 27 + 39       | c =  }} {{eqn | r = 33 + 33       | c =  }} {{eqn | l = 68       | r = 33 + 35       | c =  }} {{end-eqn}} This exhausts all possible sums of odd positive composite integers of $68$ or less. It remains to be demonstrated that all positive even integers over $68$ can be so expressed in $2$ different ways. Notice that one of $n - 15, n - 25, n - 35$ is divisible by $3$. Similarly one of $n - 9, n - 15, n - 21, n - 27, n - 33$ is divisible by $5$. For all above $n - k$ which is composite, since $k$ is composite, $\paren {n - k} + k$ is a way to express $n$ as the sum of $2$ odd positive composite integers. Since $n > 68$, these two ways are distinct unless $n - 15$ is divisible by $15$. In that case, $n - 45$ is also divisible by $15$, and $n - 45 \ne 15$. Then $n = \paren {n - 15} + 15 = \paren {n - 45} + 45$ are two ways to express $n$ as the sum of $2$ odd positive composite integers. {{qed}}
Let $n$ be an even integer greater than $18 k + 20$. Then $n - 9 \paren {2 k - 2}$ is an even integer greater than $18 k + 20 - 9 \paren {2 k - 2} = 38$. By Positive Even Integers not Expressible as Sum of 2 Composite Odd Numbers, every even integer greater than $38$ can be expressed as the sum of $2$ odd positive composite integers. Thus $n - 9 \paren {2 k - 2}$ can be expressed as the sum of $2$ odd positive composite integers. So, let $a$ and $b$ be odd composite integers such that $a + b = n - 9 \paren {2 k - 2}$. Then: :$9 \paren {2 k - 2} + a + b = n$ This is an expression for $n$ as the sum of $2 k$ odd positive composite integers, in which $2 k - 2$ of them are occurrences of $9$. Thus such an expression can always be found for $n > 18 k + 20$. {{qed|lemma}} It remains to be shown that $18 k + 20$ is not expressible as the sum of $2 k$ odd positive composite integers. The $k = 1$ case is demonstrated in Positive Even Integers not Expressible as Sum of 2 Composite Odd Numbers. Now suppose $k \ge 2$.  The two smallest odd positive composite integers are $9$ and $15$. Suppose $18 k + 20$ is expressible as the sum of $2 k$ odd positive composite integers. Then at least $2 k - 3$ of them are $9$'s, because: :$9 \paren {2 k - 4} + 4 \times 15 = 18 k + 24 > 18 k + 20$ Then the problem reduces to finding an expression of $18 k + 20 - 9 \paren {2 k - 3} = 47$ as the sum of $3$ odd positive composite integers. The first few odd positive composite integers are: :$9, 15, 21, 25, 27, 33, 35, 39, 45$ Their difference with $47$ are: :$38, 32, 26, 22, 20, 14, 12, 8, 2$ The integers above are in the set of integers not expressible as a sum of $2$ odd positive composite integers. The full set of these can be found in Positive Even Integers not Expressible as Sum of 2 Composite Odd Numbers. Thus $18 k + 20$ is not expressible as the sum of $2 k$ odd positive composite integers. This proves the result. {{qed}}
{{begin-eqn}} {{eqn | l = 78 - 72       | r = 6       | c =  }} {{eqn | l = 84 - 78       | r = 6       | c =  }} {{eqn | l = 90 - 84       | r = 6       | c =  }} {{end-eqn}} demonstrating that this is indeed an arithmetic sequence, with a common difference of $6$. Now we show: {{begin-eqn}} {{eqn | l = \map \phi {72}       | r = 24       | c = {{EulerPhiLink|72}} }} {{eqn | l = \map \phi {78}       | r = 24       | c = {{EulerPhiLink|78}} }} {{eqn | l = \map \phi {84}       | r = 24       | c = {{EulerPhiLink|84}} }} {{eqn | l = \map \phi {90}       | r = 24       | c = {{EulerPhiLink|90}} }} {{end-eqn}} {{qed|lemma}} {{ProofWanted|It remains to be shown that this is the smallest such arithmetic sequence of $4$ integers.}} {{begin-eqn}} {{eqn | l = 222 - 216       | r = 6       | c =  }} {{eqn | l = 228 - 222       | r = 6       | c =  }} {{eqn | l = 234 - 228       | r = 6       | c =  }} {{end-eqn}} demonstrating that this is indeed an arithmetic sequence, with a common difference of $6$. Now we show: {{begin-eqn}} {{eqn | l = \map \phi {216}       | r = 72       | c = {{EulerPhiLink|216}} }} {{eqn | l = \map \phi {222}       | r = 72       | c = {{EulerPhiLink|222}} }} {{eqn | l = \map \phi {228}       | r = 72       | c = {{EulerPhiLink|228}} }} {{eqn | l = \map \phi {234}       | r = 72       | c = {{EulerPhiLink|234}} }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | l = 76 \, 332 - 76 \, 326       | r = 6       | c =  }} {{eqn | l = 76 \, 338 - 76 \, 332       | r = 6       | c =  }} {{eqn | l = 76 \, 344 - 76 \, 338       | r = 6       | c =  }} {{end-eqn}} demonstrating that this is indeed an arithmetic sequence, with a common difference of $6$. Now we show: {{begin-eqn}} {{eqn | l = \map \phi {76 \, 326}       | r = 25 \, 440       | c = {{EulerPhiLink|76,326|76 \, 326}} }} {{eqn | l = \map \phi {76 \, 332}       | r = 25 \, 440       | c = {{EulerPhiLink|76,332|76 \, 332}} }} {{eqn | l = \map \phi {76 \, 338}       | r = 25 \, 440       | c = {{EulerPhiLink|76,338|76 \, 338}} }} {{eqn | l = \map \phi {76 \, 344}       | r = 25 \, 440       | c = {{EulerPhiLink|76,344|76 \, 344}} }} {{end-eqn}} {{qed}}
From Numbers with Euler Phi Value of 72, it is seen that $17$ positive integers have an Euler $\phi$ value of $72$: :$73, 91, 95, 111, 117, 135, 146, 148, 152, 182, 190, 216, 222, 228, 234, 252, 270$ {{proof wanted|boring}}