hcju/proofwikiqa huggingface dataset:
We have that: :$\dfrac {2^k} {10^{k + 1} } = \dfrac 1 {10} \paren {\dfrac 1 5}^k$ and so by Sum of Infinite Geometric Sequence: {{begin-eqn}} {{eqn | l = \sum_{k \mathop \ge 0} \dfrac {2^k} {10^{k + 1} }       | r = \dfrac {1/10} {1 - 1/5}       | c =  }} {{eqn | r = \dfrac 1 {10 \paren {4 / 5} }       | c =  }} {{eqn | r = \dfrac 5 {40}       | c =  }} {{eqn | r = \dfrac 1 8       | c =  }} {{end-eqn}} {{qed}}
From Heron's Formula, the area $A$ of $\triangle ABC$ is given by: :$A = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$ where $s = \dfrac{a + b + c} 2$ is the semiperimeter of $\triangle ABC$. For $3, 4, 5$: {{begin-eqn}} {{eqn | l = s       | r = \frac {3 + 4 + 5} 2       | c =  }} {{eqn | r = 6       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = \sqrt {6 \paren {6 - 3} \paren {6 - 4} \paren {6 - 5} }       | c =  }} {{eqn | r = \sqrt {6 \times 3 \times 2 \times 1}       | c =  }} {{eqn | r = \sqrt {6 \times 6}       | c =  }} {{eqn | r = 6       | c =  }} {{end-eqn}} For $13, 14, 15$: {{begin-eqn}} {{eqn | l = s       | r = \frac {13 + 14 + 15} 2       | c =  }} {{eqn | r = 21       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = \sqrt {21 \paren {21 - 13} \paren {21 - 14} \paren {21 - 15} }       | c =  }} {{eqn | r = \sqrt {21 \times 8 \times 7 \times 6}       | c =  }} {{eqn | r = \sqrt {\paren {3 \times 7} \times 2^3 \times 7 \times \paren {2 \times 3} }       | c =  }} {{eqn | r = 3 \times 7 \times 2^2       | c =  }} {{eqn | r = 84       | c =  }} {{end-eqn}} For $15, 28, 41$: {{begin-eqn}} {{eqn | l = s       | r = \frac {15 + 28 + 41} 2       | c =  }} {{eqn | r = 42       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = \sqrt {42 \paren {42 - 15} \paren {42 - 28} \paren {42 - 41} }       | c =  }} {{eqn | r = \sqrt {42 \times 27 \times 14 \times 1}       | c =  }} {{eqn | r = \sqrt {\paren {2 \times 3 \times 7} \times 3^3 \times \paren {2 \times 7} }       | c =  }} {{eqn | r = 2 \times 3^2 \times 7       | c =  }} {{eqn | r = 126       | c =  }} {{end-eqn}} For $15, 26, 37$: {{begin-eqn}} {{eqn | l = s       | r = \frac {15 + 26 + 37} 2       | c =  }} {{eqn | r = 39       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = \sqrt {39 \paren {39 - 15} \paren {39 - 26} \paren {39  - 37} }       | c =  }} {{eqn | r = \sqrt {39 \times 24 \times 13 \times 2}       | c =  }} {{eqn | r = \sqrt {\paren {3 \times 13} \times \paren {2^3 \times 3} \times 13 \times 2}       | c =  }} {{eqn | r = 2^2 \times 3 \times 13       | c =  }} {{eqn | r = 156       | c =  }} {{end-eqn}} {{expand|The citation below generalises this result, and this page could be turned into it.}}
=== Definition 1 implies Definition 2 === By Definition 1: :$\map E {x, \mathbf y, \mathbf z, \mathbf w} = \map F {x, \mathbf y, \mathbf w} - \map F {x, \mathbf y, \mathbf z} + \paren {\mathbf w - \mathbf z} \map {F_{\mathbf y'} } {x, \mathbf y, \mathbf z}$ By Taylor's Theorem, where expansion is done around $\mathbf w = \mathbf z$ and Lagrange form of remainder is used: :$\ds \map F {x, \mathbf y, \mathbf w} = \map F {x, \mathbf y, \mathbf z} + \frac {\partial \map F {x, \mathbf y, \mathbf z} } {\partial \mathbf y'} \paren {\mathbf w - \mathbf z} + \frac 1 2 \sum_{i, j \mathop = 1}^n \paren {w_i - z_i} \paren {w_j - z_j} \frac {\partial^2 \map F {x, \mathbf y, \mathbf z + \theta \paren {\mathbf z - \mathbf w} } } {\partial y_i'y_j'}$ where $\theta \in \R: 0 < \theta < 1$. Insertion of this expansion into the definition for Weierstrass E-Function leads to the desired result. {{qed|lemma}} === Definition 2 implies Definition 1 === {{ProofWanted}}
By definition, the increment of $J$ is: :$\ds \Delta J = \int_{\gamma^*} \map F {x, \mathbf y, \mathbf y'} \rd x - \int_\gamma \map F {x, \mathbf y, \mathbf y'} \rd x$ where $\gamma$ and $\gamma^*$ are curves described by $\paren {x, \map {\mathbf y} x}$ and $\paren {x, \paren {\mathbf y^*} x}$ respectively, such that $\map {\mathbf y^*} x - \map {\mathbf y} x = \map {\mathbf h} x$. Consider Hilbert's invariant integral: {{begin-eqn}} {{eqn | l = \map g {x, \mathbf y}       | r = \int_\Gamma \paren {-H \rd x + \mathbf p \rd \mathbf y} }} {{eqn | r = \int_\Gamma \paren {-\paren {-F + \mathbf y' F_{\mathbf y'} } \rd x + F_{\mathbf y'} \rd \mathbf y}       | c = {{Defof|Hamiltonian}} and Momentum }} {{eqn | r = \int_\Gamma \paren {F \rd x - \mathbf y' F_{\mathbf y'} \rd x + F_{\mathbf y'} \rd \mathbf y} }} {{eqn | r = \int_{\Gamma} \paren {F \rd x - \boldsymbol \psi F_{\mathbf y'} \rd x + F_{\mathbf y'} \rd \mathbf y}       | c = $\mathbf y' = \map {\boldsymbol \psi} {x, \mathbf y}$ }} {{end-eqn}} Since the integrand is full differential, the integral does not depend on the shape of $ \Gamma $, but only on its endpoints. Therefore, for $\Gamma = \gamma$ and $\Gamma = \gamma^*$ the value of the integral is the same. Since $\map {\mathbf y'} x = \map {\boldsymbol \psi} {x, \mathbf y}$ determines boundary conditions for $\gamma$, $\Gamma = \gamma$ is one of the trajectories of the field $\map {\mathbf y'} x = \map {\boldsymbol \psi} {x, \mathbf y}$. Hence, $\d \mathbf y$ is constrained by $\d \mathbf y = \boldsymbol \psi \rd x$: :$\ds \map g {x, \mathbf y} = \int_\gamma \map F {x, \mathbf y, \mathbf y'} \rd x$ Thus, $\map g {x, \mathbf y'}$ can be written in two different ways. :$\ds \int_\gamma \map F {x, \mathbf y, \mathbf y'} \rd x = \int_{\gamma^*} \paren {\paren {\map F {x, \mathbf y, \boldsymbol \psi} - \boldsymbol \psi F_{\mathbf y'} \paren {x, \mathbf y, \boldsymbol \psi} } \rd x + F_{\mathbf y'} \paren {x, \mathbf y, \boldsymbol \psi} \rd \mathbf y}$ Substitute this into the expression for $\Delta J$: {{begin-eqn}} {{eqn | l = \Delta J       | r = \int_{\gamma^*} \map F {x, \mathbf y, \mathbf y'} \rd x - \int_{\gamma^*} \paren {\paren {\map F {x, \mathbf y, \boldsymbol \psi} - \boldsymbol \psi \map {F_{\mathbf y'} } {x, \mathbf y, \boldsymbol \psi} }\rd x + \map {F_{\mathbf y'} } {x, \mathbf y, \boldsymbol \psi} \rd \mathbf y} }} {{eqn | r = \int_{\gamma^*} \paren {\map F {x, \mathbf y, \mathbf y'} - \map F {x, \mathbf y, \boldsymbol \psi} - \paren {\mathbf y'-\boldsymbol \psi} \map {F_{\mathbf y'} } {x, \mathbf y, \boldsymbol \psi} }\rd x       | c = as $\d \mathbf y = \boldsymbol \psi \rd x$ }} {{eqn | r = \int_{\gamma^*} \map E {x, \mathbf y, \boldsymbol\psi, \mathbf y'} \rd x       | c = {{Defof|Weierstrass E-Function}} }} {{end-eqn}} By assumption: :$\map E {x, \mathbf y, \boldsymbol \psi, \mathbf y'} \ge 0$ Hence the integrand is bounded below by $0$ and above by its maximum in the interval of integration. Hence, the integral is bounded below by $0$ and above by some positive number. Therefore: :$\Delta J \ge 0$ {{qed}} {{improve|Improve clarity, revisit linked pages}}
Let $q \divides M_p$. Then: :$2^p \equiv 1 \pmod q$ From Integer to Power of Multiple of Order, the multiplicative order of $2 \pmod q$ divides $p$. By Fermat's Little Theorem, the multiplicative order of $2 \pmod q$ also divides $q - 1$. Hence: :$q - 1 = 2 k p$ {{qed}}
Let $q \mathrel \backslash M_p$. From Factor of Mersenne Number $M_p$ is of form $2 k p + 1$:  :$q = 2 k p + 1$ and so by definition of congruence modulo an integer: :$q \equiv 1 \pmod p$ immediately. {{qed}}
Suppose $q \mathop \backslash M_p$. From Factor of Mersenne Number $M_p$ is of form $2 k p + 1$:  :$q - 1 = 2 k p$ From above: :$2^{\left({q - 1}\right) / 2} \equiv 2 k p \equiv 1 \pmod q$ and so $2$ is a quadratic residue $\pmod q$. From Second Supplement to Law of Quadratic Reciprocity: :$q \equiv \pm 1 \pmod 8$ {{qed}}
{{finish|a) Demonstration that these cannot be so expressed, b) demonstration that all others below 324 can be so expressed}} It will be proved that the largest integer which cannot be expressed as the sum of distinct squares is $128$. The remaining integers in the sequence can be identified by inspection. We prove this using a variant of Second Principle of Mathematical Induction. Let $\map P n$ be the proposition: :$n$ can be expressed as the sum of distinct squares. === Basis for the induction === We verify the result up to $18^2 = 324$. Then we conclude all $n$ with $129 \le n \le 324$ can be expressed as the sum of distinct squares. So $\map P n$ is true for all $129 \le n \le 324$. This is the basis for the induction. === Induction Hypothesis === Suppose for some $k > 324$, $\map P j$ is true for all $129 \le j < k$. That is, every integer between $129$ and $k - 1$ can be expressed as the sum of distinct squares. This is the induction hypothesis. === Induction Step === This is the induction step: We find the largest integer $i$ such that $i^2 < k \le \paren {i + 1}^2$. We show that $k - \paren {i - 4}^2$ can be expressed as the sum of distinct squares, and the sum does not involve $\paren {i - 4}^2$. Then $k$ can be expressed as the sum of distinct squares. Since $k > 324 = 18^2$, we must have $i \ge 18$. Hence $k > k - \paren {i - 4}^2 > i^2 - \paren {i - 4}^2 = 4 \paren {2 i - 4} \ge 4 \times 32 = 128$. Thus $k - \paren {i - 4}^2$ can be expressed as the sum of distinct squares by Induction Hypothesis. A sufficient condition such that the sum does not involve $\paren {i - 4}^2$ is $\paren {i - 4}^2 > k - \paren {i - 4}^2$. We have: {{begin-eqn}} {{eqn | l = \paren {i - 4}^2 - \paren {k - \paren {i - 4}^2 }       | o = \ge       | r = 2 \paren {i - 4}^2 - \paren {i + 1}^2       | c = From $k \le T_{i + 1}$ }} {{eqn | r = 2 i^2 - 16 i + 32 - i^2 - 2 i - 1 }} {{eqn | r = i^2 - 18 i + 31 }} {{eqn | r = i \paren {i - 18} + 31 }} {{eqn | o = \ge       | r = 31       | c = From $i \ge 18$ }} {{eqn | o = >       | r = 0 }} {{end-eqn}} Therefore we have $\paren {i - 4}^2 > k - \paren {i - 4}^2$. Hence $\map P k$ is true. By the Second Principle of Mathematical Induction, $\map P n$ is true for all $n \ge 128$. Thus every integer greater than $128$ can be expressed as the sum of distinct square numbers. {{Proofread}}
It is necessary to postulate that such (positive) integers have $3$ digits or more, as the $2$ digit solution is trivial. Let $n = \sqbrk {abc}$ be a $3$-digit number. Let $\map s n$ denote the sum of all the $2$-digit (positive) integers formed from the digits of $n$. Then: {{begin-eqn}} {{eqn | l = \map s n       | r = \sqbrk {ab} + \sqbrk {ac} + \sqbrk {bc} + \sqbrk {ba} + \sqbrk {ca} + \sqbrk {cb}       | c =  }} {{eqn | r = \paren {10 a + b} + \paren {10 a + c} + \paren {10 b + c} + \paren {10 b + a} + \paren {10 c + a} + \paren {10 c + b}       | c =  }} {{eqn | r = 2 \times \paren {10 a + a} + 2 \times \paren {10 b + b} + 2 \times \paren {10 c + c}       | c =  }} {{eqn | r = 22 \times \paren {a + b + c}       | c =  }} {{end-eqn}} So for $n = \map s n$, $n = \sqbrk {abc}$ needs to be divisible by both $22$ and $a + b + c$. {{begin-eqn}} {{eqn | l = 22 \times 4       | r = 88       | c = too small }} {{eqn | l = 22 \times 5       | r = 110       | c = but $\map s {110} = 11 + 10 + 01 = 22$ }} {{eqn | l = 22 \times 6       | r = 132       | c = and $\map s {132} = 12 + 13 + 21 + 23 + 31 + 32 = 132$ }} {{end-eqn}} and the result is apparent. {{qed}}
Single digit integers are trivial: :$d^1 = d$ for all $d \in \Z$. Then we have: {{begin-eqn}} {{eqn | l = 8^1 + 9^2       | r = 8 + 81       | rr= = 89       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 1^1 + 3^2 + 5^3       | r = 1 + 9 + 125       | rr= = 135       | c =  }} {{eqn | l = 1^1 + 7^2 + 5^3       | r = 1 + 49 + 125       | rr= = 175       | c =  }} {{eqn | l = 5^1 + 1^2 + 8^3       | r = 5 + 1 + 512       | rr= = 518       | c =  }} {{eqn | l = 5^1 + 9^2 + 8^3       | r = 5 + 81 + 512       | rr= = 598       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 1^1 + 3^2 + 0^3 + 6^4       | r = 1 + 9 + 0 + 1296       | rr= = 1306       | c =  }} {{eqn | l = 1^1 + 6^2 + 7^3 + 6^4       | r = 1 + 36 + 343 + 1296       | rr= = 1676       | c =  }} {{eqn | l = 2^1 + 4^2 + 2^3 + 7^4       | r = 2 + 16 + 8 + 2401       | rr= = 2427       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | r = 2^1 + 6^2 + 4^3 + 6^4 + 7^5 + 9^6 + 8^7        | o =        | c =  }} {{eqn | r = 2 + 36 + 64 + 1296 + 16 \, 807 + 531 \, 441 + 1 \, 097 \, 152       | c =  }} {{eqn | r = 2 \, 646 \, 798       | c =  }} {{end-eqn}} and finally: {{begin-eqn}} {{eqn | r = 1 + 2^2 + 1^3 + 5^4 + 7^5 + 6^6 + 9^7 + 2^8 + 6^9 + 2^{10} + 2^{11} + 0^{12} + 3^{13} + 9^{14} + 6^{15} + 2^{16} + 3^{17} + 5^{18} + 3^{19} + 9^{20}       | o =        | c =  }} {{eqn | r = 1 + 4 + 1 + 625 + 16 \, 807 + 46 \, 656 + 4 \, 782 \, 969 + 256 + 10 \, 077 \, 696 + 1024 + 2048 + 0 + 1 \, 594 \, 323       | c =  }} {{eqn | o =        | ro= +        | r = 22 \, 876 \, 792 \, 454 \, 961 + 470 \, 184 \, 984 \, 576 + 65 \, 536 + 129 \, 140 \, 163 + 3 \, 814 \, 697 \, 265 \, 625       | c =  }} {{eqn | o =        | ro= +        | r = 1 \, 162 \, 261 \, 467 + 12 \, 157 \, 665 \, 459 \, 056 \, 928 \, 801       | c =  }} {{eqn | r = 12 \, 157 \, 692 \, 622 \, 039 \, 623 \, 539       | c =  }} {{end-eqn}} {{ProofWanted|It remains to be proved that there are no more like this.}}
By definition of harmonic mean: :$\dfrac 1 {H \left({n}\right)} = \dfrac 1 {\tau \left({n}\right)} \left({\displaystyle \sum_{d \mathop \backslash n} \dfrac 1 d}\right)$ From Sum of Reciprocals of Divisors equals Abundancy Index: : $\displaystyle \sum_{d \mathop \backslash n} \frac 1 d = \frac {\sigma \left({n}\right)} n$ and so: :$\dfrac 1 {H \left({n}\right)} = \dfrac {\sigma \left({n}\right)} {n \tau \left({n}\right)}$ Hence the result. {{qed}}
From Harmonic Mean of Divisors in terms of Tau and Sigma, the harmonic mean of the divisors of $n$ is given by: :$H \left({n}\right) = \dfrac {n \tau \left({n}\right)} {\sigma \left({n}\right)}$ where: :$\tau \left({n}\right)$ denotes the $\tau$ (tau) function: the number of divisors of $n$ :$\sigma \left({n}\right)$ denotes the $\sigma$ (sigma) function: the sum of the divisors of $n$. Let $n$ be a perfect number. By definition of perfect number: :$\dfrac {\sigma \left({n}\right)} n = 2$ From Tau Function Odd Iff Argument is Square: :$\tau \left({n}\right) = 2 k$ for some $k \in \Z$. Hence: :$H \left({n}\right) = \dfrac {2 k} 2 = k$ Hence the result. {{qed}}
Note the integers whose harmonic mean of their divisors are integers are the Ore numbers: :$1, 6, 28, 140, 270, 496, 672, \ldots$ {{OEIS|A001599}} It remains to calculate the arithmetic mean of their divisors. Let $A \left({n}\right)$ denote the arithmetic mean of the divisors of $n$. Then we have: :$A \left({n}\right) = \dfrac {\sigma \left({n}\right)} {\tau \left({n}\right)}$ where: :$\tau \left({n}\right)$ denotes the $\tau$ (tau) function: the number of divisors of $n$ :$\sigma \left({n}\right)$ denotes the $\sigma$ (sigma) function: the sum of the divisors of $n$. {{begin-eqn}} {{eqn | l = \sigma \left({6}\right)       | r = 12       | c = {{SigmaLink|6}} }} {{eqn | l = \tau \left({6}\right)       | r = 4       | c = {{TauLink|6}} }} {{eqn | ll= \leadsto       | l = A \left({6}\right)       | r = \dfrac {12} 4 = 3       | c = and so the arithmetic mean is an integer }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \sigma \left({28}\right)       | r = 56       | c = {{SigmaLink|28}} }} {{eqn | l = \tau \left({28}\right)       | r = 4       | c = {{TauLink|28}} }} {{eqn | ll= \leadsto       | l = A \left({28}\right)       | r = \dfrac {56} 6 = 9 \cdotp \dot 3       | c = and so the arithmetic mean is not an integer }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \sigma \left({140}\right)       | r = 336       | c = {{SigmaLink|140}} }} {{eqn | l = \tau \left({140}\right)       | r = 12       | c = {{TauLink|140}} }} {{eqn | ll= \leadsto       | l = A \left({140}\right)       | r = \dfrac {336} {12} = 28       | c = and so the arithmetic mean is an integer }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \sigma \left({270}\right)       | r = 720       | c = {{SigmaLink|270}} }} {{eqn | l = \tau \left({270}\right)       | r = 16       | c = {{TauLink|270}} }} {{eqn | ll= \leadsto       | l = A \left({270}\right)       | r = \dfrac {720} {16} = 45       | c = and so the arithmetic mean is an integer }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \sigma \left({496}\right)       | r = 992       | c = {{SigmaLink|496}} }} {{eqn | l = \tau \left({496}\right)       | r = 10       | c = {{TauLink|496}} }} {{eqn | ll= \leadsto       | l = A \left({496}\right)       | r = \dfrac {992} {10} = 9 \cdotp 92       | c = and so the arithmetic mean is not an integer }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \sigma \left({672}\right)       | r = 2016       | c = {{SigmaLink|672}} }} {{eqn | l = \tau \left({672}\right)       | r = 24       | c = {{TauLink|672}} }} {{eqn | ll= \leadsto       | l = A \left({672}\right)       | r = \dfrac {2016} {24} = 84       | c = and so the arithmetic mean is an integer }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \sigma \left({1638}\right)       | r = 4368       | c = {{SigmaLink|1638}} }} {{eqn | l = \tau \left({1638}\right)       | r = 24       | c = {{TauLink|1638}} }} {{eqn | ll= \leadsto       | l = A \left({1638}\right)       | r = \dfrac {4368} {24} = 182       | c = and so the arithmetic mean is an integer }} {{end-eqn}} {{qed}}
We have that: :$1$ has no prime factors. Hence, vacuously, $1$ has no primitive prime factors. :$8 = 2^3$ and $2 \divides 2 = F_3$ :$144 = 2^4 3^2$ and: :$2 \divides 8 = F_6$ :$3 \divides 21 = F_8$ for example. {{ProofWanted|It remains to be shown that these are the only examples.}}
The smallest prime magic square whose elements are consecutive odd primes is: {{:Prime Magic Square/Examples/Order 3/Smallest with Consecutive Primes}} As can be seen by inspection, the sums of the elements in the rows, columns and Diagonal of Array is $4 \, 440 \, 084 \, 513$: {{begin-eqn}} {{eqn | l = 1 \, 480 \, 028 \, 159 + 1 \, 480 \, 028 \, 153 + 1 \, 480 \, 028 \, 201       | r = 4 \, 440 \, 084 \, 513 }} {{eqn | l = 1 \, 480 \, 028 \, 213 + 1 \, 480 \, 028 \, 171 + 1 \, 480 \, 028 \, 129       | r = 4 \, 440 \, 084 \, 513 }} {{eqn | l = 1 \, 480 \, 028 \, 141 + 1 \, 480 \, 028 \, 189 + 1 \, 480 \, 028 \, 183       | r = 4 \, 440 \, 084 \, 513 }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 1 \, 480 \, 028 \, 159 + 1 \, 480 \, 028 \, 213 + 1 \, 480 \, 028 \, 141       | r = 4 \, 440 \, 084 \, 513 }} {{eqn | l = 1 \, 480 \, 028 \, 153 + 1 \, 480 \, 028 \, 171 + 1 \, 480 \, 028 \, 189       | r = 4 \, 440 \, 084 \, 513 }} {{eqn | l = 1 \, 480 \, 028 \, 201 + 1 \, 480 \, 028 \, 129 + 1 \, 480 \, 028 \, 183       | r = 4 \, 440 \, 084 \, 513 }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 1 \, 480 \, 028 \, 159 + 1 \, 480 \, 028 \, 171 + 1 \, 480 \, 028 \, 183       | r = 4 \, 440 \, 084 \, 513 }} {{eqn | l = 1 \, 480 \, 028 \, 141 + 1 \, 480 \, 028 \, 171 + 1 \, 480 \, 028 \, 201       | r = 4 \, 440 \, 084 \, 513 }} {{end-eqn}} {{ProofWanted|It remains to be demonstrated that there are no prime magic squares that are any smaller than this.}} Category:Prime Magic Squares Category:4,440,084,513 h04gygpinc4wf0ihrubo1lf8vghj185
The smallest prime magic square whose elements are the first consecutive odd primes is: {{:Prime Magic Square/Examples/Order 12/Smallest with Consecutive Primes from 3}} The fact that this is the smallest is proven here. The sum of the first $144$ prime numbers can either be calculated or looked up: it is $54 \, 169$. We see that $2$ is not included in this prime magic square, but instead $1$ is used. So the total of all the elements of this prime magic square is $54 \, 169 - 2 + 1 = 54 \, 168$. There are $12$ rows, and $12$ columns, each with the same magic constant. This magic constant must therefore be $\dfrac {54 \, 168} {12} = 4514$. As can be seen by inspection, the sums of the elements in the rows, columns and diagonals is $4514$. {{qed}}
From examples of factorials: {{begin-eqn}} {{eqn | l = 1       | r = 1!       | c =  }} {{eqn | l = 2       | r = 2!       | c =  }} {{eqn|| l = 145       | r = 1 + 24 + 120       | c =  }} {{eqn | r = 1! + 4! + 5!       | c =  }} {{eqn|| l = 40 \, 585       | r = 24 + 1 + 120 + 40 \, 320 + 120       | c =  }} {{eqn | r = 4! + 0! + 5! + 8! + 5!       | c =  }} {{end-eqn}} A computer search can verify solutions under $2540160 = 9! \times 7$ in seconds. Let $n$ be a $7$-digit number with $n > 2540160$. Then the sum of the factorials of its digits is not more than $9! \times 7 = 2540160$. So $n$ cannot be a factorion base $10$. Now let $n$ be a $k$-digit number, for $k \ge 8$. Then the sum of the factorials of its digits is not more than $9! \times k$. But we have: {{begin-eqn}} {{eqn | l = n       | o = \ge       | r = 10^{k - 1} }} {{eqn | r = 10^7 \times 10^{k - 8} }} {{eqn | o = \ge       | r = 10^7 \times \paren {1 + 9 \paren {k - 8} }       | c = Bernoulli's Inequality }} {{eqn | o = >       | r = 8 \times 9! \times \paren {9 k - 71}       | c = $8 \times 9! = 2903040$ }} {{eqn | r = 9! \paren {72 k - 71 \times 8} }} {{eqn | o = >       | r = 9! \times k       | c = $k \ge 8$ }} {{eqn | o = \ge       | r = \text{sum of the factorials of digits of } n }} {{end-eqn}} So there are no more factorions base $10$. {{qed}}
We have that: :$145 = 5 \times 29$ Both $5$ and $29$ can be expressed as the sum of two distinct square numbers: {{begin-eqn}} {{eqn | l = 5       | r = 1^2 + 2^2 }} {{eqn | l = 29       | r = 2^2 + 5^2 }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | r = \paren {1^2 + 2^2} \paren {2^2 + 5^2}       | c =  }} {{eqn | r = \paren {1 \times 2 + 2 \times 5}^2 + \paren {1 \times 5 - 2 \times 2}^2       | c = Brahmagupta-Fibonacci Identity }} {{eqn | r = \paren {2 + 10}^2 + \paren {5 - 4}^2       | c =  }} {{eqn | r = 12^2 + 1^2       | c =  }} {{eqn | r = 144 + 1       | c =  }} {{eqn | r = 145       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | r = \paren {1^2 + 2^2} \paren {2^2 + 5^2}       | c =  }} {{eqn | r = \paren {1 \times 2 - 2 \times 5}^2 + \paren {1 \times 5 + 2 \times 2}^2       | c = Brahmagupta-Fibonacci Identity/Corollary }} {{eqn | r = \paren {2 - 10}^2 + \paren {5 + 4}^2       | c =  }} {{eqn | r = \paren {8 - 2}^2 + \paren {5 + 4}^2       | c =  }} {{eqn | r = 8^2 + 9^2       | c =  }} {{eqn | r = 64 + 81       | c =  }} {{eqn | r = 145       | c =  }} {{end-eqn}} {{qed}}
Trivially: :$1 = \dfrac 1 1$ and it follows that: $U \left({1}\right) = 1$ Also trivially: :$1 = \dfrac 1 2 + \dfrac 1 2$ and it follows that: $U \left({2}\right) = 1$ From Sum of 3 Unit Fractions that equals 1: :$U \left({3}\right) = 3$ From Sum of 4 Unit Fractions that equals 1: :$U \left({4}\right) = 14$ From Sum of 5 Unit Fractions that equals 1: :$U \left({5}\right) = 147$ {{ProofWanted|and for the rest}}