hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = 199 + 0 \times 210       | r = 199       | c = which is the $46$th prime }} {{eqn | l = 199 + 1 \times 210       | r = 409       | c = which is the $80$th prime }} {{eqn | l = 199 + 2 \times 210       | r = 619       | c = which is the $114$th prime }} {{eqn | l = 199 + 3 \times 210       | r = 829       | c = which is the $145$th prime }} {{eqn | l = 199 + 4 \times 210       | r = 1039       | c = which is the $175$th prime }} {{eqn | l = 199 + 5 \times 210       | r = 1249       | c = which is the $204$th prime }} {{eqn | l = 199 + 6 \times 210       | r = 1459       | c = which is the $232$nd prime }} {{eqn | l = 199 + 7 \times 210       | r = 1669       | c = which is the $263$rd prime }} {{eqn | l = 199 + 8 \times 210       | r = 1879       | c = which is the $289$th prime }} {{eqn | l = 199 + 9 \times 210       | r = 2089       | c = which is the $316$th prime }} {{end-eqn}} {{OEIS|A033168}} But note that $199 + 10 \times 210 = 2299 = 11^2 \times 19$ and so is not prime. {{ProofWanted|It remains to be shown that there are no smaller such APs}}
In order to make any one of these positive integers into a prime number one would have to change the last digit. Otherwise the number it was changed into would be either even, or divisible by $5$, and so not prime. But all the other integers between $200$ and $209$ are composite: {{begin-eqn}} {{eqn | l = 201       | r = 3 \times 67       | c =  }} {{eqn | l = 203       | r = 7 \times 29       | c =  }} {{eqn | l = 207       | r = 3^2 \times 23       | c =  }} {{eqn | l = 209       | r = 11 \times 19       | c =  }} {{end-eqn}} In order for there to be a smaller number with this property, a prime gap would need to be found which spans an entire decade. There are none such. On the other hand, each of $201, 203, 207, 209$ can themselves be turned into a prime number by changing the initial $2$ into a $1$: :$101, 103, 107,109$ are all prime. {{qed}}
{{begin-eqn}} {{eqn | l = 1       | r = \dfrac {1 \paren {3 \times 1 - 1} } 2       | c = Closed Form for Pentagonal Numbers }} {{eqn | r = \dfrac {1 \times \paren {1 + 1} } 2       | c = Closed Form for Triangular Numbers }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 210       | r = \dfrac {12 \paren {3 \times 12 - 1} } 2       | c = Closed Form for Pentagonal Numbers }} {{eqn | r = \dfrac {20 \times \paren {20 + 1} } 2       | c = Closed Form for Triangular Numbers }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 40 \, 755       | r = \dfrac {165 \paren {3 \times 165 - 1} } 2       | c = Closed Form for Pentagonal Numbers }} {{eqn | r = \dfrac {285 \times \paren {285 + 1} } 2       | c = Closed Form for Triangular Numbers }} {{end-eqn}} {{ProofWanted|It remains to be shown that these are the only such instances.}}
Let $n = p + q$ where $p \le q$ and both $p$ and $q$ are primes. There can be no more different representations of $n$ as $p + q$ than there are the number of possible options for $q$. As $q \ge p$, it follows that $q \ge \dfrac n 2$. Note that as $p \ge 2$, it follows that $q \le n - 2$. The number of possible values of $q$ is therefore equal to the number of primes between $\dfrac n 2$ and $n - 2$ inclusive. That is, the number of primes in the interval $\closedint {\dfrac n 2} {n - 2}$. Hence the result. {{qed}}
From Number of Representations as Sum of Two Primes, the number of ways an integer $n$ can be represented as the sum of two primes is no greater than the number of primes in the interval $\closedint {\dfrac n 2} {n - 2}$. The interval $\closedint {\dfrac {210} 2} {210 - 2}$ is $\closedint {105} {208}$. The primes in this interval can be enumerated: :$107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199$ It can be seen there are exactly $19$ of them. We have: {{begin-eqn}} {{eqn | l = 11 + 199       | r = 210 }} {{eqn | l = 13 + 197       | r = 210 }} {{eqn | l = 17 + 193       | r = 210 }} {{eqn | l = 19 + 191       | r = 210 }} {{eqn | l = 29 + 181       | r = 210 }} {{eqn | l = 31 + 179       | r = 210 }} {{eqn | l = 37 + 173       | r = 210 }} {{eqn | l = 43 + 167       | r = 210 }} {{eqn | l = 47 + 163       | r = 210 }} {{eqn | l = 53 + 157       | r = 210 }} {{eqn | l = 59 + 151       | r = 210 }} {{eqn | l = 61 + 149       | r = 210 }} {{eqn | l = 71 + 139       | r = 210 }} {{eqn | l = 73 + 137       | r = 210 }} {{eqn | l = 79 + 131       | r = 210 }} {{eqn | l = 83 + 127       | r = 210 }} {{eqn | l = 97 + 113       | r = 210 }} {{eqn | l = 101 + 109       | r = 210 }} {{eqn | l = 103 + 107       | r = 210 }} {{end-eqn}} and as can be seen, there are $19$ such representations, one for each prime in $\closedint {105} {208}$. {{ProofWanted|It remains to be shown that $210$ is the largest number that can be represented by the maximum of these.}}
We have: {{begin-eqn}} {{eqn | l = 25       | r = 4^2 + \left({2 \times 2^2}\right) + 1^2       | c = $4$ squares }} {{eqn | r = 3^2 + \left({4 \times 2^2}\right)       | c = $5$ squares }} {{eqn | r = \left({2 \times 3^2}\right) + 2^2 + \left({3 \times 1^2}\right)       | c = $6$ squares }} {{eqn | r = 4^2 + 2^2 + \left({5 \times 1^2}\right)       | c = $7$ squares }} {{eqn | r = 3^2 + \left({3 \times 2^2}\right) + \left({4 \times 1^2}\right)       | c = $8$ squares }} {{eqn | r = \left({2 \times 3^2}\right) + \left({7 \times 1^2}\right)       | c = $9$ squares }} {{eqn | r = 4^2 + \left({9 \times 1^2}\right)       | c = $10$ squares }} {{eqn | r = 3^2 + \left({2 \times 2^2}\right) + \left({8 \times 1^2}\right)       | c = $11$ squares }} {{end-eqn}} {{qed}}
Denote $\eta_{n + 1} = \eta_n + \alpha_{n + 1} \phi_{n + 1}$. For $\alpha_{n + 1} = 0$ we have: :$\eta_n = \eta_{n + 1}$ Suppose that $J \sqbrk {\eta_n}$ has been minimised {{WRT}} $\boldsymbol \alpha$. If: :$\exists \boldsymbol \alpha \in \R^n: \nexists \alpha_{n + 1} \ne 0: J \sqbrk {\eta_n} > J \sqbrk {\eta_{n + 1} }$ then $J \sqbrk {\eta_{n + 1} }$ is minimised for previously determined $\boldsymbol \alpha$ and $\alpha_{n + 1} = 0$: :$J \sqbrk {\eta_{n + 1} } = J \sqbrk {\eta_n}$ where $\eta_{n + 1} = \eta_n$. Suppose: :$\exists \boldsymbol \alpha \in \R^n: \exists \alpha_{n + 1} \ne 0: J \sqbrk {\eta_n} > J \sqbrk {\eta_{n + 1} }$ Then $\eta_{n + 1} \ne \eta_n$, because their respective $\boldsymbol \alpha$ differ by at least one value: $\alpha_{n + 1}$. Hence, for this supposition: :$J \sqbrk {\eta_{n + 1} } > J \sqbrk {\eta_n}$ Finally, both cases together imply that: :$J \sqbrk {\eta_{n + 1} } \ge J \sqbrk {\eta_n}$ or equivalently: :$\mu_n \ge \mu_{n + 1}$ {{qed}}
{{ProofWanted|Apparently it can be done using the result Cubes which are Sum of Three Cubes.}}
Suppose there were an order $2$ multiplicative magic square $M$. Let $a$ be the element of row $1$ and column $1$. Let $a b$ be the magic constant of $M$. Then $b$ is: :the element of row $1$ and column $2$, to make the product of row $1$ equal to $a b$ :the element of row $2$ and column $1$, to make the product of column $1$ equal to $a b$ and so on. But it is a principle of multiplicative magic squares that the elements are all distinct. Hence no order $2$ multiplicative magic square can exist. {{qed}} Category:Multiplicative Magic Squares by42pajlvdcgeda5on1eie60h053ebo
Let $y^*: \R \to \R$ be such that: :$\forall \epsilon > 0: J \sqbrk {y^*} < \mu + \epsilon$ By assumption of continuity of $J$: :$\forall \epsilon > 0: \exists \map \delta \epsilon > 0: \paren {\size {y - y^*} < \delta} \implies \paren {\size {J \sqbrk y - J \sqbrk {y^*} } < \epsilon}$ Let $\eta_n = \boldsymbol \alpha \boldsymbol \phi$, such that: :$\exists n \in\ N: \exists N \in \N: \paren {n > N} \implies \paren {\size {\eta_n - y^*} < \epsilon}$ where $\boldsymbol \alpha$ is an $n$-dimensional real vector. Let $y_n = \boldsymbol \alpha \boldsymbol \phi$, where $\boldsymbol \alpha$ is such that it minimises $J \sqbrk {y_n}$. Hence: :$J \sqbrk {y_n} < J \sqbrk {\eta_n} + \epsilon < \mu + 2 \epsilon$ On the other hand, by the definition of $y_n$: :$\mu \le J \sqbrk {y_n} \le J \sqbrk {\eta_n}$ Thus: :$\mu \le J \sqbrk {y_n} < \mu + 2 \epsilon$ If $J \sqbrk {y_n} = \mu$, then the inequality is satisfied, and $n$ is finite. {{explain|"finite" does not appear in the original formulation of GF; either restrict the theorem, or refine definitions used}} If $ \mu < J \sqbrk {y_n}$, then subtract $\mu$ from all the terms: :$0 < J \sqbrk {y_n} - \mu < 2 \epsilon$ We have that: :$\paren {\epsilon > 0} \implies \paren {-2 \epsilon < 0}$ Hence: :$-2 \epsilon < J \sqbrk {y_n} - \mu < 2 \epsilon$  or: :$\size {J \sqbrk {y_n} - \mu} < 2 \epsilon$ Since $\epsilon$ and $n$ inherits their constraints from the definitions of $\eta_n$ and the continuity of $J$, these together with the last inequality imply: :$\displaystyle \lim_{n \mathop \to \infty} J \sqbrk {y_n} = \mu$ where $J \sqbrk {y_n} = \mu_n$. {{qed}}
{{ProofWanted|First the concept of Fedorov group needs to be explored.}}
=== Definition 1 is equivalent to Definition 2 === Let $s \left({n}\right)$ denote the aliquot sum of (strictly) positive integer $n$. The sum of all the divisors of a (strictly) positive integer $n$ is $\sigma \left({n}\right)$, where $\sigma$ is the $\sigma$ function. The aliquot sum of $n$ is the sum of the divisors of $n$ with $n$ excluded. Thus: :$s \left({n}\right) = \sigma \left({n}\right) - n$ Suppose: :$s \left({n}\right) = m$ and: :$s \left({m}\right) = n$ Then: {{begin-eqn}} {{eqn | l = \sigma \left({n}\right) - n       | r = m       | c = {{Defof|Proper Divisor of Integer}} }} {{eqn | ll= \leadsto       | l = \sigma \left({n}\right)       | r = m + n       | c =  }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = \sigma \left({m}\right) - m       | r = n       | c = {{Defof|Proper Divisor of Integer}} }} {{eqn | ll= \leadsto       | l = \sigma \left({m}\right)       | r = m + n       | c =  }} {{end-eqn}} Thus: :$s \left({n}\right) = s \left({m}\right) = m + n$ The argument reverses. {{qed|lemma}} === Definition 1 is equivalent to Definition 3 === From the definition of definition 1 of an amicable pair: {{:Definition:Amicable Pair/Definition 1}} From the definition of a sociable chain: {{:Definition:Sociable Chain}} Here it is seen that setting $r = 2$ gives that: :$s \left({a_0}\right) = a_1$ :$s \left({a_1}\right) = a_0$ and the equivalence follows. {{qed}} Category:Amicable Pairs hj2x6uy537g5p9ne9incdjnvm7vk9x8
Let $r = 2^n a b, s = 2^n c$. Let $\map \sigma k$ denote the sigma function on an integer $k$. From Sigma Function of Power of 2: :$\map \sigma {2^n} = 2^{n + 1} - 1$ From Sigma Function of Prime Number: {{begin-eqn}} {{eqn | l = \map \sigma a       | r = 3 \times 2^n       | c =  }} {{eqn | l = \map \sigma b       | r = 3 \times 2^{n - 1}       | c =  }} {{eqn | l = \map \sigma c       | r = 9 \times 2^{2 n - 1}       | c =  }} {{end-eqn}} From Sigma Function is Multiplicative: {{begin-eqn}} {{eqn | l = \map \sigma r       | r = \map \sigma {2^n} \map \sigma a \map \sigma b       | c =  }} {{eqn | r = \paren {2^{n + 1} - 1} \paren {3 \times 2^n} \paren {3 \times 2^{n - 1} }       | c =  }} {{eqn | r = \paren {2^{n + 1} - 1} \paren {9 \times 2^{2 n - 1} }       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \map \sigma s       | r = \map \sigma {2^n} \map \sigma c       | c =  }} {{eqn | r = \paren {2^{n + 1} - 1} \paren {9 \times 2^{2 n - 1} }       | c =  }} {{eqn | r = \map \sigma r       | c =  }} {{end-eqn}} Thus it is seen that: :$\map \sigma r = \map \sigma s$ Now we have: {{begin-eqn}} {{eqn | l = r + s       | r = 2^n a b + 2^n c       | c =  }} {{eqn | r = 2^n \paren {\paren {3 \times 2^n - 1} \paren {3 \times 2^{n - 1} - 1} + 9 \times 2^{2 n - 1} - 1 }       | c =  }} {{eqn | r = 2^n \paren {9 \times 2^{2 n - 1} - 3 \times 2^n - 3 \times 2^{n - 1} + 1 + 9 \times 2^{2 n - 1} - 1}       | c =  }} {{eqn | r = 2^n \paren {2 \times 9 \times 2^{2 n - 1} - 3 \times 2 \times 2^{n - 1} - 3 \times 2^{n - 1} }       | c =  }} {{eqn | r = 2^{n + 1} \paren {9 \times 2^{2 n - 1} - 9 \times 2^{n - 2} }       | c =  }} {{eqn | r = 2^{n + 1} \paren {9 \times 2^{2 n - 1} } - 2^{n + 1} \paren {9 \times 2^{n - 2} }       | c =  }} {{eqn | r = 2^{n + 1} \paren {9 \times 2^{2 n - 1} } - \paren {9 \times 2^{2 n - 1} }       | c =  }} {{eqn | r = \paren {2^{n + 1} - 1} \paren {9 \times 2^{2 n - 1} }       | c =  }} {{end-eqn}} and so it is seen that: :$r + s = \map \sigma r = \map \sigma s$ Hence the result, by definition of amicable pair. {{qed}}
By Thabit's Rule: {{begin-eqn}} {{eqn | l = m_1       | r = 2^n \times \paren {3 \times 2^{n - 1} - 1} \times \paren {3 \times 2^n - 1}       | c =  }} {{eqn | l = m_2       | r = 2^n \times \paren {9 \times 2^{2 n - 1} - 1}       | c =  }} {{end-eqn}} for some $n \in \Z_{\ge 0}$. We have that: {{begin-eqn}} {{eqn | l = \paren {3 \times 2^{n - 1} - 1} \times \paren {3 \times 2^n - 1}       | r = 9 \times 2^{2 n - 1} - 3 \times 2^n - 3 \times 2^{n - 1} + 1       | c =  }} {{eqn | o = <       | r = 9 \times 2^{2 n - 1} - 1       | c =  }} {{end-eqn}} and so $m_1 < m_2$. Then: {{begin-eqn}} {{eqn | l = m_1       | r = 2^n \times \paren {3 \times 2^{n - 1} - 1} \times \paren {3 \times 2^n - 1}       | c =  }} {{eqn | r = \frac {2^n \times \paren {3 \times 2^{n - 1} - 1} \times \paren {3 \times 2^n - 1} \times 6} 6       | c =  }} {{eqn | r = \frac {\paren {3 \times 2^n} \times \paren {2 \times 3 \times 2^{n - 1} - 2} \times \paren {3 \times 2^n - 1} } 6       | c =  }} {{eqn | r = \frac {\paren {3 \times 2^n} \times \paren {3 \times 2^n - 1} \times \paren {3 \times 2^n - 2} } 6       | c =  }} {{eqn | r = \frac {k \times \paren {k + 1} \times \paren {k + 2} } 6       | c = where $k = 3 \times 2^n - 2$ }} {{end-eqn}} From Closed Form for Tetrahedral Numbers, this is the $k$th tetrahedral number where $k = 3 \times 2^n - 2$. {{qed}}
An amicable pair must be formed from a smaller abundant number and a larger deficient number. Suppose both $m_1, m_2$ are divisible by $3$. Since both are even, they must also be divisible by $6$. However $6$ is a perfect number. By Multiple of Perfect Number is Abundant, neither can be deficient. So $m_1, m_2$ cannot form an amicable pair. Therefore at most one of them is divisible by $3$. {{WLOG}} suppose $m_1$ is divisible by $3$. Write: :$m_1 = 2^r 3^t a, m_2 = 2^s b$ where $a, b$ are not divisible by $2$ or $3$. Then: {{begin-eqn}} {{eqn | l = m + n       | r = \map \sigma {m_2}       | c = {{Defof|Amicable Pair|subdef = Definition 2}} }} {{eqn | r = \map \sigma {2^s} \map \sigma b       | c = Sigma Function is Multiplicative }} {{eqn | r = \paren {2^{s + 1} - 1} \map \sigma b       | c = Sigma Function of Power of Prime }} {{eqn | o = \equiv       | r = \paren {\paren {-1}^{s + 1} - 1} \map \sigma b       | rr = \pmod 3       | c = Congruence of Powers }} {{end-eqn}} Since $m + n$ is not divisible by $3$, $s$ must be even. Similarly, by Sigma Function is Multiplicative, $t$ must also be even. In particular, both $s, t$ are at least $2$. Now write: :$m_1 = 2^2 \cdot 3 k, m_2 = 2^2 \cdot l$ where $k, l$ are some integers. By Multiple of Perfect Number is Abundant, $m_1$ is abundant number. Therefore $m_2 > m_1$. This leads to $l > 3 k \ge 3$. By Abundancy Index of Product is greater than Abundancy Index of Proper Factors: {{begin-eqn}} {{eqn | l = \frac {\map \sigma {m_1} } {m_1}       | o = \ge       | r = \frac {\map \sigma {12} } {12}       | c = equality occurs {{iff}} $m_1 = 12$ }} {{eqn | r = \frac 7 3 }} {{eqn | l = \frac {\map \sigma {m_2} } {m_2}       | o = >       | r = \frac {\map \sigma 4} 4 }} {{eqn | r = \frac 7 4 }} {{end-eqn}} But: {{begin-eqn}} {{eqn | l = 1       | r = \frac {m_1 + m_2} {m_1 + m_2} }} {{eqn | r = \frac {m_1} {\map \sigma {m_1} } + \frac {m_2} {\map \sigma {m_2} }       | c = {{Defof|Amicable Pair|subdef = Definition 2}} }} {{eqn | o = <       | r = \frac 3 7 + \frac 4 7 }} {{eqn | r = 1 }} {{end-eqn}} which is a contradiction. Therefore neither $m_1$ nor $m_2$ is divisible by $3$. {{qed}}
For $n \in \Z_{>0}$, let $s \left({n}\right)$ denote the aliquot sum of (strictly) positive integer $n$. The sum of all the divisors of a (strictly) positive integer $n$ is $\sigma \left({n}\right)$, where $\sigma$ is the $\sigma$ function. The aliquot sum of $n$ is the sum of the divisors of $n$ with $n$ excluded. Thus: :$s \left({n}\right) = \sigma \left({n}\right) - n$ Suppose: {{begin-eqn}} {{eqn | l = s \left({m_1}\right)       | r = m_2 + m_3 }} {{eqn | l = s \left({m_2}\right)       | r = m_1 + m_3 }} {{eqn | l = s \left({m_3}\right)       | r = m_1 + m_2 }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \sigma \left({m_1}\right) - m_1       | r = m_2 + m_3       | c = {{Defof|Proper Divisor of Integer}} }} {{eqn | ll= \leadsto       | l = \sigma \left({m_1}\right)       | r = m_1 + m_2 + m_3       | c =  }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = \sigma \left({m_2}\right) - m_2       | r = m_1 + m_3       | c = {{Defof|Proper Divisor of Integer}} }} {{eqn | ll= \leadsto       | l = \sigma \left({m_2}\right)       | r = m_1 + m_2 + m_3       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \sigma \left({m_3}\right) - m_3       | r = m_1 + m_2       | c = {{Defof|Proper Divisor of Integer}} }} {{eqn | ll= \leadsto       | l = \sigma \left({m_3}\right)       | r = m_1 + m_2 + m_3       | c =  }} {{end-eqn}} Thus: :$\sigma \left({m_1}\right) = \sigma \left({m_2}\right) = \sigma \left({m_3}\right) = m_1 + m_2 + m_3$ The argument reverses. {{qed}} Category:Amicable Triplets qc6ebahwfbfp6nhfz4z3ayqnqab1fcm
$4$ consecutive positive integers will be in the forms: {{begin-eqn}} {{eqn | l = n_0       | o = \equiv       | r = 0       | rr= \pmod 4       | c =  }} {{eqn | l = n_1       | o = \equiv       | r = 1       | rr= \pmod 4       | c =  }} {{eqn | l = n_2       | o = \equiv       | r = 2       | rr= \pmod 4       | c =  }} {{eqn | l = n_3       | o = \equiv       | r = 3       | rr= \pmod 4       | c =  }} {{end-eqn}} in some order. But from Sum of Two Squares not Congruent to 3 modulo 4, $n_3$ cannot be the sum of two squares. The result follows. {{qed}}
=== Sufficient Condition === Assume that: :$X$ is directed in $\struct {S, \preceq_1}$ Thus $X$ is non-empty. Let $x, y \in S$. By definition of directed: :$\exists z \in S: x \preceq_1 z \land y \preceq_1 z$ Thus by definition of dual ordered set: :$z \preceq_2 x \land z \preceq_2 y$ {{qed|lemma}} === Necessary Condition === Assume that: :$X$ is filtered in $\struct {S, \preceq_2}$ Thus $X$ is non-empty. Let $x, y \in S$. By definition of filtered: :$\exists z \in S: z \preceq_2 x \land z \preceq_2 y$ Thus by definition of dual ordered set: :$x \preceq_1 z \land y \preceq_1 z$ {{qed}}
Let $n \equiv 3 \pmod 4$. {{AimForCont}} $n$ can be expressed as the sum of two squares: :$n = a^2 + b^2$. From Square Modulo 4, either $a^2 \equiv 0$ or $a^2 \equiv 1 \pmod 4$. Similarly for $b^2$. So $a^2 + b^2 \not \equiv 3 \pmod 4$ whatever $a$ and $b$ are. Thus $n$ cannot be the sum of two squares. The result follows by Proof by Contradiction. {{qed}} Category:Sums of Squares jm7po2fj9tqvqo4v2sirg2sru4hrvcp
We note that for $n = 239$ we have: {{begin-eqn}} {{eqn | r = 239^2 + 1       | o =        | c =  }} {{eqn | r = 57122       | c =  }} {{eqn | r = 2 \times 13^4       | c =  }} {{end-eqn}} Thus the largest prime factor of $239^2 + 1$ is $13$. {{ProofWanted|Now the rest of the result is to be shown.}}
Setting $x = 1$: {{begin-eqn}} {{eqn | r = 1^2 + 1       | o =        | c =  }} {{eqn | r = 2       | c =  }} {{eqn | r = 2 \times 1^4       | c =  }} {{end-eqn}} and so $y = 1$. Setting $x = 239$: {{begin-eqn}} {{eqn | r = 239^2 + 1       | o =        | c =  }} {{eqn | r = 57122       | c =  }} {{eqn | r = 2 \times 13^4       | c =  }} {{end-eqn}} and so $y = 13$. {{ProofWanted|It remains to be shown this is the only solution. Perhaps trial and error for $y$ going from $0$ up to $13$ and then using Largest Prime Factor of $n^2 + 1$?}}
Let $\map \tau n$ denote the $\tau$ (tau) function of $n$: :$\map \tau n$ is the number of divisors of $n$. Then: {{begin-eqn}} {{eqn | l = \map \tau {720 \, 720}       | r = 240       | c = {{TauLink|720,720|720 \, 720}} }} {{eqn | l = \map \tau {831 \, 600}       | r = 240       | c = {{TauLink|831,600|831 \, 600}} }} {{eqn | l = \map \tau {942 \, 480}       | r = 240       | c = {{TauLink|942,480|942 \, 480}} }} {{eqn | l = \map \tau {982 \, 800}       | r = 240       | c = {{TauLink|982,800|982 \, 800}} }} {{eqn | l = \map \tau {997 \, 920}       | r = 240       | c = {{TauLink|997,920|997 \, 920}} }} {{end-eqn}} {{ProofWanted|Prove that these are indeed the smallest}}
{{begin-eqn}} {{eqn | l = p_n       | r = \dfrac {p_{n - 1} + p_{n + 1} } 2       | c = Definition 1 }} {{eqn | ll= \iff       | l = 2 p_n       | r = p_{n - 1} + p_{n + 1}       | c =  }} {{eqn | ll= \iff       | l = p_n - p_{n - 1}       | r = p_{n + 1} - p_n       | c = Definition 3: equal prime gaps }} {{eqn | r = d       | c = for some $d \in \Z$ }} {{eqn | ll= \iff       | l = p_{n - 1} + d       | r = p_n       | c = Definition 2 }} {{eqn | l = p_{n - 1} + 2 d       | r = p_{n + 1}       | c =  }} {{end-eqn}} {{qed}} Category:Balanced Primes a8nmmveebhimoee5r36swim77fm2unp
By Dual of Dual Ordering: :$\left({S, \preceq_1}\right)$ is dual of $\left({S, \preceq_2}\right)$ Thus by Directed iff Filtered in Dual Ordered Set: :$X$ is filtered in $\left({S, \preceq_1}\right)$ {{iff}} $X$ is directed in $\left({S, \preceq_2}\right)$ {{qed}}
Let $x$ be a set. By definition of $\mathit{Filt}$: :$x \in \mathit{Filt}\left({L_1}\right) \iff x$ is a filter on $L_1$. By Filter is Ideal in Dual Ordered Set: :$x \in \mathit{Filt}\left({L_1}\right) \iff x$ is an ideal in $L_2$. By definition of $\mathit{Ids}$: :$x \in \mathit{Filt}\left({L_1}\right) \iff x \in \mathit{Ids}\left({L_2}\right)$ Hence by definition of set equality: :$\mathit{Filt}\left({L_1}\right) = \mathit{Ids}\left({L_2}\right)$ {{qed}}
=== Sufficient Condition === Let this congruence hold: :$3^{\paren {F_n - 1} / 2} \equiv -1 \pmod {F_n}$ Then: :$3^{F_n - 1} \equiv 1 \pmod {F_n}$ Thus the multiplicative order of $3$ modulo $F_n$ is a divisor of $F_n - 1 = 2^{2^n}$. This is a power of $2$. On the other hand, the multiplicative order of $3$ modulo $F_n$ is not a divisor of $\dfrac {F_n - 1} 2$. Therefore it must be equal to $F_n - 1$. In particular, there are at least $F_n - 1$ numbers below $F_n$ which are coprime to $F_n<$. This can happen only if $F_n$ is prime. {{qed|lemma}} === Necessary Condition === Let $F_n$ be prime. By Euler's Criterion: :$3^{\paren {F_n - 1} / 2} \equiv \paren {\dfrac 3 {F_n} } \pmod {F_n}$ $\left({\dfrac 3 {F_n} }\right)$ denotes the Legendre symbol. By repeated squaring: :$2^{2^n} \equiv 1 \pmod 3$ Thus: :$F_n \equiv 2 \pmod 3$ and so: :$\paren {\dfrac {F_n} 3} = -1$ As $F_n \equiv 1 \pmod 4$, it follows from the Law of Quadratic Reciprocity that: :$\paren {\dfrac 3 {F_n} } = -1$ {{qed}} {{Namedfor|Jean François Théophile Pépin|cat = Pépin}}