avewright/proofwiki-math huggingface dataset:
Theorem: 91 is Pseudoprime to 35 Bases less than 91

$91$ is a Fermat pseudoprime in $35$ bases less than itself:
:$3, 4, 9, 10, 12, 16, 17, 22, 23, 25, 27, 29, 30, 36, 38, 40, 43, 48, 51, 53, 55, 61, 62, 64, 66, 68, 69, 74, 75, 79, 81, 82, 87, 88, 90$

Proof:
By definition of a Fermat pseudoprime, we need to check for $a < 91$:
:$a^{90} \equiv 1 \pmod {91}$
is satisfied or not.
By Chinese Remainder Theorem, this is equivalent to checking whether:
:$a^{90} \equiv 1 \pmod 7$
and:
:$a^{90} \equiv 1 \pmod {13}$
are both satisfied.
If $a$ is a multiple of $7$ or $13$, $a^{90} \not \equiv 1 \pmod {91}$.
Therefore we consider $a$ not divisible by $7$ or $13$.
By Fermat's Little Theorem, we have:
:$a^6 \equiv 1 \pmod 7$
and thus:
:$a^{90} \equiv 1^{15} \equiv 1 \pmod 7$
Now by Fermat's Little Theorem again:
:$a^{12} \equiv 1 \pmod {13}$
and thus:
:$a^{90} \equiv a^6 \paren{1^7} \equiv a^6 \pmod {13}$
We have:
{{begin-eqn}}
{{eqn | l = \paren {\pm 1}^6
      | o = \equiv
      | r = 1
      | rr= \pmod {13}
}}
{{eqn | l = \paren {\pm 2}^6
      | o = \equiv
      | r = -1
      | rr= \pmod {13}
}}
{{eqn | l = \paren {\pm 3}^6
      | o = \equiv
      | r = 1
      | rr= \pmod {13}
}}
{{eqn | l = \paren {\pm 4}^6
      | o = \equiv
      | r = 1
      | rr= \pmod {13}
}}
{{eqn | l = \paren {\pm 5}^6
      | o = \equiv
      | r = -1
      | rr= \pmod {13}
}}
{{eqn | l = \paren {\pm 6}^6
      | o = \equiv
      | r = -1
      | rr= \pmod {13}
}}
{{end-eqn}}
and thus $a$ must be equivalent to $1, 3, 4, 9, 10, 12 \pmod {13}$.
This gives $1$ and the $35$ bases less than $91$ listed above.
{{qed}}
Theorem: Even Integers not Sum of 2 Twin Primes

The following even integers cannot be expressed as the sum of $2$ prime numbers which are each one of a pair of twin primes:
:$2, 4, 94, 96, 98, 400, 402, 404, \ldots$
{{OEIS|A007534}}
{{expand|add a page to the effect that it is conjectured that this list is complete.}}

Proof:
{{ProofWanted|Can be demonstrated by brute force. A pseudocode program and a printout may be adequate, but first we need to develop a standard language-agnostic {{ProofWiki}} pseudocode.}}
Theorem: Reciprocal of 97

:$\dfrac 1 {97} = 0 \cdotp \dot 01030 \, 92783 \, 50515 \, 46391 \, 75257 \, 73195 \, 87628 \, 86597 \, 93814 \, 43298 \, 96907 \, 21649 \, 48453 \, 60824 \, 74226 \, 80412 \, 37113 \, 40206 \, 18556 \, \dot 7$

Proof:
Performing the calculation using long division:
<pre>
   0.01030927835051546391752577319587628865979381443298969072164948453608247422680412371134020618556701...
  ------------------------------------------------------------------------------------------------------
97)1.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
     97      485     873    679    582    679    291     679    776     388     388    291      485
     --      ---     ---    ---    ---    ---    ---     ---    ---     ---     ---    ---      ---
      300      500    170    310    280    910    290     210    440     720     120    390      650
      291      485     97    291    194    873    194     194    388     679      97    388      582
      ---      ---    ---    ---    ---    ---    ---     ---    ---     ---     ---    ---      ---
        900     150    730    190    860    370    960     160    520     410     230     200     680
        873      97    679     97    776    291    983      97    485     388     194     194     679
        ---     ---    ---    ---    ---    ---    ---     ---    ---     ---     ---     ---     ---
         270     530    510    930    840    790    870     630    350     220     360      600     100
         194     485    485    873    776    776    776     582    291     194     291      582      97
         ---     ---    ---    ---    ---    ---    ---     ---    ---     ---     ---      ---     ---
          760     450    250    570    640    140    940     480    590     260     690      180    ...
          679     388    194    485    582     97    873     388    582     194     679       97
          ---     ---    ---    ---    ---    ---    ---     ---    ---     ---     ---      ---
           810     620    560    850    580    430    670     920     800    660     110      830
           776     582    485    776    485    388    582     873     776    582      97      776
           ---     ---    ---    ---    ---    ---    ---     ---     ---    ---     ---      ---
            340     380    750    740    950    420    880     470     240    780     130      540
            291     291    679    679    873    388    873     388     194    776      97      485
            ---     ---    ---    ---    ---    ---    ---     ---     ---    ---     ---      ---
             490     890    710    610    770    320     700    820     460     400    330      550
             485     873    679    582    679    291     679    776     388     388    291      485
</pre>
{{qed}}
Theorem: Central Field is Field of Functional

Let $\mathbf y$ be an $N$-dimensional vector.
Let $J$ be a functional, such that:
:$\ds J \sqbrk {\mathbf y} = \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$
Let the following be a central field:
:$\map {\mathbf y'} x = \map {\boldsymbol \psi} {x, \mathbf y}$
Then this central field is a field of functional $J$.

Proof:
Suppose:
:$\ds \map g {x, \mathbf y} = \int_c^{\paren {x, \mathbf y} } \map F {x, \hat {\mathbf y}, \hat {\mathbf y}'} \rd x$
where $\hat {\mathbf y}$ is an extremal of $J$ connecting points $c$ and $\tuple {x, \mathbf y}$.
{{explain|Extremum or extremal? Can this be made consistent?}}
Define a field of directions in $R$ by the following:
:$\map {\mathbf p} {x, \mathbf y, \mathbf y'} = \map {g_{\mathbf y} } {x, \mathbf y}$
where $\mathbf p$ stands for momentum.
Note, that it does not depend on the path defined by $\hat {\mathbf y}$, only on its endpoint at $\tuple {x, \mathbf y}$.
By definition, $ \map g {x, \mathbf y}$ is a geodetic distance $S$.
Hence, $g_\mathbf y$ does not explicitly depend on $\mathbf y'$.
Then:
:$\map {g_{\mathbf y} } {x, \mathbf y} = \map {\mathbf p} {x, \mathbf y, \mathbf z}$
where $\mathbf z = \map {\mathbf z} {x, \mathbf y}$ denotes slope of the curve joining $c$ and $\tuple {x, \mathbf y}$ at the point $\tuple {x, \mathbf y}$.
Furthermore, since $g$ is a geodetic distance, it satisfies Hamilton-Jacobi equation:
:$\ds \frac {\partial \map g {x, \mathbf y} } {\partial x} + \map H {x, \mathbf y, \map {g_{\mathbf y} } {x, \mathbf y} } = 0$
which together with the previous relation results in a system of equations:
$\quad \begin {cases}
\dfrac {\partial \map g {x, \mathbf y} } {\partial x} + \map H {x, \mathbf y, \map {\mathbf p} {x, \mathbf y, \mathbf z} } = 0 \\
\map {\mathbf z} {x, \mathbf y} = \map{\mathbf y'} x 
\end {cases}$
Differentiation {{WRT|Differentiation}} $\mathbf y$ yields:
$\quad \begin {cases}
\dfrac {\partial \mathbf p} {\partial x} = -\dfrac {\partial H} {\partial \mathbf y} \\
\map {\mathbf z} {x,\mathbf y} = \map{\mathbf y'} x 
\end {cases}$
These are Euler's equations in canonical coordinates with a constraint on derivative.
{{explain|the last part is not exactly clear; where is the second Euler equation?}}
Hence, the previously defined field of directions coincides with the field of functional.
{{qed}}
Theorem: Reciprocal of 98

:$\dfrac 1 {98} = 0 \cdotp 0 \dot 1020 \, 40816 \, 32653 \, 06122 \, 44897 \, 95918 \, 36734 \, 69387 \, 75 \dot 5$

Proof:
Performing the calculation using long division:
<pre>
   0.01020408163265306122448979591836734693877551...
  --------------------------------------------------
98)1.00000000000000000000000000000000000000000000000
     98       196     196    490    294    686
     --       ---     ---    ---    ---    ---
      200      640     440    900    460    540
      196      588     392    882    392    490
      ---      ---     ---    ---    ---    ---
        400     520     480    180    680    500
        392     490     392     98    588    490
        ---     ---     ---    ---    ---    ---
          800    300     880    820    920    100
          784    294     784    784    882     98
          ---    ---     ---    ---    ---    ---
           160     600    960    360    380   ...
            98     588    882    294    294
           ---     ---    ---    ---    ---
            620     120    780    660    860
            588      98    686    588    784
            ---     ---    ---    ---    ---
             320     220    940    720    760
             294     196    882    686    686
             ---     ---    ---    ---    ---
              260     240    580    340    740
              196     196    490    294    686
</pre>
We have that:
:$0 \cdotp 01 + 0 \cdotp 0002 + 0 \cdotp 000004 + 0 \cdotp 00000008 + \cdots$
is nothing else but:
:$\ds \dfrac 1 2 \paren {\dfrac 1 {50} + \paren {\dfrac 1 {50} }^2 + \paren {\dfrac 1 {50} }^3 + \paren {\dfrac 1 {50} }^4 + \cdots} = \dfrac 1 2 \sum_{k \mathop \ge 1} \paren {\dfrac 1 {50} }^k$
Hence:
{{begin-eqn}}
{{eqn | l = \dfrac 1 2 \sum_{k \mathop \ge 1} \paren {\dfrac 1 {50} }^k
      | r = \dfrac 1 2 \dfrac {1 / 50} {1 - 1 / 50}
      | c = Sum of Infinite Geometric Sequence: Corollary 1
}}
{{eqn | r = \dfrac 1 2 \dfrac 1 {50 - 1}
      | c = 
}}
{{eqn | r = \dfrac 1 2 \dfrac 1 {49}
      | c = 
}}
{{eqn | r = \dfrac 1 {98}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Even Integers not Expressible as Sum of 3, 5 or 7 with Prime

The even integers that cannot be expressed as the sum of $2$ prime numbers where one of those primes is $3$, $5$ or $7$ begins:
:$98, 122, 124, 126, 128, 148, 150, \ldots$
{{OEIS|A283555}}

Proof:
These are the primes which coincide with the upper end of a prime gap greater than $6$.
These can be found at:
:$89$ to $97$: prime gap of $8$
:$113$ to $127$: prime gap of $14$
:$139$ to $149$: prime gap of $10$
and so on.
We have that:
{{begin-eqn}}
{{eqn | l = 98
      | r = 19 + 79
}}
{{eqn | l = 122
      | r = 13 + 109
}}
{{eqn | l = 124
      | r = 11 + 113
}}
{{eqn | l = 126
      | r = 13 + 113
}}
{{eqn | l = 128
      | r = 19 + 109
}}
{{end-eqn}}
and so on.
{{qed}}
Theorem: Reciprocal of 99

:$\dfrac 1 {99} = 0 \cdotp \dot 0 \dot 1$

Proof:
Performing the calculation using long division:
<pre>
   0.0101...
  ----------
99)1.0000000
     99
     --
      100
       99
      ---
      ...
</pre>
{{qed}}
Theorem: Repdigit Number consisting of Instances of 9 is Kaprekar

A repdigit number that consists entirely of the digit $9$ is a Kaprekar number.

Proof:
We note as examples:
{{begin-eqn}}
{{eqn | l = 9^2
      | r = 81
}}
{{eqn | l = 8 + 1
      | r = 9
}}
{{eqn | l = 99^2
      | r = 9801
}}
{{eqn | l = 98 + 01
      | r = 99
}}
{{eqn | l = 999^2
      | r = 998001
}}
{{eqn | l = 998 + 001
      | r = 999
}}
{{end-eqn}}
Now we consider:
{{begin-eqn}}
{{eqn | l = \paren {\sum_{k \mathop = 0}^n 9 \times 10^k}^2
      | r = \paren {10^{n + 1} - 1}^2
      | c = Sum of Geometric Sequence
}}
{{eqn | r = 10^{n + 1} \paren {10^{n + 1} - 2} + 1
}}
{{end-eqn}}
So $\paren {10^{n + 1} - 1}^2$ can be split as:
:$\paren {10^{n + 1} - 2} + 1 = 10^{n + 1} - 1 = \ds \sum_{k \mathop = 0}^n 9 \times 10^k$
Thus $\ds \sum_{k \mathop = 0}^n 9 \times 10^k$ is a Kaprekar number.
{{qed}}
Theorem: Numbers equal to Sum of Primes not Greater than its Prime Counting Function Value

Let $\map \pi n: \Z_{\ge 0} \to \Z_{\ge 0}$ denote the prime-counting function:
:$\map \pi n =$ the count of the primes less than $n$
Consider the equation:
:$\ds n = \sum_{p \mathop \le \map \pi n} p$
where $p \le \pi \left({n}\right)$ denotes the primes not greater than $\pi \left({n}\right)$.
Then $n$ is one of:
:$5, 17, 41, 77, 100$
{{OEIS|A091864}}

Proof:
We have that:
{{begin-eqn}}
{{eqn | l = \map \pi 5
      | r = 3
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2 + 3
      | r = 5
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map \pi {17}
      | r = 7
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2 + 3 + 5 + 7
      | r = 17
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map \pi {41}
      | r = 13
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2 + 3 + 5 + 7 + 11 + 13
      | r = 41
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map \pi {77}
      | r = 21
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19
      | r = 77
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map \pi {100}
      | r = 25
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23
      | r = 100
      | c = 
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown that there are no others.}}
Theorem: Smallest Seventh Power which is Sum of 8 other Seventh Powers

The smallest seventh power that can be expressed as the sum of $8$ other seventh powers is $102^7$:
:$102^7 = 12^7 + 35^7 + 53^7 + 58^7 + 64^7 + 83^7 + 85^7 + 90^7$

Proof:
{{ProofWanted|Brute force}}
Theorem: 103 is Smallest Prime whose Period of Reciprocal is One Third of Maximal

:$\dfrac 1 {103} = 0 \cdotp \dot 00970 \, 87378 \, 64077 \, 66990 \, 29126 \, 21359 \, 223 \dot 3$

Proof:
From Reciprocal of $103$:
{{:Reciprocal of 103}}
and so by counting it can be seen that its period of recurrence is $34$.
By Maximum Period of Reciprocal of Prime, the maximum period of recurrence of the reciprocal of $p$ when expressed in decimal notation is $p - 1$.
Therefore in order for a prime number to have its period of its reciprocal to equal $\dfrac 1 3$ of its maximum length, we must have:
:$3 \divides p - 1$
The prime numbers less than $103$ with this property are $7$, $13$, $19$, $31$, $37$, $43$, $61$, $67$, $73$, $79$ and $97$.
We have:
* $7$ is Long Period Prime
* $19$ is Long Period Prime
* $61$ is Long Period Prime
* $97$ is Long Period Prime
which shows that each of these primes has maximum period $p - 1$.
We have:
:Reciprocal of $13$: $\dfrac 1 {13} = 0 \cdotp \dot 07692 \dot 3$: recurring with period $6 = \dfrac {p - 1} 2$.
:Reciprocal of $31$: $\dfrac 1 {31} = 0 \cdotp \dot 03225 \, 80645 \, 1612 \dot 9$: recurring with period $15 = \dfrac {p - 1} 2$.
:Reciprocal of $37$: $\dfrac 1 {37} = 0 \cdotp \dot 02 \dot 7$: recurring with period $3 = \dfrac {p - 1} {12}$.
:Reciprocal of $43$: $\dfrac 1 {43} = 0 \cdotp \dot 02325 \, 58139 \, 53488 \, 37209 \dot 3$: recurring with period $21 = \dfrac {p - 1} 2$.
:Reciprocal of $67$: $\dfrac 1 {67} = 0 \cdotp \dot 01492 \, 53731 \, 34328 \, 35820 \, 89552 \, 23880 \, 59 \dot 7$: recurring with period $33 = \dfrac {p - 1} 2$.
:Reciprocal of $73$: $\dfrac 1 {73} = 0 \cdotp \dot 01369 \, 86 \dot 3$: recurring with period $8 = \dfrac {p - 1} 9$.
:Reciprocal of $79$: $\dfrac 1 {79} = 0 \cdotp \dot 01265 \, 82278 \, 48 \dot 1$: recurring with period $13 = \dfrac {p - 1} 6$.
Thus $103$ is the smallest prime number with this property.
{{qed}}
Theorem: Odd Integers whose Smaller Odd Coprimes are Prime

Let $n \in \Z_{>0}$ be an odd positive integer such that all smaller odd integers greater than $1$ which are coprime to it are prime.
The complete list of such $n$ is as follows:
:$1, 3, 5, 7, 9, 15, 21, 45, 105$
{{OEIS|A327823}}

Proof:
First it is demonstrated that $105$ itself satisfies this property.
Let $d \in \Z_{> 1}$ be odd and coprime to $105$.
Then $d$ does not have $3$, $5$ or $7$ as a prime factor.
Thus $d$ must have at least one odd prime as a divisor which is $11$ or greater.
The smallest such composite number is $11^2$.
But $11^2 = 121 > 105$.
Thus $d$ must be prime.
Thus it has been demonstrated that all odd integers greater than $1$ and smaller than $105$ which are coprime to $105$ are prime.
{{qed|lemma}}
Using an argument similar to the above, we see that for an integer to have this property, 
if it is greater than $p^2$ for some odd prime $p$, then it must be divisible by $p$.
If not, it will be coprime to $p^2$, a composite number.
Let $p_n$ denote the $n$th prime.
Suppose $N$ has this property.
By the argument above, if $p_{n + 1}^2 \ge N > p_n^2$, we must have $p_2 p_3 \cdots p_n \divides N$.
By Absolute Value of Integer is not less than Divisors, we have $p_2 p_3 \cdots p_n \le N$.
Bertrand-Chebyshev Theorem asserts that there is a prime between $p_n$ and $2 p_n$.
Thus we have $2 p_n > p_{n + 1}$.
Hence for $n \ge 5$:
{{begin-eqn}}
{{eqn | l = N
      | o = \ge
      | r = p_2 p_3 \cdots p_n
}}
{{eqn | r = 3 \times 5 p_4 \cdots p_n
}}
{{eqn | o = >
      | r = 8 p_{n - 1} p_n
}}
{{eqn | o = >
      | r = 4 p_n^2
      | c = Bertrand-Chebyshev Theorem
}}
{{eqn | o = >
      | r = p_{n + 1}^2
      | c = Bertrand-Chebyshev Theorem
}}
{{eqn | o = \ge
      | r = N
      | c = From assumption
}}
{{end-eqn}}
This is a contradiction. 
Hence we must have $N \le p_5^2 = 121$.
From the argument above we also have:
:$3 \divides N$ for $9 < N \le 25$
:$3, 5 \divides N$ for $25 < N \le 49$
:$3, 5, 7 \divides N$ for $49 < N \le 121$
So we end up with the list $N = 1, 3, 5, 7, 9, 15, 21, 45, 105$.
{{qed}}
Theorem: Sequences of 3 Consecutive Integers with Rising Phi

The following ordered triples of consecutive integers have $\phi$ values which are strictly increasing:
:$105, 106, 107$
:$165, 166, 167$
:$315, 316, 317$

Proof:
{{begin-eqn}}
{{eqn | l = \map \phi {105}
      | r = 48
      | c = {{EulerPhiLink|105}}
}}
{{eqn | l = \map \phi {106}
      | r = 52
      | c = {{EulerPhiLink|106}}
}}
{{eqn | l = \map \phi {107}
      | r = 106
      | c = Euler Phi Function of Prime: $107$ is prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map \phi {165}
      | r = 80
      | c = {{EulerPhiLink|165}}
}}
{{eqn | l = \map \phi {166}
      | r = 82
      | c = {{EulerPhiLink|166}}
}}
{{eqn | l = \map \phi {167}
      | r = 166
      | c = Euler Phi Function of Prime: $167$ is prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map \phi {315}
      | r = 144
      | c = {{EulerPhiLink|315}}
}}
{{eqn | l = \map \phi {316}
      | r = 156
      | c = {{EulerPhiLink|316}}
}}
{{eqn | l = \map \phi {317}
      | r = 316
      | c = Euler Phi Function of Prime: $317$ is prime
}}
{{end-eqn}}
{{qed}}
Theorem: Integers whose Divisor Sum equals Half Phi times Divisor Count

The following positive integers $n$ have the property where:
:$\map {\sigma_1} n = \dfrac {\map \phi n \times \map {\sigma_0} n} 2$
where:
:$\map {\sigma_1} n$ denotes the divisor sum function: the sum of the divisors of $n$
:$\map \phi n$ denotes the Euler $\phi$ function: the count of positive integers smaller than of $n$ which are coprime to $n$
:$\map {\sigma_0} n$ denotes the divisor count function: the count of the divisors of $n$:
These positive integers are:
:$35, 105, \ldots$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \map \phi {35}
      | r = 24
      | c = {{EulerPhiLink|35}}
}}
{{eqn | l = \map {\sigma_0} {35}
      | r = 4
      | c = {{DCFLink|35}}
}}
{{eqn | ll= \leadsto
      | l = \map \phi {35} \times \map {\sigma_0} {35}
      | r = \dfrac {24 \times 4} 2
      | c = 
}}
{{eqn | r = 48
      | c = 
}}
{{eqn | r = \map {\sigma_1} {35}
      | c = {{DSFLink|35}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map \phi {105}
      | r = 48
      | c = {{EulerPhiLink|105}}
}}
{{eqn | l = \map {\sigma_0} {105}
      | r = 8
      | c = {{DCFLink|105}}
}}
{{eqn | ll= \leadsto
      | l = \map \phi {105} \times \map {\sigma_0} {105}
      | r = \dfrac {48 \times 8} 2
      | c = 
}}
{{eqn | r = 192
      | c = 
}}
{{eqn | r = \map {\sigma_1} {105}
      | c = {{DSFLink|105}}
}}
{{end-eqn}}
{{qed}}
Theorem: Smallest Polyomino with Hole

The smallest polyomino with a hole is the heptomino in the form of a $3 \times 3$ 
square with the center $1 \times 1$ square and a corner $1 \times 1$ square missing:
:200px

Proof:
From 35 Hexominoes and by inspection, none of the $35$ hexominoes has a hole.
From Number of Heptominoes and by inspection, exactly one of the $108$ heptominoes has a hole.
This is the smallest polyomino with a hole.
{{qed}}
Theorem: Magic Constant of Smallest Prime Magic Square

The magic constant of the smallest prime magic square is $111$.

Proof:
The smallest prime magic square (including $1$) is:
{{:Prime Magic Square/Examples/Order 3/Smallest}}
As can be seen by inspection, the sums of the elements in the rows, columns and diagonals are $111$:
{{begin-eqn}}
{{eqn | l = 67 + 1 + 43
      | r = 111
}}
{{eqn | l = 13 + 37 + 61
      | r = 111
}}
{{eqn | l = 31 + 73 + 7
      | r = 111
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 67 + 13 + 31
      | r = 111
}}
{{eqn | l = 1 + 37 + 73
      | r = 111
}}
{{eqn | l = 43 + 61 + 7
      | r = 111
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 67 + 37 + 7
      | r = 111
}}
{{eqn | l = 43 + 37 + 31
      | r = 111
}}
{{end-eqn}}
{{qed}}
Theorem: Sequence of Palindromic Lucky Numbers

The The sequence of lucky numbers which are also palindromic begins:
:$1, 3, 7, 9, 33, 99, 111, 141, 151, 171, \ldots$
{{OEIS|A031161}}

Proof:
{{:Definition:Lucky Number/Sequence}}
Of these, the palindromic ones can be picked out by inspection.
{{qed}}
Theorem: Sequence of Square Lucky Numbers

The sequence of lucky numbers which are also square begins:
:$1, 9, 25, 49, 169, 289, 361, 529, \ldots$
{{OEIS|A031162}}

Proof:
{{:Definition:Lucky Number/Sequence}}
Of these, the square ones can be picked out by inspection.
{{qed}}
Category:Lucky Numbers
Category:Square Numbers
926u5u5kr0ql3e1xd83oaoyatuxjcbw
Theorem: Conditions for Extremal Embedding in Field of Functional

Let $J$ be a functional such that:
:$\ds J \sqbrk {\mathbf y} = \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$
Let $\gamma$ be an extremal of $J$, defined by $\mathbf y = \map {\mathbf y} x$ for $x \in \closedint a b$.
Suppose:
:$\forall x \in \closedint a b: \map \det {F_{\mathbf y' \mathbf y'} } \ne 0$
Suppose no points conjugate to $\paren {a, \map {\mathbf y} a}$ lie on $\gamma$.
Then $\gamma$ can be embedded in a field.

Proof:
Let $c \in \R$ be conjugate to $a$, such that $c < a$.
By assumption:
:$c \notin \closedint a b$
Hence, there exists a set $\closedint c b$ such that:
:$\closedint c b = \closedint c a \cup \closedint a b$
where $\size {c - a} > 0$.
By there exists a real point between two real points:
:$\exists \epsilon: \size {c - a} > \epsilon > 0$
Therefore, there exists $\epsilon > 0$ such that:
:$\gamma$ can be extended onto the whole interval $\closedint {a - \epsilon} b$, where extension means definition of some mapping in $\hointr {a - \epsilon} a$
{{explain|the author provides this as a condition, but does a definability affect existence ? }}
:the interval $\closedint {a - \epsilon} b$ contains no points conjugate to $a$.
Consider a family of extremals leaving the point $\tuple {a - \epsilon, \map {\mathbf y} {a - \epsilon} }$.
There are no points conjugate to $a - \epsilon$ in $\closedint {a - \epsilon} b$.
Hence, for $x \in \closedint a b$ no two extremals in this family which are sufficiently close to the original extremal $\gamma$ can intersect.
{{explain|Follows from the definition, but a picture or a better explanation would help}}
Since all the functions are extremals, they satisfy same differential equations.
Lack of intersection implies different boundary conditions.
Denote these conditions collectively as $\map {\boldsymbol \psi} {x, \mathbf y} = \map {\mathbf y'} x$.
Thus, in some region $R$ containing $\gamma$ extremals sufficiently close to $\gamma$ define a central field in which $\gamma$ is embedded.
By Central Field is Field of Functional, $\gamma$ can be embedded in the field of functional.
{{qed}}
Theorem: Sequence of Smallest Consecutive Composite Numbers longer than 100

The $1$st prime gap greater than $100$ is between $370 \, 261$ and $370 \, 373$, of length $112$.
That is, the sequence of the smallest consecutive composite positive integers longer than $100$ is that of $111$ such, from  $370 \, 262$ to $370 \, 372$.

Proof:
{{ProofWanted|Brute force?}}