avewright/proofwiki-math huggingface dataset:
Theorem: Summation over Interval equals Indexed Summation

Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $a, b \in \Z$ be integers.
Let $\closedint a b$ be the integer interval between $a$ and $b$.
Let $f: \closedint a b \to \mathbb A$ be a mapping.
Then the summation over the finite set $\closedint a b$ equals the indexed summation from $a$ to $b$:
:$\ds \sum_{k \mathop \in \closedint a b} \map f k = \sum_{k \mathop = a}^b \map f k$

Proof:
By Cardinality of Integer Interval, $\closedint a b$ has cardinality $b - a + 1$.
By Translation of Integer Interval is Bijection, the mapping $T : \closedint 0 {b - a} \to \closedint a b$ defined as:
:$\map T k = k + a$
is a bijection.
By definition of summation:
:$\ds \sum_{k \mathop \in \closedint a b} \map f k = \sum_{k \mathop = 0}^{b - a} \map f {k + a}$
By Indexed Summation over Translated Interval:
:$\ds \sum_{k \mathop = 0}^{b - a} \map f {k + a} = \sum_{k \mathop = a}^b \map f k$
{{qed}}
Category:Summations
3vk6eiqlzglmh7kezwms7af2mfn8tff
Theorem: Hardy-Ramanujan Number/Examples/1729

The $2$nd Hardy-Ramanujan number $\map {\operatorname {Ta}} 2$ is $1729$:
{{begin-eqn}}
{{eqn | l = 1729
      | r = 12^3 + 1^3
      | c = 
}}
{{eqn | r = 10^3 + 9^3
      | c = 
}}
{{end-eqn}}

Proof:
We wish to demonstrate that $1729$ is the $2$nd '''Hardy-Ramanujan number''' making it the smallest positive integer which can be expressed as the sum of $2$ positive cubes in $2$ different ways.
To accomplish this, we will need to inspect $66$ sums of $a^3 + b^3$ starting with $1^3 + 1^3$ and ending with $11^3 + 11^3$
Recall from Integer Addition is Commutative, $a^3 + b^3 = b^3 + a^3$, therefore we only need to inspect one of these pairs.
{{WLOG}}, we will choose to inspect $a \ge b$ in the table below.
The reason that we don't need to inspect $a \ge 12$ is that $12^3 + 1^3 = 1729$ and adding anything larger than $1$ to $12^3$ will be larger than $1729$ and the cube of any number larger than $12$ will be larger than $1729$
The reason there are $66$ sums to inspect is that the $1$st row will have $1$ sum, the $2$nd row will have $2$ sums, all the way up to the $11$th row having $11$ sums. 
From Closed Form for Triangular Numbers, this sum is known to be:
:$\ds T_{11} = \sum_{i \mathop = 1}^{11} i = \frac {11 \paren {11 + 1} } 2 = 66$
Finally, we note that the only other strictly positive integer(s) that "almost" took this honor from $1729$ are $854$ and $855$ which are only apart by $1$.
:800px
{{qed}}
Category:Hardy-Ramanujan Numbers
Category:Taxicab Numbers
1vbjotsazuxot1el37cw4ejs4857z4h
Theorem: Cardinality of Integer Interval

Let $a, b \in \Z$ be integers.
Let $\left[{a \,.\,.\, b}\right]$ denote the integer interval between $a$ and $b$.
Then $\left[{a \,.\,.\, b}\right]$ is finite and its cardinality equals:
:$\begin{cases} 
b - a + 1 & : b \ge a - 1 \\ 
0 & : b \le a - 1 
\end{cases}$

Proof:
Let $b < a$.
Then $\left[{a \,.\,.\, b}\right]$ is empty.
By Empty Set is Finite, $\left[{a \,.\,.\, b}\right]$ is finite.
By Cardinality of Empty Set, $\left[{a \,.\,.\, b}\right]$ has cardinality $0$.
Let $b \ge a$.
By Translation of Integer Interval is Bijection, there exists a bijection between $\left[{a \,.\,.\, b}\right]$ and $\left[{0 \,.\,.\, b - a}\right]$.
Thus $\left[{a \,.\,.\, b}\right]$ is finite of cardinality $b - a + 1$.
{{qed}}
Category:Set Theory
tr6bchkgtxm9a40pp9dzcux54f7jo80
Theorem: Indexed Summation of Sum of Mappings

Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $a, b$ be integers.
Let $\closedint a b$ denote the integer interval between $a$ and $b$.
Let $f, g: \closedint a b \to \mathbb A$ be mappings.
Let $h = f + g$ be their pointwise sum.
Then we have the equality of indexed summations:
:$\ds \sum_{i \mathop = a}^b \map h i = \sum_{i \mathop = a}^b \map f i + \sum_{i \mathop = a}^b \map g i$

Proof:
The proof proceeds by induction on $b$.
For all $b \in \Z_{\ge 0}$, let $\map P b$ be the proposition:
:$\ds \sum_{i \mathop = a}^b \map h i = \sum_{i \mathop = a}^b \map f i + \sum_{i \mathop = a}^b \map g i$
Theorem: Summation of Sum of Mappings on Finite Set

Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ be a finite set.
Let $f, g: S \to \mathbb A$ be mappings.
Let $h = f + g$ be their sum.
Then we have the equality of summations on finite sets:
:$\ds \sum_{s \mathop \in S} \map h s = \sum_{s \mathop \in S} \map f s + \sum_{s \mathop \in S} \map g s$

Proof:
Let $n$ be the cardinality of $S$.
Let $\sigma: \N_{< n} \to S$ be a bijection, where $\N_{< n}$ is an initial segment of the natural numbers.
By definition of summation, we have to prove the following equality of indexed summations:
:$\ds \sum_{i \mathop = 0}^{n - 1} \map h {\map \sigma i} = \sum_{i \mathop = 0}^{n - 1} \map f {\map \sigma i} + \sum_{i \mathop = 0}^{n - 1} \map g {\map \sigma i}$
By Sum of Mappings Composed with Mapping, $h \circ \sigma = f \circ \sigma + g \circ \sigma$.
The above equality now follows from Indexed Summation of Sum of Mappings.
{{qed}}
Theorem: Indexed Summation of Multiple of Mapping

Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $a, b$ be integers.
Let $\closedint a b$ denote the integer interval between $a$ and $b$.
Let $f: \closedint a b \to \mathbb A$ be a mapping.
Let $\lambda \in \mathbb A$.
Let $g = \lambda \cdot f$ be the product of $f$ with $\lambda$.
Then we have the equality of indexed summations:
:$\ds \sum_{i \mathop = a}^b \map g i = \lambda \cdot \sum_{i \mathop = a}^b \map f i$

Proof:
The proof goes by induction on $b$.
Theorem: Summation of Multiple of Mapping on Finite Set

Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ be a finite set.
Let $f: S \to \mathbb A$ be a mapping.
Let $\lambda \in \mathbb A$.
Let $g = \lambda \cdot f$ be the product of $f$ with $\lambda$.
Then we have the equality of summations on finite sets:
:$\ds \sum_{s \mathop \in S} \map g s = \lambda \cdot \sum_{s \mathop \in S} \map f s$

Proof:
Let $n$ be the cardinality of $S$.
Let $\sigma: \N_{< n} \to S$ be a bijection, where $\N_{< n}$ is an initial segment of the natural numbers.
By definition of summation, we have to prove the following equality of indexed summations:
:$\ds \sum_{i \mathop = 0}^{n - 1} \map g {\map \sigma i} = \lambda \cdot \sum_{i \mathop = 0}^{n - 1} \map f {\map \sigma i}$
By Multiple of Mapping Composed with Mapping, $g \circ \sigma = \lambda \cdot \paren {f \circ \sigma}$.
The above equality now follows from Indexed Summation of Multiple of Mapping.
{{qed}}
Theorem: Linear Combination of Indexed Summations

Let $\mathbb A$ be one of the standard number systems $\N,\Z,\Q,\R,\C$.
Let $a,b$ be integers.
Let $\closedint a b$ denote the integer interval between $a$ and $b$.
Let $f, g : \closedint a b \to \mathbb A$ be mappings.
Let $\lambda, \mu \in \mathbb A$.
Let $\lambda \cdot f + \mu \cdot g$ be the sum of the product of $f$ with $\lambda$ and the product of $g$ with $\mu$.
Then we have the equality of indexed summations:
:$\ds \sum_{i \mathop = a}^b \paren {\lambda \cdot \map f i + \mu \cdot \map g i} = \lambda \cdot \sum_{i \mathop = a}^b \map f i + \mu \cdot \sum_{i \mathop = a}^b \map g i$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = a}^b \paren {\lambda \cdot \map f i + \mu \cdot \map g i}
      | r = \sum_{i \mathop = a}^b \paren {\lambda \cdot \map f i} + \sum_{i \mathop = a}^b \paren {\mu \cdot \map g i}
      | c = Indexed Summation of Sum of Mappings
}}
{{eqn | r = \lambda \cdot \sum_{i \mathop = a}^b \map f i + \mu \cdot \sum_{i \mathop = a}^b \map g i
      | c = Indexed Summation of Multiple of Mapping
}}
{{end-eqn}}
{{qed}}
Theorem: Triangle Inequality for Indexed Summations

Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $a,b$ be integers.
Let $\closedint a b$ denote the integer interval between $a$ and $b$.
Let $f : \closedint a b \to \mathbb A$ be a mapping.
Let $\size {\, \cdot \,}$ denote the standard absolute value.
Let $\vert f \vert$ be the absolute value of $f$.
Then we have the inequality of indexed summations:
:$\ds \size {\sum_{i \mathop = a}^b \map f i} \le \sum_{i \mathop = a}^b \size {\map f i}$

Proof:
The proof goes by induction on $b$.
Theorem: Triangle Inequality for Summation over Finite Set

Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ be a finite set.
Let $f : S \to \mathbb A$ be a mapping.
Let $\size {\, \cdot\,}$ denote the standard absolute value.
Let $\size f$ be the absoute value of $f$.
Then we have the inequality of summations on finite sets:
:$\ds \size {\sum_{s \mathop \in S} \map f s} \le \sum_{s \mathop \in S} \size {\map f s}$

Proof:
Let $n$ be the cardinality of $S$.
Let $\sigma: \N_{< n} \to S$ be a bijection, where $\N_{<n}$ is an initial segment of the natural numbers.
By definition of summation, we have to prove the following inequality of indexed summations:
:$\ds \size {\sum_{i \mathop = 0}^{n - 1} \map f {\map \sigma i} } \le \sum_{i \mathop = 0}^{n - 1} \map {\paren {\size f \circ \sigma} } i$
By Absolute Value of Mapping Composed with Mapping:
:$\size f \circ \sigma = \size {f \circ \sigma}$
The above equality now follows from Triangle Inequality for Indexed Summations.
{{qed}}
Theorem: Exchange of Order of Indexed Summations/Rectangular Domain

Let $D = \closedint a b \times \closedint c d$ be the cartesian product.
Let $f: D \to \mathbb A$ be a mapping
Then we have an equality of indexed summations:
:$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = c}^d \map f {i, j} = \sum_{j \mathop = c}^d \sum_{i \mathop = a}^b \map f {i, j}$

Proof:
The proof proceeds by induction on $d$.
Theorem: Exchange of Order of Summations over Finite Sets/Cartesian Product

Let $f: S \times T \to \mathbb A$ be a mapping.
Then we have an equality of summations over finite sets:
:$\ds \sum_{s \mathop \in S} \sum_{t \mathop \in T} \map f {s, t} = \sum_{t \mathop \in T} \sum_{s \mathop \in S} \map f {s, t}$

Proof:
Let $n$ be the cardinality of $T$.
The proof goes by induction on $n$.
=== Basis for the Induction ===
Let $n = 0$.
{{finish}}
=== Induction Step ===
Let $n > 0$.
Let $t \in T$.
Use Cardinality of Set minus Singleton
{{ProofWanted}}
Theorem: Exchange of Order of Summations over Finite Sets

Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S, T$ be finite sets.
Let $S \times T$ be their cartesian product.

Proof:
Let $n$ be the cardinality of $T$.
The proof goes by induction on $n$.
=== Basis for the Induction ===
Let $n = 0$.
{{finish}}
=== Induction Step ===
Let $n > 0$.
Let $t \in T$.
Use Cardinality of Set minus Singleton
{{ProofWanted}}
Theorem: Sum over Complement of Finite Set

Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ be a finite set.
Let $f: S \to \mathbb A$ be a mapping.
Let $T \subseteq S$ be a subset.
Let $S \setminus T$ be its relative complement.
Then we have the equality of summations over finite sets:
:$\ds \sum_{s \mathop \in S \setminus T} \map f s = \sum_{s \mathop \in S} \map f s - \sum_{t \mathop \in T} \map f t$

Proof:
Note that by Subset of Finite Set is Finite, $T$ is indeed finite.
By Set is Disjoint Union of Subset and Relative Complement, $S$ is the disjoint union of $S \setminus T$ and $T$.
The result now follows from Sum over Disjoint Union of Finite Sets.
{{qed}}
Theorem: Mapping Defines Additive Function of Subalgebra of Power Set

Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ be a finite set.
Let $f: S \to \mathbb A$ be a mapping.
Let $B$ be an algebra of sets over $S$.
Define $\Sigma: B \to \mathbb A$ using summation as:
:$\ds \map \Sigma T = \sum_{t \mathop \in T} \map f t$
for $T\subseteq S$.
Then $\Sigma$ is an additive function on $B$.

Proof:
Note that by Subset of Finite Set is Finite, $B$ consists of finite sets.
The result now follows from Sum over Disjoint Union of Finite Sets.
{{qed}}
Theorem: Sum over Union of Finite Sets

Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ and $T$ be finite sets.
Let $f: S \cup T \to \mathbb A$ be a mapping.
Then we have the equality of summations over finite sets:
:$\ds \sum_{u \mathop \in S \mathop \cup T} \map f u = \sum_{s \mathop \in S} \map f s + \sum_{t \mathop \in T} \map f t - \sum_{v \mathop \in S \mathop \cap T} \map f v$

Proof:
Follows from:
:Mapping Defines Additive Function of Subalgebra of Power Set
:Power Set is Algebra of Sets
:Inclusion-Exclusion Principle
{{qed}}
Theorem: Summation over Finite Set Equals Summation over Support

Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ be a finite set.
Let $f: S \to \mathbb A$ be a mapping.
Let $\map \supp f$ be its support.
Then we have an equality of summations over finite sets:
:$\ds \sum_{s \mathop \in S} \map f s = \sum_{s \mathop \in \map \supp f} \map f s$

Proof:
Note that by Subset of Finite Set is Finite, $\map \supp f$ is indeed finite.
The result now follows from:
* Sum over Complement of Finite Set
* Sum of Zero over Finite Set
* Identity Element of Addition on Numbers
{{qed}}
Theorem: Summation of Zero/Indexed Summation

Let $a, b$ be integers.
Let $\closedint a b$ denote the integer interval between $a$ and $b$.
Let $f_0 : \closedint a b \to \mathbb A$ be the zero mapping.
Then the indexed summation of $0$ from $a$ to $b$ equals zero:
:$\ds \sum_{i \mathop = a}^b \map {f_0} i = 0$

Proof:
At least three proofs are possible:
* by induction, using Identity Element of Addition on Numbers
* using Indexed Summation of Multiple of Mapping
* using Indexed Summation of Sum of Mappings
{{ProofWanted}}
Category:Summations
drp3mu7oymvpogx6al6gdb0lpczczok
Theorem: Summation of Zero/Finite Set

Let $S$ be a finite set.
Let $0 : S \to \mathbb A$ be the zero mapping.
{{explain|Presumably the above is a constant mapping on $0$ -- needs to be made explicit.}}
Then the summation of $0$ over $S$ equals zero:
:$\ds \sum_{s \mathop \in S} 0 \left({s}\right) = 0$

Proof:
At least three proofs are possible:
:using the definition of summation and Indexed Summation of Zero
:using Indexed Summation of Sum of Mappings
:using Summation of Multiple of Mapping on Finite Set.
{{ProofWanted}}
Category:Summations
h5knszknmzvjg820yqugvmdcw8um0sv
Theorem: Summation of Zero/Set

Let $S$ be a set.
Let $0: S \to \mathbb A$ be the zero mapping.
Then the summation with finite support of $0$ over $S$ equals zero:
:$\ds \sum_{s \mathop \in S} \map 0 s = 0$

Proof:
By Support of Zero Mapping, the support of $0$ is empty.
By Empty Set is Finite, the support of $0$ is indeed finite.
By Summation over Empty Set:
:$\ds \sum_{s \mathop \in S} \map 0 s = \sum_{s \mathop \in \O} \map 0 s = 0$
{{qed}}
Category:Summations
7euo1h4t8r8qpdusxsnkr2bmrj3ezg6
Theorem: Exchange of Order of Summations over Finite Sets/Subset of Cartesian Product

Let $D\subset S \times T$ be a subset.
Let $\pi_1 : D \to S$ and $\pi_2 : D \to T$ be the restrictions of the projections of $S\times T$.
Then we have an equality of summations over finite sets:
:$\ds \sum_{s \mathop \in S} \sum_{t \mathop \in \map {\pi_2} {\map {\pi_1^{-1} } s} } \map f {s, t} = \sum_{t \mathop \in T} \sum_{s \mathop \in \map {\pi_1} {\map {\pi_2^{-1} } t} } \map f {s, t}$

Proof:
Define an extension $\overline f$ of $f$ to $S \times T$ by:
:$\map {\overline f} {s, t} = \begin{cases}
\map f {s, t} & : \tuple {s, t} \in D \\
0 & : \tuple {s, t} \notin D
\end{cases}$
Then for all $s \in S$, by:
:Preimage of Disjoint Union is Disjoint Union
:Sum over Disjoint Union of Finite Sets
:Summation over Finite Set of Zero:
:$\ds \sum_{t \mathop \in \map {\pi_2} {\map {\pi_1^{-1} } s} } \map f {s, t} = \sum_{t \mathop \in T} \map {\overline f} {s, t}$
Thus:
:$\ds \sum_{s \mathop \in S} \sum_{t \mathop \in \map {\pi_2} {\map {\pi_1^{-1} } s} } \map f {s, t} = \sum_{s \mathop \in S} \sum_{t \mathop \in T} \map {\overline f} {s, t}$
Similarly:
:$\ds \sum_{t \mathop \in T} \sum_{s \mathop \in \map {\pi_1} {\map {\pi_2^{-1} } t} } \map f {s, t} = \sum_{t \mathop \in T} \sum_{s \mathop \in S} \map {\overline f} {s, t}$
By Exchange of Order of Summation over Cartesian Product of Finite Sets, the result follows.
{{qed}}
Category:Summations
plkzn6lij7ibt6uq3x1g1ysgm5desxd
Theorem: Algebra Defined by Ring Homomorphism on Ring with Unity is Unitary

Let $R$ be a commutative ring.
Let $\struct {S, +, *}$ be a ring with unity.
Let $f: R \to S$ be a ring homomorphism.
Let $\struct {S_R, *}$ be the algebra defined by the ring homomorphism $f$.
Then $\struct {S_R, *}$ is a unitary algebra.

Proof:
By definition, the multiplication of $\struct {S_R, *}$ is the ring product of $S$.
Thus it follows immediately from the fact that $S$ is a ring with unity, that $\struct {S_R, *}$ is a unitary algebra.
{{qed}}
Category:Algebras
Category:Unital Algebras
3ecweu0z3ty2ohlojp54ic7qg4wpq4n
Theorem: Algebra Defined by Ring Homomorphism is Associative

Let $R$ be a commutative ring.
Let $\struct {S, +, *}$ be a ring with unity.
Let $f: R \to S$ be a ring homomorphism.
Let the image of $f$ be a subset of the center of $S$.
Let $\struct {S_R, *}$ be the algebra defined by the ring homomorphism $f$.
Then $\struct {S_R, *}$ is an associative algebra.

Proof:
By definition, the multiplication of $\struct {S_R, *}$ is the ring product of $S$.
Thus it follows immediately from the fact that $S$ is a ring, that $\struct {S_R, *}$ is an associative algebra.
{{qed}}
Theorem: Algebra Defined by Ring Homomorphism on Commutative Ring is Commutative

Let $R$ be a commutative ring.
Let $\struct {S, +, *}$ be a commutative ring.
Let $f: R \to S$ be a ring homomorphism.
Let $\struct {S_R, *}$ be the algebra defined by the ring homomorphism $f$.
Then $\struct {S_R, *}$ is a commutative algebra.

Proof:
Note that by Center of Commutative Ring, the image of $f$ is indeed a subset of the center of $S$.
By definition, the multiplication of $\struct {S_R, *}$ is the ring product of $S$.
Thus it follows immediately from the fact that $S$ is a ring, that $\struct {S_R, *}$ is a commutative algebra.
{{qed}}
Theorem: Smallest Integer which is Sum of 3 Sixth Powers in 2 Ways

The smallest positive integer which can be expressed as the sum of $3$ sixth powers in $2$ different ways is:
{{begin-eqn}}
{{eqn | l = 160 \, 426 \, 514
      | r = 3^6 + 19^6 + 22^6
      | c = 
}}
{{eqn | r = 10^6 + 15^6 + 23^6
      | c = 
}}
{{end-eqn}}
Also note that:
{{begin-eqn}}
{{eqn | l = 854
      | r = 3^2 + 19^2 + 22^2
      | c = 
}}
{{eqn | r = 10^2 + 15^2 + 23^2
      | c = 
}}
{{end-eqn}}

Proof:
We have that:
{{begin-eqn}}
{{eqn | l = 160 \, 426 \, 514
      | r = 729 + 47 \, 045 \, 881 + 113 \, 379 \, 904
      | c = 
}}
{{eqn | r = 3^6 + 19^6 + 22^6
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 160 \, 426 \, 514
      | r = 1 \, 000 \, 000 + 11 \, 390 \, 625 + 148 \, 035 \, 889
      | c = 
}}
{{eqn | r = 10^6 + 15^6 + 23^6
      | c = 
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown there are no smaller}}
Theorem: Infinite Number of Integers which are Sum of 3 Sixth Powers in 2 Ways

There exist an infinite number of positive integers which can be expressed as the sum of $3$ sixth powers in $2$ different ways.

Proof:
There are many parametric solutions to $x^6 + y^6 + z^6 = u^6 + v^6 + w^6$. One is given by:
{{begin-eqn}}
{{eqn | l = x
      | r = 2 m^4 + 4 m^3 n - 5 m^2 n^2 - 12 m n^3 - 9 n^4
}}
{{eqn | l = y
      | r = 3 m^4 + 9 m^3 n + 18 m^2 n^2 + 21 m n^3 + 9 n^4
}}
{{eqn | l = z
      | r = -m^4 - 10 m^3 n - 17 m^2 n^2 - 12 m n^3
}}
{{eqn | l = u
      | r = m^4 - 3 m^3 n - 14 m^2 n^2 - 15 m n^3 - 9 n^4
}}
{{eqn | l = v
      | r = 3 m^4 + 8 m^3 n + 9 m^2 n^2
}}
{{eqn | l = w
      | r = 2 m^4 + 12 m^3 n + 19 m^2 n^2 + 18 m n^3 + 9 n^4
}}
{{end-eqn}}
This set of solutions also satisfy:
{{begin-eqn}}
{{eqn | l = x^2 + y^2 + z^2 
      | r = u^2 + v^2 + w^2
}}
{{eqn | l = 3 x + y + z
      | r = 3 u + v + w
}}
{{end-eqn}}
{{finish|For people who enjoy degree $24$ polynomials in $2$ unknowns}}
Theorem: Consecutive Primes of form 4n+1

The sequence of $16$ consecutive prime numbers beginning from $207 \, 622 \, 273$ are all of the form $4 n + 1$.

Proof:
{{begin-eqn}}
{{eqn | n = 1
      | l = 207 \, 622 \, 273
      | r = 4 \times 51 \, 905 \, 568 + 1
      | c = and is the $11 \, 477 \, 482$nd prime
}}
{{eqn | n = 2
      | l = 207 \, 622 \, 297
      | r = 4 \times 51 \, 905 \, 574 + 1
      | c = and is the $11 \, 477 \, 483$rd prime
}}
{{eqn | n = 3
      | l = 207 \, 622 \, 301
      | r = 4 \times 51 \, 905 \, 575 + 1
      | c = and is the $11 \, 477 \, 484$th prime
}}
{{eqn | n = 4
      | l = 207 \, 622 \, 313
      | r = 4 \times 51 \, 905 \, 578 + 1
      | c = and is the $11 \, 477 \, 485$th prime
}}
{{eqn | n = 5
      | l = 207 \, 622 \, 321
      | r = 4 \times 51 \, 905 \, 580 + 1
      | c = and is the $11 \, 477 \, 486$th prime
}}
{{eqn | n = 6
      | l = 207 \, 622 \, 381
      | r = 4 \times 51 \, 905 \, 595 + 1
      | c = and is the $11 \, 477 \, 487$th prime
}}
{{eqn | n = 7
      | l = 207 \, 622 \, 409
      | r = 4 \times 51 \, 905 \, 602 + 1
      | c = and is the $11 \, 477 \, 488$th prime
}}
{{eqn | n = 8
      | l = 207 \, 622 \, 417
      | r = 4 \times 51 \, 905 \, 604 + 1
      | c = and is the $11 \, 477 \, 489$th prime
}}
{{eqn | n = 9
      | l = 207 \, 622 \, 421
      | r = 4 \times 51 \, 905 \, 605 + 1
      | c = and is the $11 \, 477 \, 490$th prime
}}
{{eqn | n = 10
      | l = 207 \, 622 \, 489
      | r = 4 \times 51 \, 905 \, 622 + 1
      | c = and is the $11 \, 477 \, 491$st prime
}}
{{eqn | n = 11
      | l = 207 \, 622 \, 501
      | r = 4 \times 51 \, 905 \, 625 + 1
      | c = and is the $11 \, 477 \, 492$nd prime
}}
{{eqn | n = 12
      | l = 207 \, 622 \, 517
      | r = 4 \times 51 \, 905 \, 629 + 1
      | c = and is the $11 \, 477 \, 493$rd prime
}}
{{eqn | n = 13
      | l = 207 \, 622 \, 537
      | r = 4 \times 51 \, 905 \, 634 + 1
      | c = and is the $11 \, 477 \, 494$th prime
}}
{{eqn | n = 14
      | l = 207 \, 622 \, 549
      | r = 4 \times 51 \, 905 \, 637 + 1
      | c = and is the $11 \, 477 \, 495$th prime
}}
{{eqn | n = 15
      | l = 207 \, 622 \, 553
      | r = 4 \times 51 \, 905 \, 638 + 1
      | c = and is the $11 \, 477 \, 496$th prime
}}
{{eqn | n = 16
      | l = 207 \, 622 \, 561
      | r = 4 \times 51 \, 905 \, 640 + 1
      | c = and is the $11 \, 477 \, 497$th prime
}}
{{end-eqn}}
Note that the $11 \, 477 \, 481$st prime:
:$207 \, 622 \, 271 = 4 \times 51 \, 905 \, 568 - 1$
and the $11 \, 477 \, 498$th prime:
:$207 \, 622 \, 567 = 4 \times 51 \, 905 \, 642 - 1$
and so are not of the form $4 n + 1$.
{{qed}}
Theorem: Integer which is Sum of 3 Fourth Powers in 2 Ways and Products of Those Roots

The positive integer $256 \, 103 \, 393$ can be expressed as the sum of $3$ fourth powers in $2$ different ways:
{{begin-eqn}}
{{eqn | l = 256 \, 103 \, 393
      | r = 22^4 + 93^4 + 116^4
      | c = 
}}
{{eqn | r = 29^4 + 66^4 + 124^4
      | c = 
}}
{{end-eqn}}
Also note that:
{{begin-eqn}}
{{eqn | l = 237 \, 336
      | r = 22 \times 93 \times 116
      | c = 
}}
{{eqn | r = 29 \times 66 \times 124
      | c = 
}}
{{end-eqn}}

Proof:
We have that:
{{begin-eqn}}
{{eqn | l = 256 \, 103 \, 393
      | r = 234 \, 256 + 74 \, 805 \, 201 + 181 \, 063 \, 936
      | c = 
}}
{{eqn | r = 22^4 + 93^4 + 116^4
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 256 \, 103 \, 393
      | r = 707 \, 281 + 18 \, 974 \, 736 + 236 \, 421 \, 376
      | c = 
}}
{{eqn | r = 29^4 + 66^4 + 124^4
      | c = 
}}
{{end-eqn}}
Then we have:
{{begin-eqn}}
{{eqn | l = 22 \times 93 \times 116
      | r = \left({2 \times 11}\right) \times \left({3 \times 31}\right) \times \left({2^2 \times 29}\right)
      | c = 
}}
{{eqn | r = 2^3 \times 3 \times 11 \times 29 \times 31
      | c = 
}}
{{eqn | r = 237 \, 336
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 29 \times 66 \times 124
      | r = 29 \times \left({2 \times 3 \times 11}\right) \times \left({2^2 \times 31}\right)
      | c = 
}}
{{eqn | r = 2^3 \times 3 \times 11 \times 29 \times 31
      | c = 
}}
{{eqn | r = 237 \, 336
      | c = 
}}
{{end-eqn}}
{{qed}}