avewright/proofwiki-math huggingface dataset:
Theorem: Triangular Numbers which are also Pentagonal

The sequence of triangular numbers which are also pentagonal begins:
:$1, 210, 40 \, 755, 7 \, 906 \, 276, 1 \, 533 \, 776 \, 805, 297 \, 544 \, 793 \, 910, \ldots$
{{OEIS|A014979}}

Proof:
{{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.}}
Theorem: Numbers such that Divisor Count divides Phi divides Divisor Sum

The sequence of integers $n$ with the property that:
:$\map {\sigma_0} n \divides \map \phi n \divides \map {\sigma_1} n$
where:
:$\divides$ denotes divisibility
:$\sigma_0$ denotes the divisor count of $n$
:$\phi$ denotes the Euler $\phi$ (phi) function: the count of smaller integers coprime to $n$
:$\sigma_1$ denotes the divisor sum of $n$
begins:
:$1, 3, 15, 30, 35, 56, 70, 78, 105, 140, 168, 190, 210, 248, 264, \ldots$

Proof:
By inspection and investigation.
{{qed}}
Theorem: 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 $\closedint {\dfrac n 2} {n - 2}$.

Proof:
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}}
Theorem: Integers whose Number of Representations as Sum of Two Primes is Maximum

$210$ is the largest integer which can be represented as the sum of two primes in the maximum number of ways.
The full list of such numbers is as follows:
:$1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 16, 18, 24, 30, 36, 42, 48, 60, 90, 210$
{{OEIS|A141340}}
The list contains:
:$n \le 8$
:$n \le 18$ where $2 \divides n$
:$n \le 48$ where $2 \times 3 \divides n$
:$n \le 90$ where $2 \times 3 \times 5 \divides n$
:$210 = 2 \times 3 \times 5 \times 7$

Proof:
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.}}
Theorem: 25 as Sum of 4 to 11 Squares

$25$ can be expressed as the sum of $n$ non-zero squares for all $n$ from $4$ to $11$.

Proof:
We have:
{{begin-eqn}}
{{eqn | l = 25
      | r = 4^2 + \paren {2 \times 2^2} + 1^2
      | c = $4$ squares
}}
{{eqn | r = 3^2 + \paren {4 \times 2^2}
      | c = $5$ squares
}}
{{eqn | r = \paren {2 \times 3^2} + 2^2 + \paren {3 \times 1^2}
      | c = $6$ squares
}}
{{eqn | r = 4^2 + 2^2 + \paren {5 \times 1^2}
      | c = $7$ squares
}}
{{eqn | r = 3^2 + \paren {3 \times 2^2} + \paren {4 \times 1^2}
      | c = $8$ squares
}}
{{eqn | r = \paren {2 \times 3^2} + \paren {7 \times 1^2}
      | c = $9$ squares
}}
{{eqn | r = 4^2 + \paren {9 \times 1^2}
      | c = $10$ squares
}}
{{eqn | r = 3^2 + \paren {2 \times 2^2} + \paren {8 \times 1^2}
      | c = $11$ squares
}}
{{end-eqn}}
{{qed}}
Theorem: Ritz Method implies Not Worse Approximation with Increased Number of Functions

Consider the Ritz method.
Let $\eta_n = \boldsymbol \alpha \boldsymbol \phi$.
Let $J \sqbrk {\eta_n} = \mu_n$.
Then:
:$\mu_n \ge \mu_{n + 1}$

Proof:
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}}
Theorem: Dissection of Cube into 3 Cubes using 8 Pieces

A cube can be dissected into $3$ smaller cubes by cutting it into $8$ pieces and reassembling them.

Proof:
{{ProofWanted|Apparently it can be done using the result Cubes which are Sum of Three Cubes.}}
Theorem: Smallest Magic Constant of Order 3 Multiplicative Magic Square

The magic constant of the smallest possible multiplicative magic square with the smallest magic constant is as follows.
{{:Multiplicative Magic Square/Examples/Order 3/Smallest}}
</onlyinclude>

Proof:
From Smallest Multiplicative Magic Square is of Order 3, the smallest possible multiplicative magic square is of order $3$.
Let $C$ be a magic constant for an order $3$ multiplicative magic square.
We prove that $C \ge 216$ is smallest in $2$ steps:
:Step $1$: If $C$ is a prime power $p^k$, then $k \ge 8$.
:Step $2$: For each prime $p$ dividing $C$, we must have $p^3$ dividing $C$ as well.
Then $216 = 2^3 \cdot 3^3$ is the smallest such constant.
Theorem: Smallest Multiplicative Magic Square is of Order 3

The order of the smallest multiplicative magic square is $3$, for example:
{{:Multiplicative Magic Square/Examples/Order 3/Smallest}}

Proof:
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
Theorem: For Complete Ritz Sequence Continuous Functional approaches its Minimal Value

Let $J$ be a continuous functional.
Let $\sequence {\phi_n}$ be a complete Ritz sequence.
{{explain|The concept of $\sequence {\phi_n}$ does not appear to be related in any way to the statement of the theorem.}}
Then:
:$\ds \lim_{n \mathop \to \infty} \mu_n = \mu$
where $\ds \mu = \inf_y J \sqbrk y$.

Proof:
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 $\eta_n = \bsalpha \bsphi$, such that:
:$\exists n \in\ N: \exists N \in \N: \paren {n > N} \implies \paren {\size {\eta_n - y^*} < \epsilon}$
where $\bsalpha$ is an $n$-dimensional real vector.
Let $y_n = \bsalpha \bsphi$, where $\bsalpha$ 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:
:$\ds \lim_{n \mathop \to \infty} J \sqbrk {y_n} = \mu$
where $J \sqbrk {y_n} = \mu_n$.
{{qed}}
Theorem: 219 Fedorov Groups

There are $219$ Fedorov groups, if chiral copies are considered the same.

Proof:
{{ProofWanted|First the concept of Fedorov group needs to be explored.}}
Theorem: 17 Wallpaper Groups

There are $17$ wallpaper groups.

Proof:
{{ProofWanted|First the concept of wallpaper group needs to be explored.}}
Theorem: 230 Fedorov Groups where Chiral Pairs are Distinct

There are $230$ Fedorov groups, if chiral copies are considered distinct.

Proof:
{{ProofWanted|First the concept of Fedorov group needs to be explored.}}
Theorem: Equivalence of Definitions of Amicable Pair

Let $m \in \Z_{>0}$ and $n \in \Z_{>0}$ be (strictly) positive integers.
{{TFAE|def = Amicable Pair}}

Proof:
=== Definition 1 is equivalent to Definition 2 ===
Let $\map s n$ denote the aliquot sum of (strictly) positive integer $n$.
The sum of all the divisors of a (strictly) positive integer $n$ is $\map {\sigma_1} n$, where $\sigma_1$ is the divisor sum function.
The aliquot sum of $n$ is the sum of the divisors of $n$ with $n$ excluded.
Thus:
:$\map s n = \map {\sigma_1} n - n$
Suppose:
:$\map s n = m$
and:
:$\map s m = n$
Then:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} n - n
      | r = m
      | c = {{Defof|Proper Divisor of Integer}}
}}
{{eqn | ll= \leadsto
      | l = \map {\sigma_1} n
      | r = m + n
      | c = 
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} m - m
      | r = n
      | c = {{Defof|Proper Divisor of Integer}}
}}
{{eqn | ll= \leadsto
      | l = \map {\sigma_1} m
      | r = m + n
      | c = 
}}
{{end-eqn}}
Thus:
:$\map s n = \map s m = m + n$
The argument reverses.
{{qed|lemma}}
Theorem: Thabit's Rule

Let $n$ be a positive integer such that:
{{begin-eqn}}
{{eqn | l = a
      | r = 3 \times 2^n - 1
      | c = 
}}
{{eqn | l = b
      | r = 3 \times 2^{n - 1} - 1
      | c = 
}}
{{eqn | l = c
      | r = 9 \times 2^{2 n - 1} - 1
      | c = 
}}
{{end-eqn}}
are all prime.
Then:
:$\tuple {2^n a b, 2^n c}$
forms an amicable pair.

Proof:
Let $r = 2^n a b, s = 2^n c$.
Let $\map {\sigma_1} k$ denote the divisor sum of an integer $k$.
From Divisor Sum of Power of 2:
:$\map {\sigma_1} {2^n} = 2^{n + 1} - 1$
From Divisor Sum of Prime Number:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} a
      | r = 3 \times 2^n
      | c = 
}}
{{eqn | l = \map {\sigma_1} b
      | r = 3 \times 2^{n - 1}
      | c = 
}}
{{eqn | l = \map {\sigma_1} c
      | r = 9 \times 2^{2 n - 1}
      | c = 
}}
{{end-eqn}}
From Divisor Sum Function is Multiplicative:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} r
      | r = \map {\sigma_1} {2^n} \map {\sigma_1} a \map {\sigma_1} 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_1} s
      | r = \map {\sigma_1} {2^n} \map {\sigma_1} c
      | c = 
}}
{{eqn | r = \paren {2^{n + 1} - 1} \paren {9 \times 2^{2 n - 1} }
      | c = 
}}
{{eqn | r = \map {\sigma_1} r
      | c = 
}}
{{end-eqn}}
Thus it is seen that:
:$\map {\sigma_1} r = \map {\sigma_1} s$
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:
:$r + s = \map {\sigma_1} r = \map {\sigma_1} s$
Hence the result, by definition of amicable pair.
{{qed}}
Theorem: Smaller of Thabit Pair is Tetrahedral

Let $\tuple {m_1, m_2}$ be a Thabit pair such that $m_1 < m_2$.
Then $m_1$ is a tetrahedral number.

Proof:
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}}
Theorem: Numbers in Even-Even Amicable Pair are not Divisible by 3

Let $\tuple {m_1, m_2}$ be an amicable pair such that both $m_1$ and $m_2$ are even.
Then neither $m_1$ nor $m_2$ is divisible by $3$.

Proof:
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_1} {m_2}
      | c = {{Defof|Amicable Pair|subdef = Definition 2}}
}}
{{eqn | r = \map {\sigma_1} {2^s} \map {\sigma_1} b
      | c = Divisor Sum Function is Multiplicative
}}
{{eqn | r = \paren {2^{s + 1} - 1} \map {\sigma_1} b
      | c = Divisor Sum of Power of Prime
}}
{{eqn | o = \equiv
      | r = \paren {\paren {-1}^{s + 1} - 1} \map {\sigma_1} b
      | rr = \pmod 3
      | c = Congruence of Powers
}}
{{end-eqn}}
Since $m + n$ is not divisible by $3$, $s$ must be even.
Similarly, by Divisor Sum 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_1} {m_1} } {m_1}
      | o = \ge
      | r = \frac {\map {\sigma_1} {12} } {12}
      | c = equality occurs {{iff}} $m_1 = 12$
}}
{{eqn | r = \frac 7 3
}}
{{eqn | l = \frac {\map {\sigma_1} {m_2} } {m_2}
      | o = >
      | r = \frac {\map {\sigma_1} 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_1} {m_1} } + \frac {m_2} {\map {\sigma_1} {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}}
Theorem: Equivalence of Definitions of Amicable Triplet

Let $m_1, m_2, m_3 \in \Z_{>0}$ be (strictly) positive integers.
{{TFAE|def = Amicable Triplet}}

Proof:
For $n \in \Z_{>0}$, let $\map s n$ denote the aliquot sum of (strictly) positive integer $n$.
The sum of all the divisors of a (strictly) positive integer $n$ is $\map {\sigma_1} n$, where $\sigma_1$ is the divisor sum function.
The aliquot sum of $n$ is the sum of the divisors of $n$ with $n$ excluded.
Thus:
:$\map s n = \map {\sigma_1} n - n$
Suppose:
{{begin-eqn}}
{{eqn | l = \map s {m_1}
      | r = m_2 + m_3
}}
{{eqn | l = \map s {m_2}
      | r = m_1 + m_3
}}
{{eqn | l = \map s {m_3}
      | r = m_1 + m_2
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {m_1} - m_1
      | r = m_2 + m_3
      | c = {{Defof|Proper Divisor of Integer}}
}}
{{eqn | ll= \leadsto
      | l = \map {\sigma_1} {m_1}
      | r = m_1 + m_2 + m_3
      | c = 
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {m_2} - m_2
      | r = m_1 + m_3
      | c = {{Defof|Proper Divisor of Integer}}
}}
{{eqn | ll= \leadsto
      | l = \map {\sigma_1} {m_2}
      | r = m_1 + m_2 + m_3
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {m_3} - m_3
      | r = m_1 + m_2
      | c = {{Defof|Proper Divisor of Integer}}
}}
{{eqn | ll= \leadsto
      | l = \map {\sigma_1} {m_3}
      | r = m_1 + m_2 + m_3
      | c = 
}}
{{end-eqn}}
Thus:
:$\map {\sigma_1} {m_1} = \map {\sigma_1} {m_2} = \map {\sigma_1} {m_3} = m_1 + m_2 + m_3$
The argument reverses.
{{qed}}
Category:Amicable Triplets
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Theorem: Largest Number not Expressible as Sum of Less than 37 Positive Fifth Powers

The largest positive integer which cannot be expressed as the sum of less than $37$ positive fifth powers is $223$:
:$223 = 31 \times 1^5 + 6 \times 2^5$

Proof:
{{ProofWanted|It needs to be shown that there are no larger numbers with this property.}}
Theorem: Smallest Triple of Consecutive Sums of Squares

The smallest triple of consecutive positive integers each of which is the sum of two squares is:
:$\tuple {232, 233, 234}$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = 232
      | r = 14^2 + 6^2
      | c = 
}}
{{eqn | l = 233
      | r = 13^2 + 8^2
      | c = 
}}
{{eqn | l = 234
      | r = 15^2 + 3^2
      | c = 
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown this is the smallest such triple.}}
Theorem: No Quadruple of Consecutive Sums of Squares Exists

It is not possible for a quadruple of consecutive positive integers each of which is the sum of two squares.

Proof:
$4$ 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}}
Theorem: Directed iff Filtered in Dual Ordered Set

Let $\struct {S, \preceq_1}$ be an ordered set.
Let $\struct {S, \preceq_2}$ be a dual ordered set of $\struct {S, \preceq_1}$
Let $X \subseteq S$.
Then:
:$X$ is directed in $\struct {S, \preceq_1}$
{{iff}}:
:$X$ is filtered in $\struct {S, \preceq_2}$.

Proof:
=== 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}}
Theorem: Sum of Two Squares not Congruent to 3 modulo 4

Let $n \in \Z$ such that $n = a^2 + b^2$ where $a, b \in \Z$.
Then $n$ is not congruent modulo $4$ to $3$.

Proof:
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
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Theorem: Numbers not Expressible as Sum of Less than 9 Positive Cubes

The following are the only positive integers cannot be expressed as the sum of less than $9$ positive cubes:
{{begin-eqn}}
{{eqn | l = 23
      | r = 2^3 + 2^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3
      | c = 
}}
{{eqn | l = 239
      | r = 4^3 + 4^3 + 3^3 + 3^3 + 3^3 + 3^3 + 1^3 + 1^3 + 1^3
      | c = 
}}
{{end-eqn}}

Proof:
{{ProofWanted|It needs to be demonstrated that these are the only ones.}}
Theorem: Largest Prime Factor of n squared plus 1

Let $n \in \Z$ be greater than $239$.
Then the largest prime factor of $n^2 + 1$ is at least $17$.

Proof:
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.}}
Theorem: Solution of Ljunggren Equation

The only solutions of the Ljunggren equation:
:$x^2 + 1 = 2 y^4$
are:
:$x = 1,  y = 1$
:$x = 239,  y = 13$
{{OEIS|A229384}}

Proof:
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 $n^2 + 1$?}}