hcju/proofwikiqa huggingface dataset:
Let $F_n$ be a Fermat number which is triangular. We use the weaker form of Divisor of Fermat Number: :Every divisor of $F_n$ is of the form $k \times 2^{n + 1} + 1$ as this formulation is valid for all $n \in \N$. From Closed Form for Triangular Numbers: :$F_n = \dfrac {m \paren {m + 1} } 2$ Either $m$ or $m + 1$ is even. Suppose $m$ is even. Then: {{begin-eqn}} {{eqn | ll = \exists k \in \N:       | l = \frac m 2       | r = k \times 2^{n + 1} + 1       | c = Divisor of Fermat Number }} {{eqn | lll = \leadsto        | l = m + 1       | r = 2 \paren {k \times 2^{n + 1} + 1} + 1 }} {{eqn | r = k \times 2^{n + 2} + 3 }} {{eqn | ll = \exists j \in \N:       | l = m + 1       | r = j \times 2^{n + 1} + 1       | c = Divisor of Fermat Number }} {{eqn | lll = \leadsto       | l = k \times 2^{n + 2} + 2       | r = j \times 2^{n + 1} }} {{eqn | lll = \leadsto       | l = k \times 2^{n + 1} + 1       | r = j \times 2^n }} {{end-eqn}} As the {{LHS}} is always odd, we must have $2^n = 1$. This forces $n = 0$, with $F_n = 3$. We check that $3$ is indeed a triangular number. Suppose $m + 1$ is even. Then: {{begin-eqn}} {{eqn | ll = \exists k \in \N:       | l = \frac {m + 1} 2       | r = k \times 2^{n + 1} + 1       | c = Divisor of Fermat Number }} {{eqn | lll = \leadsto        | l = m       | r = 2 \paren {k \times 2^{n + 1} + 1} - 1 }} {{eqn | r = k \times 2^{n + 2} + 1 }} {{eqn | l = F_n       | r = \frac {m \paren {m + 1} } 2 }} {{eqn | lll = \leadsto       | l = 2^{2^n} + 1       | r = \paren {k \times 2^{n + 2} + 1} \paren {k \times 2^{n + 1} + 1} }} {{eqn | r = k^2 \times 2^{2 n + 3} + k \paren {2^{n + 2} + 2^{n + 1} } + 1 }} {{eqn | lll = \leadsto       | l = 2^{2^n}       | r = k^2 \times 2^{2 n + 3} + k \paren {2^{n + 2} + 2^{n + 1} } }} {{eqn | r = k \times 2^{n + 1} \times \paren {k \times 2^{n + 2} + 3 } }} {{end-eqn}} As $k \times 2^{n + 2} + 3 > 1$ and is odd, $k \times 2^{n + 2} + 3$ cannot be a power of $2$. So $m + 1$ cannot be even, and we have exhausted all possibilities. {{qed}}
Recall the definition of Fermat numbers: :$F_n = 2^{(2^n)}+1$, where $n = 0, 1, 2, \ldots$ === Marginal Case === $F_0 = 3$ is not a square. === General Case === It will be shown that Fermat numbers lie between $2$ consecutive squares, thus cannot be itself a square: {{begin-eqn}} {{eqn | l = \left({2^{ \left({2^{n-1} }\right)} }\right)^2       | r = 2^{\left({2^n}\right)}       | c = Index Laws/Product of Indices }} {{eqn | o = <       | r = 2^{\left({2^n}\right)} + 1       | c = adding a positive term }} {{eqn | r = F_n       | c =  }} {{eqn | o = <       | r = 2^{\left({2^n}\right)} + 2^{\left({2^{n-1} }\right) + 1} + 1       | c = adding another positive term }} {{eqn | r = \left({2^{\left({2^{n-1} }\right)} + 1}\right)^2       | c = Completing the Square }} {{end-eqn}} {{qed}}
From Construction of Regular Prime $p$-Gon Exists iff $p$ is Fermat Prime it is known that this construction is possible. {{ProofWanted}}
Let $x$ be a set. By definition of $\mathit{Filt}$: :$x \in \mathit{Filt}\left({L_2}\right) \iff x$ is a filter on $L_2$. By Ideal is Filter in Dual Ordered Set: :$x \in \mathit{Filt}\left({L_2}\right) \iff x$ is an ideal in $L_1$. By definition of $\mathit{Ids}$: :$x \in \mathit{Filt}\left({L_2}\right) \iff x \in \mathit{Ids}\left({L_1}\right)$ Hence by definition of set equality: :$\mathit{Ids}\left({L_1}\right) = \mathit{Filt}\left({L_2}\right)$ {{qed}}
=== Filtered === By Set is Element of its Power Set: :$X \in \mathcal P\left({X}\right)$ Thus by definition: :$\mathcal P\left({X}\right)$ is a non-empty set. Let $x, y \in \mathcal P\left({X}\right)$. By Intersection is Subset: :$x \cap y \subseteq x$ and $x \cap y \subseteq y$ By Subset Relation is Transitive: :$x \cap y \in \mathcal P\left({X}\right)$ Thus :$\exists z \in \mathcal P\left({X}\right): z \subseteq x \land z \subseteq y$ {{qed|lemma}} === Upper === Thus we have: :$\forall x, y \in \mathcal P\left({X}\right): x \subseteq y \implies y \in \mathcal P\left({X}\right)$ {{qed|lemma}} Thus by definition of filter in ordered set: :$\mathcal P\left({X}\right)$ is a filter on $L$. {{qed}}
We have that: {{begin-eqn}} {{eqn | l = 295 - 267       | r = 28       | c =  }} {{eqn | l = 323 - 295       | r = 28       | c =  }} {{end-eqn}} demonstrating that $267, 295, 323$ are in arithmetic sequence with common difference $28$. Then: {{begin-eqn}} {{eqn | l = \map \sigma {267}       | r = 360       | c = {{SigmaLink|267}} }} {{eqn | l = \map \sigma {295}       | r = 360       | c = {{SigmaLink|295}} }} {{eqn | l = \map \sigma {323}       | r = 360       | c = {{SigmaLink|323}} }} {{end-eqn}} {{ProofWanted|It remains to be shown that these are the smallest.}}
By Singleton is Directed and Filtered Subset: :$\left\{ {X}\right\}$ is filtered. We will prove that :$\left\{ {X}\right\}$ is an upper set. Let $x \in \left\{ {X}\right\}$, $y \in \mathcal P\left({X}\right)$ such that :$x \subseteq y$ By definition of singleton: :$x = X$ By definition of power set: :$y \subseteq X$ By definition of set equality: :$y = X$ Thus: :$y \in \left\{ {X}\right\}$ {{qed|lemma}} Thus by definition of filter in ordered set: :$\left\{ {X}\right\}$ is a filter. {{qed}}
Let $p_i$ be the $i$th prime. Construct the sequence $\sequence {t_i}_{t \mathop = 1}^\infty$ of natural numbers $t_i$ recursively as follows: :$t_1 = 2$ :$t_{i + 1} = \map \phi {p_i^{t_i + 1} } - 1 = p_i^{t_i} \paren {p_i - 1} - 1$ for $i \ge 1$ We claim that this sequence has the property: :$(1): \quad p_i^{t_i + 1} \divides \map \sigma {p_{i + 1}^{t_{i + 1} } }$ for $i \ge 1$ For $i = 1$, $p_1 = 2$ and $t_1 = 2$. :$t_2 = 2^2 \paren {2 - 1} - 1 = 3$ :$\map \sigma {p_2^{t_2} } = \map \sigma {3^3} = 40$ which is a multiple of $2^3 = 8$. For $i > 1$: {{begin-eqn}} {{eqn | l = \paren {p_{i + 1} - 1} \map \sigma {p_{i + 1}^{t_{i + 1} } }       | r = p_{i + 1}^{t_{i + 1} + 1} - 1       | c = Sigma Function of Power of Prime }} {{eqn | r = p_{i + 1}^{\map \phi {p_i^{t_i + 1} } } - 1 }} {{eqn | r = 0       | rr = \pmod {p_i^{t_i + 1} }       | c = Euler's Theorem }} {{end-eqn}} For $i > 1$, $p_{i + 1} - 1 > 2$. Hence $p_{i + 1} - 1$ is even and composite. Since $p_i$ is odd and $p_i > \dfrac {p_{i + 1} - 1} 2$ by Bertrand-Chebyshev Theorem: :$p_{i + 1} - 1$ and $p_i^{t_i + 1}$ are coprime. By Euclid's Lemma, $p_i^{t_i + 1} \divides \map \sigma {p_{i + 1}^{t_{i + 1} } }$. We define for $l \ge 1$: $A_l = \set {m \in \N: p_i^{t_i} \divides m, p_i^{t_i + 1} \nmid m, \forall i: 1 \le i \le l}$. Denote $\map s n$ the aliquot sum of $n$. Then for $m \in A_l$ where $l \ge 2$, we claim: {{begin-eqn}} {{eqn | n = 2       | l = \map s m       | o = >       | r = m }} {{eqn | n = 3       | l = \map s m       | o = \in       | r = A_{l - 1} }} {{end-eqn}} Let $m \in A_l$ where $l \ge 2$. From definition we have: {{begin-eqn}} {{eqn | l = p_1^{t_1}       | r = 4 \divides m }} {{eqn | l = p_2^{t_2}       | r = 27 \divides m }} {{end-eqn}} Hence $m$ is a multiple of $27 \times 4 = 108$. Since $108$ is abundant, $m$ is also abundant by Multiple of Abundant Number is Abundant. $\map s m > m$ follows from definition of abundant numbers. Write $m = p_1^{t_1} \cdots p_l^{t_l} k$, where $k$ is not divisible by $p_1, \dots, p_l$. Then: {{begin-eqn}} {{eqn | l = \map s m       | r = \map \sigma m - m }} {{eqn | r = \map \sigma {p_1^{t_1} } \map \sigma {p_2^{t_2} } \cdots \map \sigma {p_l^{t_l} } \map \sigma k - m       | c = Sigma Function is Multiplicative }} {{eqn | r = K p_1^{t_1 + 1} \cdots p_{l - 1}^{t_{l - 1} + 1} - p_1^{t_1} \cdots p_l^{t_l} k       | c = for some $K \in \N$, by $(1)$ }} {{eqn | r = p_1^{t_1} \cdots p_{l - 1}^{t_{l - 1} } \paren {K p_1 \cdots p_{l - 1} - p_l^{t_l} k} }} {{end-eqn}} Since $K p_1 \cdots p_{l - 1}$ is divisible by $p_1, \dots, p_{l - 1}$ while $p_l^{t_l} k$ isn't: :$p_1, \dots, p_{l - 1} \nmid \paren {K p_1 \cdots p_{l - 1} - p_l^{t_l} k}$ Hence for each $i$ where $1 \le i \le l - 1$: :$p_i^{t_i} \divides \map s m, p_i^{t_i + 1} \nmid \map s m$ This shows that $\map s m \in A_{l - 1}$. Hence to create an aliquot sequence of length $l$, one picks an element $m$ from $A_l$. Applying $(2)$ and $(3)$ repeatedly, we see that: :$m, \map s m, \map s {\map s m}, \dots, \map {s^l} m$ is a monotonically increasing aliquot sequence of length $l$. {{qed}}
By the brute force technique: {{begin-eqn}} {{eqn | l = 1 \times 9       | r = 9 }} {{eqn | l = 2 \times 9       | r = 18 }} {{eqn | l = 3 \times 9       | r = 27 }} {{eqn | l = 4 \times 9       | r = 36 }} {{eqn | l = 5 \times 9       | r = 45 }} {{eqn | l = 6 \times 9       | r = 54 }} {{eqn | l = 7 \times 9       | r = 63 }} {{eqn | l = 8 \times 9       | r = 72 }} {{eqn | l = 9 \times 9       | r = 81 }} {{eqn | l = 10 \times 9       | r = 90 }} {{eqn | l = 11 \times 9       | r = 99 }} {{eqn | l = 12 \times 9       | r = 108 }} {{end-eqn}} All integer multiples of $9$ $k \times 9$ for $k = 12$ through to $k = 22$ begin with $1$, and so have at least one odd digit. Then: {{begin-eqn}} {{eqn | l = 22 \times 9       | r = 198 }} {{eqn | l = 23 \times 9       | r = 207 }} {{eqn | l = 24 \times 9       | r = 216 }} {{eqn | l = 25 \times 9       | r = 225 }} {{eqn | l = 26 \times 9       | r = 234 }} {{eqn | l = 27 \times 9       | r = 243 }} {{eqn | l = 28 \times 9       | r = 252 }} {{eqn | l = 39 \times 9       | r = 261 }} {{eqn | l = 30 \times 9       | r = 270 }} {{eqn | l = 31 \times 9       | r = 279 }} {{eqn | l = 32 \times 9       | r = 288 }} {{end-eqn}} Hence the result. {{qed}}
Let $n$ be the smallest integer multiple of $9$ all of whose digits are even. From Divisibility by 9, the digits of $n$ must add to an integer multiple of $9$. But from Sum of Even Integers is Even, the digits of $n$ must add to an even integer multiple of $9$: $18, 36, 54$, etc. There is only $1$ integer multiple of $9$ with $2$ digits whose digits add to an even integer multiple of $9$, and that is $99$. Thus we have to look at $3$-digit integers. The following sets of $3$ even digits add to $18$: :$\set {2, 8, 8}$ :$\set {4, 6, 8}$ :$\set {6, 6, 6}$ and that seems to be about it. There are no sets of $3$ even digits which add up to $36$ or higher. The result follows by inspection. {{qed}}
Since $n$ is a Kaprekar number of $D$ digits: :$\begin {cases} n^2 = a \times 10^D + b \\ n = a + b \end {cases}$ for some positive integers $a$ and $b$, $b < 10^D$. Hence: {{begin-eqn}} {{eqn | l = \paren {10^D - n}^2       | r = 10^{2 D} - 2 n \times 10^D + n^2 }} {{eqn | r = 10^{2 D} - 2 \paren {a + b} 10^D + a \times 10^D + b }} {{eqn | r = 10^D \paren {10^D - a - 2 b} + b }} {{end-eqn}} and we have: :$\paren {10^D - a - 2 b} + b = 10^D - a - b = 10^D - n$ Finally we check that $10^D - a - 2 b \ge 0$: {{AimForCont}} $10^D - a - 2 b \le -1$. Then: {{begin-eqn}} {{eqn | l = \paren {10^D - n}^2       | r = 10^D \paren {10^D - a - 2 b} + b }} {{eqn | o = \le       | r = -10^D + b }} {{eqn | o = <       | r = 0 }} {{end-eqn}} but squares are positive, a contradiction. Hence $10^D - n$ is also a Kaprekar number of $D$ digits. {{qed}} {{Namedfor|Dattathreya Ramchandra Kaprekar|cat = Kaprekar}}
Since $n$ is a Kaprekar number of $m$ digits: :$\begin {cases} n^2 = a \times 10^m + b \\ n = a + b \end {cases}$ for some positive integers $a$ and $b$. Hence: {{begin-eqn}} {{eqn | l = n^2 - n       | r = a \times 10^m + b - a - b }} {{eqn | r = a \paren {10^m - 1} }} {{eqn | r = 9 a R_m }} {{end-eqn}} {{qed}}
By Power Set is Filter in Lattice of Power Set: :$\mathcal P\left({X}\right)$ is a filter on $L$. Let $A \in \mathit{Filt}\left({L}\right)$. Thus by definition of filter: :$A \subseteq \mathcal P\left({X}\right)$ Thus by definitions: :$F$ is bounded above and $\top_F = \mathcal P\left({X}\right)$ {{qed}}
By Singleton of Set is Filter in Lattice of Power Set: :$\left\{{X}\right\}$ is a filter on $L$. Let $A \in \mathit{Filt}\left({L}\right)$. By definition of non-empty set: :$\exists x: x \in A$ By definition of power set: :$x \subseteq X$ By definition of upper set: :$X \in A$ Thus by definitions of singleton and subset: :$\left\{{X}\right\} \subseteq A$ Thus by definitions: :$F$ is bounded below and $\bot_F = \left\{{X}\right\}$ {{qed}}
:{| border="1" |- ! align="center" style = "padding: 2px 10px" | $n_{10}$  ! align="center" style = "padding: 2px 10px" | $n_2$  |- | align="right" style = "padding: 2px 10px" | $3$ | align="right" style = "padding: 2px 10px" | $11$ |- | align="right" style = "padding: 2px 10px" | $5$ | align="right" style = "padding: 2px 10px" | $101$ |- | align="right" style = "padding: 2px 10px" | $7$ | align="right" style = "padding: 2px 10px" | $111$ |- | align="right" style = "padding: 2px 10px" | $313$ | align="right" style = "padding: 2px 10px" | $100 \, 111 \, 001$ |- | align="right" style = "padding: 2px 10px" | $7 \, 284 \, 717 \, 174 \, 827$ | align="right" style = "padding: 2px 10px" | $1 \, 101 \, 010 \, 000 \, 000 \, 011 \, 010 \, 111 \, 110 \, 101 \, 100 \, 000 \, 000 \, 101 \, 011$ |- | align="right" style = "padding: 2px 10px" | $390 \, 714 \, 505 \, 091 \, 666 \, 190 \, 505 \, 417 \, 093$ | align="right" style = "padding: 2px 10px" | $10 \, 100 \, 001 \, 100 \, 110 \, 001 \, 000 \, 000 \, 111 \, 100 \, 001 \, 100 \, 011 \, 000 \, 111 \, 011 \, 100 \, 011 \, 000 \, 110 \, 000 \, 111 \, 100 \, 000 \, 010 \, 001 \, 100 \, 110 \, 000 \, 101$ |} The last two numbers have $43$ and $89$ digits (in binary) respectively . {{qed}}
Let $X \subseteq \mathit{Ids}\left({L}\right)$. By Intersection of Semilattice Ideals is Ideal/Set of Sets: :$\bigcap X$ is an ideal. By Intersection is Largest Subset/General Result: :$\forall A \in \mathit{Ids}\left({L}\right):\left({\forall I \in X: A \subseteq I}\right) \iff A \subseteq \bigcap X$ Thus by definition: :$X$ admits an infimum. Thus by dual to Lattice is Complete iff it Admits All Suprema: :$\mathcal I$ is complete lattice. {{qed}}
Define $L^{-1} := \left({S, \succeq}\right)$ be a dual ordered set of $L$. By Dual Pairs (Order Theory): :$L^{-1}$ is a bounded below meet semilattice. By Filters equal Ideals in Dual Ordered Set: :$\mathit{Filt}\left({L}\right) = \mathit{Ids}\left({L^{-1} }\right)$ where $\mathit{Ids}\left({L^{-1} }\right)$ denotes the set of all ideals in $L^{-1}$. Thus by Ideals form Complete Lattice: :$F$ is a complete lattice. {{qed}}
First note that $5^4 = 625 > 319$. Then note that $2 \times 4^4 = 512 > 319$. Hence any expression of $319$ as fourth powers uses no $n^4$ for $n \ge 5$, and uses not more than $1$ instance of $4^4$. For the remainder, using $2^4$ uses fewer fourth powers than $16$ instances of $1^4$. Now we have: {{begin-eqn}} {{eqn | l = 319       | r = 4^4 + 3 \times 2^4 + 15 \times 1^4 }} {{eqn | r = 3 \times 3^4 + 4 \times 2^4 + 12 \times 1^4 }} {{eqn | r = 2 \times 3^4 + 9 \times 2^4 + 13 \times 1^4 }} {{eqn | r = 1 \times 3^4 + 14 \times 2^4 + 14 \times 1^4 }} {{eqn | r = 0 \times 3^4 + 19 \times 2^4 + 15 \times 1^4 }} {{end-eqn}} and it can be seen that the first two use the least number of fourth powers, at $19$. {{qed}}
{{ProofWanted|Arises during the course of Not Every Number is the Sum or Difference of Two Prime Powers and may be either a lemma or corollary -- hence may require a rename.  Work is needed to analyse that result.}}
By definition of completely irreducible: :$p^\succeq \setminus \set p$ admits a minimum. Then: :$p^\succeq \setminus \set p$ admits a infimum and $\map \inf {p^\succeq \setminus \set p} \in p^\succeq \setminus \set p$ By definition of difference: :$\map \inf {p^\succeq \setminus \set p} \notin \set p$ Thus by definition of singleton: :$\map \inf {p^\succeq \setminus \set p} \ne p$ {{qed}}
What is to be demonstrated is that there exist odd integers which cannot be expressed as $2^m \pm q^n$. {{ProofWanted|I have the paper downloaded and I am studying it.}}
Assume that :$p$ is completely irreducible. By Completely Irreducible Element iff Exists Element that Strictly Succeeds First Element :$\exists q \in S: p \prec q \land \left({\forall s \in S: p \prec s \implies q \preceq s}\right) \land p^\succeq = \left\{ {p}\right\} \cup q^\succeq$ where $p^\succeq$ denotes the upper closure of $p$. Let $a, b \in S$ such that :$p = a \wedge b$ Aiming for a contradiction suppose that :$a \ne p$ and $b \ne p$ By Meet Precedes Operands: :$p \preceq b$ By definition of strictly precede: :$p \prec b$ Then :$q \preceq b$ By definition of strictly precede: :$p \preceq q$ By Meet Precedes Operands: :$p \preceq a$ By definition of strictly precede: :$p \prec a$ Then :$q \preceq a$ By Meet Semilattice is Ordered Structure: :$q \preceq p$ By definition of antisymmetry: :$p = q$ Thus this contradicts $p \prec q$ by definition of strictly precede. {{qed}}
By definition, the length $Y_T$ of the tropical year is defined as $\approx 365 \cdotp 24219 \, 878$ days By definition of the Julian year: :$Y_J = \begin{cases} 366 \, \text {days} & : 4 \divides y \\ 365 \, \text {days} & : 4 \nmid y \end{cases}$ where: :$Y_J$ denotes the length of the Julian year in days :$y$ denotes the number of the year :$4 \divides y$ denotes that $y$ is divisible by $4$ :$4 \nmid y$ denotes that $y$ is not divisible by $4$. Thus: {{begin-eqn}} {{eqn | l = 4 Y_J       | r = 3 \times 365 + 366       | c = considering $4$ consecutive Julian years }} {{eqn | ll= \leadsto       | l = Y_J       | r = \frac {1461} 4       | c =  }} {{eqn | r = 365 \cdotp 25       | c =  }} {{eqn | ll= \leadsto       | l = Y_J - Y_T       | o = \approx       | r = 365 \cdotp 25 - 365 \cdotp 24219 \, 878       | c =  }} {{eqn | o = \approx       | r = 0 \cdotp 0078013       | c =  }} {{eqn | o = \approx       | r = \frac 1 {128 \cdotp 18}       | c =  }} {{end-eqn}} Thus the mean Julian year is $\dfrac 1 {128 \cdotp 18}$ days longer than the tropical year. This means that after approximately $128$ Julian years, the Julian calendar starts one day later relative to the tropical year. {{qed}}
By definition, the length $Y_T$ of the tropical year is defined as $\approx 365 \cdotp 24219 \, 878$ days By definition of the Gregorian year: :$Y_G = \begin{cases} 366 \, \text {days} & : 400 \divides y \\ 365 \, \text {days} & : 400 \nmid  y \text { and } 100 \divides y\\ 366 \, \text {days} & : 100 \nmid  y \text { and } 4 \divides y\\ 365 \, \text {days} & : 4 \nmid y \end{cases}$ where: :$Y_G$ denotes the length of the Gregorian year in days :$y$ denotes the number of the year :$4 \divides y$ denotes that $y$ is divisible by $4$ :$4 \nmid y$ denotes that $y$ is not divisible by $4$. Thus in every $400$ years, there are: :$300$ years which are not divisible by $4$, which have $365$ days :$1$ year which is divisible by $400$, which has $366$ days :$3$ years which are divisible by $100$ but not $400$, which have $365$ days :The remaining $96$ years which are divisible by $4$ but not $100$, which have $366$ days. Thus: {{begin-eqn}} {{eqn | l = 400 Y_G       | r = 303 \times 365 + 97 \times 366       | c = considering $400$ consecutive Gregorian years }} {{eqn | ll= \leadsto       | l = Y_G       | r = \frac {110 \, 595 + 35 \, 502} {400}       | c =  }} {{eqn | r = \frac {146 \, 097} {400}       | c =  }} {{eqn | r = 365 \cdotp 2425       | c =  }} {{eqn | ll= \leadsto       | l = Y_G - Y_T       | o = \approx       | r = 365 \cdotp 2425 - 365 \cdotp 24219 \, 878       | c =  }} {{eqn | o = \approx       | r = 0 \cdotp 0003013       | c =  }} {{eqn | o = \approx       | r = \frac 1 {3318 \cdotp 95}       | c =  }} {{end-eqn}} Thus the mean Gregorian year is $\dfrac 1 {3318 \cdotp 95}$ days longer than the tropical year. This means that after approximately $3318$ Gregorian years, the Gregorian calendar starts one day later relative to the tropical year. {{qed}}
Aiming for a contradiction suppose that :$x \notin X$ By Completely Irreducible Element iff Exists Element that Strictly Succeeds First Element :$\exists q \in S: x \prec q \land \left({\forall s \in S: x \prec s \implies q \preceq s}\right) \land x^\succeq = \left\{ {x}\right\} \cup q^\succeq$ where $x^\succeq$ denotes the upper closure of $x$. We will prove that :$X \subseteq q^\succeq$ Let $y \in X$. By definitions of infimum and lower bound: :$x \preceq y$ By definition of strictly precede: :$x \prec y$ By existence of $q$: :$q \preceq y$ Thus by definition of upper closure of element: :$y \in q^\succeq$ {{qed|lemma}} By Infimum of Upper Closure of Element: :$\inf\left({q^\succeq}\right) = q$ By Infimum of Subset: :$q \preceq x$ Thus this contrsdicts $x \prec q$ {{qed}}
By definition of order generating: :$x^\succeq \cap X$ admits an infimum and $x = \inf\left({x^\succeq \cap X}\right)$ By Completely Irreducible and Subset Admits Infimum Equals Element implies Element Belongs to Subset: :$x \in x^\succeq \cap X$ Thus by definition of intersection: :$x \in X$ {{qed}}
First note that $5^4 = 625 > 399$. Then note that $2 \times 4^4 = 512 > 399$. Hence any expression of $399$ as fourth powers uses no $n^4$ for $n \ge 5$, and uses not more than $1$ instance of $4^4$. For the remainder, using $2^4$ uses fewer fourth powers than $16$ instances of $1^4$ does. Now we have: {{begin-eqn}} {{eqn | l = 399       | r = 4^4 + 3^4 + 3 \times 2^4 + 14 \times 1^4 }} {{eqn | r = 4^4 + 8 \times 2^4 + 15 \times 1^4 }} {{eqn | r = 4 \times 3^4 + 4 \times 2^4 + 11 \times 1^4 }} {{eqn | r = 3 \times 3^4 + 9 \times 2^4 + 12 \times 1^4 }} {{eqn | r = 2 \times 3^4 + 14 \times 2^4 + 13 \times 1^4 }} {{eqn | r = 1 \times 3^4 + 19 \times 2^4 + 14 \times 1^4 }} {{eqn | r = 0 \times 3^4 + 24 \times 2^4 + 15 \times 1^4 }} {{end-eqn}} and it can be seen that the first and the third uses the least number of fourth powers, at $19$. {{qed}}
By definition of maximal: :$x \in S \setminus y^\succeq$ By definition of difference: :$x \notin y^\succeq$ By definition of upper closure of element: :$y \npreceq x$ We will prove that :$x^\succeq \setminus \left\{ {x}\right\} \subseteq x^\succeq \cap y^\succeq$ Let $a \in x^\succeq \setminus \left\{ {x}\right\}$ By definition of difference: :$a \in x^\succeq$ and $a \notin \left\{ {x}\right\}$ By definitions of upper closure of element and singleton: :$x \preceq a$ and $a \ne x$ By definition of strictly precede: :$x \prec a$ By definition of maximal: :$a \notin S \setminus y^\succeq$ By definition of difference: :$a \in y^\succeq$ Thus by definition of intersection: :$a \in x^\succeq \cap y^\succeq$ {{qed|lemma}} We will prove that :$x^\succeq \cap y^\succeq \subseteq x^\succeq \setminus \left\{ {x}\right\}$ Let $a \in x^\succeq \cap y^\succeq$ By definition of intersection: :$a \in x^\succeq$ and $a \in y^\succeq$ By definition of upper closure of element: :$y \preceq a$ Then :$a \ne x$ By definition of singleton: :$a \notin \left\{ {x}\right\}$ Thus by definition of difference: :$a \in x^\succeq \setminus \left\{ {x}\right\}$ {{qed|lemma}} Thus by definition of set equality: :$x^\succeq \setminus \left\{ {x}\right\} = x^\succeq \cap y^\succeq$ {{qed}}
The case where $n = 1$ is trivial, as there are no prime numbers less than $1$. Let $n, m \in \Z$ be integers such that $n > 1$. Let $n^2 = m^2 + p$ where $p$ is prime. Then: {{begin-eqn}} {{eqn | l = p       | r = n^2 - m^2       | c =  }} {{eqn | r = \left({n + m}\right) \left({n - m}\right)       | c = Difference of Two Squares }} {{eqn | ll= \leadsto       | l = n - m       | r = 1       | c = {{Defof|Prime Number}} }} {{eqn | ll= \leadsto       | l = m       | r = n - 1       | c = {{Defof|Prime Number}} }} {{eqn | ll= \leadsto       | l = n + m       | r = 2 n - 1       | c =  }} {{eqn | r = p       | c =  }} {{end-eqn}} So if $2 n - 1$ is composite, there exists no prime $p$ such that $n^2 = m^2 + p$. Further, if such a $p$ does exist, then $m = n - 1$, and so: :$n^2 = \left({n - 1}\right)^2 + p$ {{qed}} Category:Numbers not Sum of Square and Prime qphnse71dg6iklevcnpp6qqp071fatz