hcju/proofwikiqa huggingface dataset:
Performing the calculation using long division: <pre>   0.1428571  ---------- 7)1.0000000     7   ---     30     28     --      20      14      --       60       56       --        40        35        --         50         49         --          10           7          --          ..... </pre> The reciprocals of $1$, $2$, $4$ and $5$ do not recur: {{begin-eqn}} {{eqn | l = \frac 1 1       | r = 1       | c =  }} {{eqn | l = \frac 1 2       | r = 0 \cdotp 5       | c =  }} {{eqn | l = \frac 1 4       | r = 0 \cdotp 25       | c =  }} {{eqn | l = \frac 1 5       | r = 0 \cdotp 2       | c =  }} {{end-eqn}} while those of $3$ and $6$ do recur, but with the non-maximum period of $1$: {{begin-eqn}} {{eqn | l = \frac 1 3       | r = 0 \cdotp \dot 3       | c =  }} {{eqn | l = \frac 1 6       | r = 0 \cdotp 1 \dot 6       | c =  }} {{end-eqn}} {{qed}}
Demonstrated by inspection and calculation: {{begin-eqn}} {{eqn | l = \frac 1 1       | r = 1 \cdotp 0       | c = hence has a period of $0$ }} {{eqn | l = \frac 1 3       | r = 0 \cdotp \dot 3       | c =  }} {{eqn | l = \frac 1 {11}       | r = 0 \cdotp \dot 0 \dot 9       | c =  }} {{eqn | l = \frac 1 {27}       | r = 0 \cdotp \dot 03 \dot 7       | c = Period of Reciprocal of 27 is Smallest with Length 3 }} {{eqn | l = \frac 1 {101}       | r = 0 \cdotp \dot 009 \dot 9       | c =  }} {{eqn | l = \frac 1 {41}       | r = 0 \cdotp \dot 0243 \dot 9       | c =  }} {{eqn | l = \frac 1 7       | r = 0 \cdotp \dot 14285 \dot 7       | c = Period of Reciprocal of 7 is of Maximal Length }} {{eqn | l = \frac 1 {239}       | r = 0 \cdotp \dot 00418 \, 4 \dot 1       | c =  }} {{eqn | l = \frac 1 {73}       | r = 0 \cdotp \dot 01369 \, 86 \dot 3       | c =  }} {{eqn | l = \frac 1 {81}       | r = 0 \cdotp \dot 01234 \, 567 \dot 9       | c =  }} {{end-eqn}}
Let $m$ be a multiple of $n$ with $3$ digits. Then we have: {{begin-eqn}} {{eqn | l = n \times c       | r = a_2 \times 10^2 + a_1 \times 10^1 + a_0 }} {{end-eqn}} Let us now cyclically permute the digits of $m$ by multiplying by $10$. Then we have: {{begin-eqn}} {{eqn | l = 10 \times n \times c        | r = 10 \times \paren {a_2 \times 10^2 + a_1 \times 10^1 + a_0}       | c = multiply original number by $10$ }} {{eqn | r =  a_2 \times 10^3 + a_1 \times 10^2 + a_0 \times 10^1       | c =  }} {{eqn | r =  a_1 \times 10^2 + a_0 \times 10^1 + a_2 \times 10^0       | c = $10^3$ and $10^0 \equiv 1 \pmod {n}$ }} {{end-eqn}} From the above, we see that: :$n$ is a divisor of a cyclic permutation of $m$ {{iff}}: :$n \divides \paren {10^3 - 1 }$ We now note that: :$10^3 - 1 = 37 \times 27 = 37 \times 3^3$ Upon inspection, we see that the only $2$-digit factors are $27$ and $37$. {{qed}}
By Boundary of Set is Closed, $\partial K$ is closed in $X$. By Set is Closed iff it Contains its Boundary, $\partial K \subset K$. By Closed Set in Topological Subspace, $\partial K$ is closed in $K$. By Closed Subspace of Compact Space is Compact, $\partial K$ is compact. {{qed}}
By Subset of Real Numbers is Interval iff Connected, $\closedint a b$ is connected. {{AimForCont}} there is no fixed point. Then $\map f a > a$ and $\map f b < b$. Let: :$U = \set {x \in \closedint a b: \map f x > x}$ :$V = \set {x \in \closedint a b: \map f x < x}$ Then $U$ and $V$ are open in $\closedint a b$. Because $a \in U$ and $b\in V$, $U$ and $V$ are non-empty. By assumption: :$U \cup V = \closedint a b$ Thus $\closedint a b$ is not connected, which is a contradiction. Thus, by Proof by Contradiction, there exists at least one fixed point. {{qed}}
Let $g \in \overline H$ be in the closure of $H$. We will show that $g \in H$. {{AimForCont}} $g \notin H$. Let $e$ be the identity of $G$. Because $H$ is discrete, there exists an open set $U \subset G$ such that $U \cap H = \left\{{e}\right\}$. Then $V = U \cap U^{-1}$ is an open neighborhood of $e$ in $G$. By Right and Left Regular Representations in Topological Group are Homeomorphisms, $g V$ is a neighborhood of $g$ in $G$. By definition of closure, $g V \cap H \ne \varnothing$. Let $h \in g V \cap H$. Because $g \ne h$ and $G$ is Hausdorff, there exists a neighborhood $W$ of $g$ in $G$ with $h \notin W$. Then $g V \cap W$ is a neighborhood of $g$. Let $k \in g V \cap W \cap H$. Then $k \ne h$. Then {{begin-eqn}} {{eqn | l = k^{-1} h       | o = \in       | r = \left({g V \cap H}\right)^{-1} \cdot \left({g V \cap H}\right)       | c =  }} {{eqn | r = \left({g V}\right)^{-1}g V \cap H       | c =  }} {{eqn | r = V^{-1} V \cap H       | c =  }} {{eqn | r = V \cap H       | c =  }} {{eqn | o = \subset       | r =  U \cap H       | c =  }} {{end-eqn}} Bu assumption, $k^{-1} h = e$, so $k = h$, which is a contradiction. Thus, by Proof by Contradiction, $g \in H$. Thus $\overline H = H$. Thus $H$ is closed. {{qed}}
By Compact Subspace of Hausdorff Space is Closed, $K$ is closed in $X$. By Boundary of Compact Closed Set is Compact, $\partial K$ is compact. {{qed}} Category:Boundaries Category:Compact Spaces Category:Closed Sets Category:Hausdorff Spaces b4itq4dij2w350awtk2k5uws7xldzx0
Follows directly from: : Discrete Subgroup of Hausdorff Group is Closed : Topological Group is Hausdorff iff Identity is Closed {{qed}}
=== Necessary Condition === Follows directly from Topological Group is Hausdorff iff Identity is Closed. {{explain|How?}} {{qed|lemma}} === Sufficient Condition === Let $G$ be a Hausdorff topological group. Let $H \leq G$ be a closed discrete subgroup. Let $e$ be the identity of $G$. By Set in Discrete Topology is Clopen, $\left\{{e}\right\}$ is closed in $H$. By Closed Set in Topological Subspace: Corollary, $\left\{{e}\right\}$ is closed in $G$. By Topological Group is Hausdorff iff Identity is Closed, $G$ is Hausdorff. {{qed}}
We use the One-Step Subgroup Test. Because $H \subset \overline H$, $\overline H$ is non-empty. Let $a, b \in \overline H$. Let $U$ be a neighborhood of $a b^{-1}$. Let the mapping $f: G\times G \to G$ be defined as: :$f\left({x, y}\right) = x y^{-1}$ By definition of topological group, $f$ is continuous. By definition of product space, there exist neighborhoods $A, B$ of $a, b$ respectively such that: :$A \times B \subset f^{-1} \left({U}\right)$ By assumption, there exist $x \in A \cap H$ and $y \in B \cap H$. Then: :$x y^{-1} \subset U \cap H$ Because $U$ was arbitrary: :$a b^{-1} \in \overline H$ By One-Step Subgroup Test, $\overline H$ is a subgroup of $G$. {{qed}} Category:Topological Groups 7m4erc7c3mcochdu8lpsmvvk66mdo2b
By Image of Group Homomorphism is Subgroup, $\Img f$ is a group. Let $e$ be the identity of $H$. By Topological Group is Hausdorff iff Identity is Closed, $\set e$ is closed in $\Img f$. Because $f$ is continuous, $\map \ker f = \map {f^{-1} } e$ is closed in $G$. {{qed}} Category:Topological Groups Category:Hausdorff Spaces 5zyfd3xvxrgt9mw0fn3u8ovtws5b1kz
Let $U \subset X$ be open. We have to show that $\pi \left({U}\right)$ is open. By definition of quotient topology, this is the case {{iff}} $\pi^{-1} \left({\pi \left({U}\right)}\right)$ is open. By definition of saturation under group action: :$\displaystyle \pi^{-1} \left({\pi \left({U}\right)}\right) = \bigcup_{g \mathop \in G} g U$ Because $G$ acts by homeomorphisms, $\pi^{-1} \left({\pi \left({U}\right)}\right)$ is open. {{qed}}
=== Definition 1 implies Definition 2 === {{:Equivalence of Definitions of Locally Connected Space/Definition 1 implies Definition 2}}{{qed|lemma}} === Definition 2 implies Definition 1 === {{:Equivalence of Definitions of Locally Connected Space/Definition 2 implies Definition 1}}{{qed|lemma}} === Definition 1 implies Definition 3 === {{:Equivalence of Definitions of Locally Connected Space/Definition 1 implies Definition 3}}{{qed|lemma}} === Definition 3 implies Definition 1 === {{:Equivalence of Definitions of Locally Connected Space/Definition 3 implies Definition 1}}{{qed|lemma}} === Definition 3 implies Definition 4 === {{:Equivalence of Definitions of Locally Connected Space/Definition 3 implies Definition 4}}{{qed|lemma}} === Definition 4 implies Definition 3 === {{:Equivalence of Definitions of Locally Connected Space/Definition 4 implies Definition 3}}{{qed}}
Follows directly from the definition of neighborhood and Subset Relation is Transitive. {{qed}}
=== Definition 1 implies Definition 2 === {{:Equivalence of Definitions of Locally Path-Connected Space/Definition 1 implies Definition 2}}{{qed|lemma}} === Definition 2 implies Definition 1 === {{:Equivalence of Definitions of Locally Path-Connected Space/Definition 2 implies Definition 1}}{{qed|lemma}} === Definition 1 implies Definition 3 === {{:Equivalence of Definitions of Locally Path-Connected Space/Definition 1 implies Definition 3}}{{qed|lemma}} === Definition 3 implies Definition 1 === {{:Equivalence of Definitions of Locally Path-Connected Space/Definition 3 implies Definition 1}}{{qed|lemma}} === Definition 3 implies Definition 4 === {{:Equivalence of Definitions of Locally Path-Connected Space/Definition 3 implies Definition 4}}{{qed|lemma}} === Definition 4 implies Definition 3 === {{:Equivalence of Definitions of Locally Path-Connected Space/Definition 4 implies Definition 3}}{{qed}}
By the Theorem of Even Perfect Numbers: :$n = 2^{p - 1} \paren {2^p - 1}$ where $p$ is prime. With the exception of $6 = 2^1 \paren {2^2 - 1}$ and $28 = 2^2 \paren {2^3 - 1}$: :$p$ is odd and $p > 4$. We claim that: :$n$ ends in $\phantom 0 6$ preceded by an odd digit if $p \equiv 1 \pmod 4$ :$n$ ends in $28$ preceded by an odd digit if $p \equiv 3 \pmod 4$ These statements are equivalent to: :$n \equiv \phantom 0 16 \pmod {\phantom 0 20}$ if $p \equiv 1 \pmod 4$ :$n \equiv 128 \pmod {200}$ if $p \equiv 3 \pmod 4$ By Powers of 16 Modulo 20, we have: :$2^{4 n} = 16^n \equiv 16 \pmod {20}$ for $n \ge 1$ === Case $1$: $p \equiv 1 \pmod 4$ === Write $p = 4 n + 1$. Then: {{begin-eqn}} {{eqn | l = n       | r = 2^{p - 1} \paren {2^p - 1} }} {{eqn | r = 2^{4 n} \paren {2 \times 2^{4 n} - 1} }} {{eqn | o = \equiv       | r = 16 \paren {2 \times 16 - 1}       | rr = \pmod {20}       | c = Powers of 16 Modulo 20 }} {{eqn | o = \equiv       | r = 496       | rr = \pmod {20} }} {{eqn | o = \equiv       | r = 16       | rr = \pmod {20} }} {{end-eqn}} showing our first claim. {{qed|lemma}} === Case $2$: $p \equiv 3 \pmod 4$ === Write $p = 4 n + 3$. By Powers of 16 Modulo 20, we can write $2^{4 n} = 20 K + 16$ for some $K \in \Z$. Then: {{begin-eqn}} {{eqn | l = n       | r = 2^{p - 1} \paren {2^p - 1} }} {{eqn | r = 4 \times 2^{4 n} \paren {8 \times 2^{4 n} - 1} }} {{eqn | r = 4 \paren {20 K + 16} \paren {8 \paren {20 K + 16} - 1} }} {{eqn | r = \paren {80 K + 64} \paren {160 K + 127} }} {{eqn | r = 12800 K^2 + 20400 K + 8128 }} {{eqn | o = \equiv       | r = 128       | rr = \pmod {200} }} {{end-eqn}} showing our second claim. {{qed}}
By definition of locally connected space, $T$ has a basis of connected sets in $T$. Thus $S$ is a union of open connected sets in $T$. By Components are Open iff Union of Open Connected Sets, the components of $T$ are open. {{qed}}
By definition of locally path-connected, $T$ has a basis of path-connected set. Thus $S$ is a union of open path-connected sets of $T$. By Path Components are Open iff Union of Open Path-Connected Sets, the path components of $T$ are open in $T$. {{qed}}
By: : Path Component is Closed : Path Component of Locally Path-Connected Space is Open the path components of $T$ are clopen. Because $T$ is connected, every path component equals $T$. That is, $T$ is path-connected. {{qed}} Category:Connected Spaces Category:Path-Connected Spaces Category:Locally Path-Connected Spaces f6shk5etxlx4a0u9zwacyh86wofjkn0
=== Condition (1) implies Condition (2) === {{:Components are Open iff Union of Open Connected Sets/Components are Open implies Space is Union of Open Connected Sets}}{{qed|lemma}} === Condition (2) implies Condition (1) === {{:Components are Open iff Union of Open Connected Sets/Space is Union of Open Connected Sets implies Components are Open}}{{qed}}
=== Condition (1) implies Condition (2) === {{:Path Components are Open iff Union of Open Path-Connected Sets/Path Components are Open implies Space is Union of Open Path-Connected Sets}}{{qed|lemma}} === Condition (2) implies Condition (1) === {{:Path Components are Open iff Union of Open Path-Connected Sets/Space is Union of Open Path-Connected Sets implies Path Components are Open}}{{qed}}
Let $a$ be an even perfect number. From the Theorem of Even Perfect Numbers, $a$ is in the form $2^{p - 1} \paren {2^p - 1}$ where $2^p - 1$ is prime. Thus: {{begin-eqn}} {{eqn | l = a       | r = \paren {2^p - 1} 2^{p - 1}       | c =  }} {{eqn | r =  2^{p - 1} \paren {2 \times 2^{p - 1} - 1}       | c =  }} {{eqn | r = n \paren {2 n - 1}       | c = where $n = 2^{p - 1}$ }} {{end-eqn}} The result follows from Closed Form for Hexagonal Numbers. {{qed}}
{{begin-eqn}} {{eqn | l = H_n       | r = n \paren {2 n - 1}       | c = Closed Form for Hexagonal Numbers }} {{eqn | r = \frac {2 n \paren {2 n - 1} } 2       | c =  }} {{eqn | r = \frac {m \paren {m + 1} } 2       | c = for $m = 2 n - 1$ }} {{end-eqn}} Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{d \mathop \divides n} d       | r = \map \sigma n       | c = {{Defof|Sigma Function}} }} {{eqn | ll= \leadsto       | l = \dfrac 1 n \sum_{d \mathop \divides n} d       | r = \dfrac {\map \sigma n} n       | c =  }} {{eqn | ll= \leadsto       | l = \sum_{d \mathop \divides n} \frac d n       | r = \dfrac {\map \sigma n} n       | c =  }} {{eqn | ll= \leadsto       | l = \sum_{d \mathop \divides n} \frac 1 d       | r = \dfrac {\map \sigma n} n       | c =  }} {{end-eqn}} The result follows by definition of perfect number: {{:Definition:Perfect Number/Definition 4}}{{qed}}
From Theorem of Even Perfect Numbers: :$n = 2^{p - 1} \paren {2^p - 1} = \dfrac {2^p \paren {2^p - 1} } 2$ where $p$ is prime. From Odd Power of 2 is Congruent to 2 Modulo 3: :$2^p \equiv 2 \pmod 3$ for odd $p$. Thus: {{begin-eqn}} {{eqn | l = n       | r = \dfrac {\paren {3 k + 2} \paren {3 k + 1} } 2       | c = for some $k \in \Z_{>0}$ }} {{eqn | r = \frac {9 k^2 + 9 k + 2} 2       | c =  }} {{eqn | r = 9 \frac {k \paren {k + 1} } 2 + 1       | c =  }} {{eqn | r = 9 T_k + 1       | c = Closed Form for Triangular Numbers }} {{end-eqn}} So $n$ is $1$ more than $9$ times the $k$th triangular number for some $k$. That is: :$n \equiv 1 \pmod 9$ When $n = 6$ the situation is different. We have: {{begin-eqn}} {{eqn | l = n       | r = \dfrac {2^2 \paren {2^2 - 1} } 2       | c =  }} {{eqn | r = \dfrac {\paren {3 k + 1} \paren {3 k} } 2       | c = where $k = 1$ }} {{eqn | o = \equiv       | r = 0 \pmod 3       | c =  }} {{end-eqn}} and so the result does not hold. {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$2^n \equiv \paren {-1}^n \pmod 3$ $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = 2^0       | r = 1       | c =  }} {{eqn | r = \paren {-1}^0       | c =  }} {{eqn | o = \equiv       | r = \paren {-1}^0       | rr= \pmod 3 }} {{end-eqn}} Thus $\map P 0$ is seen to hold. === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = 2^1       | r = 2       | c =  }} {{eqn | r = 3 + \paren {-1}       | c =  }} {{eqn | r = 3 + \paren {-1}^1       | c =  }} {{eqn | o = \equiv       | r = \paren {-1}^1       | rr= \pmod 3 }} {{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: :$2^k \equiv \paren {-1}^k \pmod 3$ from which it is to be shown that: :$2^{k + 1} \equiv \paren {-1}^{k + 1} \pmod 3$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = 2^{k + 1}       | r = 2 \times 2^k       | c =  }} {{eqn | r = 2 \times \paren {3 q + \paren {-1}^k}       | c =  }} {{eqn | r = \paren {3 \paren {2 q} + 2 \paren {-1}^k}       | c =  }} {{eqn | r = \paren {3 \paren {2 q} + 2 \paren {-1}^k}       | c =  }} {{end-eqn}} If $k$ is odd, this means: {{begin-eqn}} {{eqn | l = 2^{k + 1}       | o = \equiv       | r = -2       | rr= \pmod 3       | c =  }} {{eqn | o = \equiv       | r = 1       | rr= \pmod 3       | c =  }} {{eqn | o = \equiv       | r = \paren {-1}^{k + 1}       | rr= \pmod 3       | c =  }} {{end-eqn}} If $k$ is even, this means: {{begin-eqn}} {{eqn | l = 2^{k + 1}       | o = \equiv       | r = 2       | rr= \pmod 3       | c =  }} {{eqn | o = \equiv       | r = -1       | rr= \pmod 3       | c =  }} {{eqn | o = \equiv       | r = \paren {-1}^{k + 1}       | rr= \pmod 3       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 0}: 2^n \equiv \paren {-1}^n \pmod 3$ {{qed}} Category:Number Theory Category:2 Category:3 6danxn4orx1nl2k4oellhusxp275x6k
From Congruence Modulo 3 of Power of 2: :$2^n \equiv \paren {-1}^n \pmod 3$ We have that $n$ is odd. Hence: {{begin-eqn}} {{eqn | l = 2^n       | o = \equiv       | r = -1       | rr= \pmod 3       | c =  }} {{eqn | o = \equiv       | r = 3 - 1       | rr= \pmod 3       | c =  }} {{eqn | o = \equiv       | r = 2       | rr= \pmod 3       | c =  }} {{end-eqn}} {{qed}} Category:Number Theory Category:2 Category:3 7z9oqafotjpzea7v785tizd4gmegimi
Let $V$ be open in $X$. By Intersection with Subset is Subset, $V\cap U = V$. By definition of topological subspace, $V$ is open in $U$. Let $V$ be open in $U$. By definition of topological subspace, there exists an open subset $W\subset X$ with $V=U\cap W$. Because $U$ and $W$ are open in $X$, $V=U\cap W$ is open in $X$. {{qed}}
=== $1$ implies $2$ === Let $x \in S$. Let $K$ be a compact neighborhood of $x$. Let $\mathcal B$ be the set of compact neighborhoods of $x$. It is shown that $\mathcal B$ is a neighborhood basis of $x$. Let $U$ be a neighborhood of $x$. We have to show that $U$ contains a compact neighborhood of $x$. By Neighborhood in Topological Subspace, $U\cap K$ is a neighborhood of $x$ in $K$. By Neighborhood in Compact Hausdorff Space Contains Compact Neighborhood, there is a compact neighborhood $V$ of $x$ in $K$ such that $V \subset U \cap K$. From Neighborhood in Topological Subspace there exists a neighborhood $W$ of $x$ in $T$ such that $W \cap K = V$.  From Intersection of Neighborhoods in Topological Space is Neighborhood we have $W \cap K = V$ is a neighborhood of $x$ in $T$. From Compact in Subspace is Compact in Topological Space then $V$ is compact in $T$. Hence $V \in \mathcal B$. The result follows. {{qed|lemma}} === $2$ implies $1$ === Follows from Locally Compact Space is Weakly Locally Compact. {{qed}} Category:Locally Compact Hausdorff Spaces tjc3a5mu3eowtzmv1hszqwlqcxy24s9
By definition of neighborhood, there exists an open set $V$ with $x\in V\subset U$. Then $X \setminus V$ is closed. By Compact Hausdorff Space is T4, there exist disjoint open sets $A,B$ such that: :$X\setminus V\subset A$ :$x\in B$ Then: :$X \setminus A$ is compact by Closed Subspace of Compact Space is Compact :$B \subset X \setminus A$ by Empty Intersection iff Subset of Relative Complement :$X \setminus A \subset X \setminus \paren{X \setminus V}$ by Relative Complement inverts Subsets :$X \setminus \paren{X \setminus V} = V$ by Relative Complement of Relative Complement Thus $x\in B \subset X\setminus A \subset V$, so $X\setminus A$ is a compact neighborhood of $x$ contained in $V$. {{qed}}
We will prove that :$F$ inherits directed suprema. Let $D$ be a directed subset of $f \sqbrk S$ such that :$D$ admits a supremum in $L$. By definition of ordered subset: :$D$ is directed in $L$. By definition of mapping preserves directed suprema: :$\map {\sup_L} {f \sqbrk D} = \map f {\sup_L D}$ By definition of idempotent mapping: :$f \sqbrk D = D$ Thus by definition of image of set: :$\sup_L D \in f \sqbrk S$ {{qed|lemma}} By definition of idempotent mapping: :$f \sqbrk S = \set {x \in S: \map f x = x}$ By Directed Suprema Preserving Mapping is Increasing: :$f$ is an increasing mapping. Thus by Image under Increasing Mapping equal to Special Set is Complete Lattice: :$F$ is complete lattice. {{qed}}
Let $n = a^4 + b^4 + c^4$ for $a, b, c \in \Z$. === Necessary Condition === Let $a$, $b$ and $c$ all be divisible by either $5$ or $29$. That is: :$a, b, c \equiv 0 \pmod 5$ or: :$a, b, c \equiv 0 \pmod {29}$ We have that for integer $p$: {{begin-eqn}} {{eqn | l = a, b, c       | o = \equiv       | r = 0       | rr= \pmod p       | c =  }} {{eqn | ll= \leadsto       | l = a^4, b^4, c^4       | o = \equiv       | r = 0       | rr= \pmod p       | c = Congruence of Powers }} {{eqn | o = \equiv       | r = 0       | rr= \pmod p       | c = Modulo Addition is Well-Defined }} {{eqn | ll= \leadsto       | l = p       | o = \divides       | r = a^4 + b^4 + c^4       | c = where $\divides$ denotes divisibility }} {{end-eqn}} Hence if all three addends are divisible by either $5$ or $29$, then so is the sum. {{qed|lemma}} === Sufficient Condition === Let $n$ be divisible by $5$. Then: :$a^4 + b^4 + c^4 \equiv 0 \mod 5$ From Fourth Power Modulo 5: :$x \in \Z_5 \implies x^4 \in \set {0, 1}$ So: :$a^4 + b^4 + c^4 \equiv 0 \mod 5 \implies a, b, c \equiv 0 \mod 5$ {{qed|lemma}} Let $n$ be divisible by $29$. Then $a^4 + b^4 + c^4 \equiv 0 \mod {29}$. But for $x \in \Z_{29}$, $x^4 \in \set {0, 1, 7, 16, 20, 23, 24, 25} = R$ necessarily. {{explain|Demonstrate the above}} So: :$a^4 + b^4 + c^4 \equiv 0 \mod {29} \implies a, b, c \equiv 0 \mod {29}$ as $29$ and its non-zero multiples are not partitioned by any 3 elements in $R$ {{explain|The above statement needs to be gone into in more detail, probably on a separate page.}} {{qed}}
{{ProofWanted|Follows somehow from the fact that these numbers are the hypotenuses of Definition:Almost Isosceles Pythagorean Triangles.}}
=== 3 === {{:Prime-Generating Quadratics of form 2 a squared plus p/3}} === 5 === {{:Prime-Generating Quadratics of form 2 a squared plus p/5}} === 11 === {{:Prime-Generating Quadratics of form 2 a squared plus p/11}} === 29 === {{:Prime-Generating Quadratics of form 2 a squared plus p/29}} When $x = p$ we have: {{begin-eqn}} {{eqn | l = 2 p^2 + p       | r = p \left({2 p + 1}\right)       | c =  }} {{end-eqn}} which has divisors $p$ and $2 p + 1$ and so is not prime. {{qed}}
{{begin-eqn}} {{eqn | l = 2 \times 0^2 + 3       | r = 0 + 3 }} {{eqn | r = 3       | c = which is prime }} {{eqn | l = 2 \times 1^2 + 3       | r = 2 + 3       | c =  }} {{eqn | r = 5       | c = which is prime }} {{eqn | l = 2 \times 2^2 + 3       | r = 2 \times 4 + 3       | c =  }} {{eqn | r = 8 + 3       | c =  }} {{eqn | r = 11       | c = which is prime }} {{end-eqn}}
{{begin-eqn}} {{eqn | l = 2 \times 0^2 + 5       | r = 0 + 5 }} {{eqn | r = 5       | c = which is prime }} {{eqn | l = 2 \times 1^2 + 5       | r = 2 + 5       | c =  }} {{eqn | r = 7       | c = which is prime }} {{eqn | l = 2 \times 2^2 + 5       | r = 2 \times 4 + 5       | c =  }} {{eqn | r = 8 + 5       | c =  }} {{eqn | r = 13       | c = which is prime }} {{eqn | l = 2 \times 3^2 + 5       | r = 2 \times 9 + 5       | c =  }} {{eqn | r = 18 + 5       | c =  }} {{eqn | r = 23       | c = which is prime }} {{eqn | l = 2 \times 4^2 + 5       | r = 2 \times 16 + 5       | c =  }} {{eqn | r = 32 + 5       | c =  }} {{eqn | r = 37       | c = which is prime }} {{end-eqn}}