nyuuzyou/wikis huggingface dataset:
Integer Powers of 2 under Multiplication form Infinite Abelian Group

== Theorem ==
Let $S$ be the set of integers defined as:
:$S = \set {2^k: k \in \Z}$
Then $\struct {S, \times}$ is an infinite abelian group.
== Proof ==
It can be seen by inspection that $S \subseteq \Q_{>0}$.
That is, all the elements of $S$ are strictly positive rational numbers.
From Strictly Positive Rational Numbers under Multiplication form Countably Infinite Abelian Group:
:$\struct {\Q_{>0}, \times}$ is an infinite abelian group.
It is noted that $S$ is an infinite set and so trivially $S \ne \O$.
Let $a, b \in S$.
Then:
:$\exists k_1 \in \Z: a = 2^{k_1}$
:$\exists k_2 \in \Z: b = 2^{k_2}$
We have that:
:$\dfrac 1 {2^{k_2} } 2^{k_2} = 1$
and so $\dfrac 1 {2^{k_2} } = 2^{-k_2} = b^{-1}$ is the inverse of $b \in \struct {\Q_{>0}, \times}$.
Then we have:
We have that:
:$k_1 + k_2 \in \Z$
and so $a \times b^{-1} \in \S$.
Hence by the One-Step Subgroup Test, $\struct {S, \times}$ is a subgroup of $\struct {\Q_{>0}, \times}$.
It has been established that $S$ is an infinite set.
Hence by definition $\struct {S, \times}$ is an infinite group.
Finally, from Subgroup of Abelian Group is Abelian, $\struct {S, \times}$ is an abelian group.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: Examples: $(1) \ \text {(a)}$
Category:Examples of Abelian Groups
Category:Examples of Infinite Groups
Integer Reciprocal Space contains Cauchy Sequence with no Limit

== Theorem ==
Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $\struct {A, \tau_d}$ be the integer reciprocal space under the usual (Euclidean) topology.
Then $A$ has a Cauchy sequence which has no limit in $A$.
== Proof ==
Let $\sequence {x_n}$ be the sequence $1, \dfrac 1 2, \dfrac 1 3, \ldots$
Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.
By the Axiom of Archimedes:
:$\exists \N \in n: n > \dfrac 1 \epsilon$
and so:
:$\exists \N \in n: \dfrac 1 n < \epsilon$
As:
:$0 < \dfrac 1 {n + 1} < \dfrac 1 n$
it follows that:
:$\size {\dfrac 1 n - \dfrac 1 {n + 1} } < \epsilon$
and so $\sequence {x_n}$ is a Cauchy sequence.
From Sequence of Powers of Reciprocals is Null Sequence, $\sequence {x_n}$ is a basic null sequence.
That is:
:$\ds \lim_{n \mathop \to \infty} \frac 1 n = 0$
As $0 \notin A$, the result follows.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $1 \ \text{(c)}$
Category:Integer Reciprocal Space
Integer Reciprocal Space is Topological Space

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line $\R$ under the usual (Euclidean) topology $\tau_d$.
Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set {\dfrac 1 n: n \in \Z_{>0} }$
Then the integer reciprocal space $\struct {A, \tau_d}$ is a topological space.
== Proof ==
We have that $A \subseteq \R$.
By definition, $\struct {A, \tau_d}$ is a subspace of $\struct {\R, \tau_d}$.
Hence the result from Topological Subspace is Topological Space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $1$
Category:Integer Reciprocal Space
Integer Reciprocal Space with Zero is Totally Separated

== Theorem ==
Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set 0 \cup \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $\struct {A, \tau_d}$ be the integer reciprocal space with zero under the usual (Euclidean) topology.
Then $A$ is totally separated.
== Proof ==
Let $a, b \in A$ such that $a < b$.
From Between two Rational Numbers exists Irrational Number:
:$\exists \alpha \in \R \setminus \Q: a < \alpha < b$
Because $\forall x \in A: x \in \Q$ it follows that $\alpha \notin A$.
Consider the half-open intervals $S = \hointr 0 \alpha$ and $T = \hointl \alpha 1$
Let:
:$U := S \cap A, V := T \cap A$
Let $\beta \in A$.
Then either:
:$(1): \quad \beta < \alpha$
in which case:
:$\beta \in U$
or:
:$(2): \quad \beta > \alpha$
in which case:
:$\beta \in V$
Thus $U \cup V = A$.
Let $a \in U$.
Then $a < \alpha$ and so $a \notin V$.
Similarly, let $b \in V$.
Then $b > \alpha$ and so $b \notin U$.
Then note that $x \in U$ and $y \in V$.
Thus $U \ne \O$ and $V \ne \O$.
Thus, by definition, $U$ and $V$ constitute a separation of $A$ such that $x \in U$ and $y \in V$.
Hence the result by definition of totally separated.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $2 \ \text{(c)}$
Category:Integer Reciprocal Space
Integer Reciprocal Space with Zero is not Extremally Disconnected

== Theorem ==
Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set 0 \cup \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $\struct {A, \tau_d}$ be the integer reciprocal space with zero under the usual (Euclidean) topology.
Then $A$ is not extremally disconnected.
== Proof ==
$\struct {A, \tau_d}$ is a metric space.
We have:
:Extremally Disconnected Metric Space is Discrete
We also have:
:Topological Space is Discrete iff All Points are Isolated
:Zero is Limit Point of Integer Reciprocal Space
From definition of limit point:
:$0$ is not an isolated point of $A$
Hence integer reciprocal space with zero is not the discrete space, and the result follows.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $2 \ \text{(d)}$
Category:Integer Reciprocal Space
Category:Reciprocals
Category:Extremally Disconnected Spaces
Integer Reciprocal Space with Zero is not Locally Connected

== Theorem ==
Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set 0 \cup \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $\struct {A, \tau_d}$ be the integer reciprocal space with zero under the usual (Euclidean) topology.
Then $A$ is not locally connected.
== Proof ==
We have:
:Integer Reciprocal Space with Zero is Totally Separated
:Totally Separated Space is Totally Disconnected
:Totally Disconnected and Locally Connected Space is Discrete
We also have:
:Topological Space is Discrete iff All Points are Isolated
:Zero is Limit Point of Integer Reciprocal Space
From definition of limit point:
:$0$ is not an isolated point of $A$
Hence integer reciprocal space with zero is not the discrete space, and the result follows.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $2 \ \text{(a)}$
Category:Integer Reciprocal Space
Integer Square Function is not Injective

== Example of Mapping which is Not an Injection ==
Let $f: \Z \to \Z$ be the real square function:
:$\forall x \in \Z: \map f x = x^2$
Then $f$ is not an injection.
== Proof ==
For $f$ to be an injection, it would be necessary that:
:$\forall x_1, x_2 \in \R: \map f {x_1} = \map f {x_2} \implies x_1 = x_2$
By definition of the squaring operation, we have:
:$\map f x = \map f {-x}$
But unless $x = 0$ it is not the case that $x = -x$.
Hence $f$ is not an injection.
== Sources ==
* : Chapter $3$. Mappings: Exercise $2 \ \text {(i) (a)}$
Category:Examples of Injections
Category:Square Function
Integer Square Function is not Surjective

== Example of Mapping which is Not a Surjection ==
Let $f: \Z \to \Z$ be the real square function:
:$\forall x \in \Z: \map f x = x^2$
Then $f$ is not a surjection.
== Proof ==
For $f$ to be a surjection, it would be necessary that:
:$\forall y \in \Z: \exists x \in \Z: \map f x = y$
However from Square of Real Number is Non-Negative:
:$\forall y \in \Z_{< 0}: \nexists x \in \Z: \map f x = y$
We also have that, for example, Square Root of Prime is Irrational.
So for all $p \in \Z$ such that $p$ is prime:
:$\nexists x \in \Z: \map f x = p$
Hence $f$ is not a surjection.
== Sources ==
* : Chapter $3$. Mappings: Exercise $2 \ \text {(i) (b)}$
Category:Examples of Surjections
Category:Square Function
Integer Subtraction is Closed

== Theorem ==
The set of integers is closed under subtraction:
:$\forall a, b \in \Z: a - b \in \Z$
== Proof ==
From the definition of subtraction:
:$a - b := a + \paren {-b}$
where $-b$ is the inverse for integer addition.
From Integers under Addition form Abelian Group, the algebraic structure $\struct {\Z, +}$ is a group.
Thus:
:$\forall a, b \in \Z: a + \paren {-b} \in \Z$
Therefore integer subtraction is closed.
== Sources ==
* : Chapter $1$: Integral Domains: $\S 2$. Operations: Example $1$
* : $1$: Complex Numbers: The Real Number System: $2$
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers
Category:Integer Subtraction
Category:Algebraic Closure
Integer and Fifth Power have same Last Digit

== Theorem ==
Let $n \in \Z$ be an integer.
Then $n^5$ has the same last digit as $n$ when both are expressed in conventional decimal notation.
== Proof ==
From Fermat's Little Theorem: Corollary 1:
:$n^5 \equiv n \pmod 5$
Suppose $n \equiv 1 \pmod 2$.
Then from Congruence of Powers:
:$n^5 \equiv 1^5 \pmod 2$
and so:
:$n^5 \equiv 1 \pmod 2$
Similarly, suppose $n \equiv 0 \pmod 2$.
Then from Congruence of Powers:
:$n^5 \equiv 0^5 \pmod 2$
and so:
:$n^5 \equiv 0 \pmod 2$
Hence:
:$n^5 \equiv n \pmod 2$
So we have, by Chinese Remainder Theorem:
:$n^5 \equiv n \pmod {2 \times 5}$
and the result follows.
== Sources ==
* : $\text {3-2}$ Fermat's Little Theorem: Exercise $3$
Category:Fifth Powers
Integer as Difference between Two Squares/Formulation 1

== Theorem ==
Let $n$ be a positive integer.
Then $n$ can be expressed as:
:$n = a^2 - b^2$
 $n$ has at least two distinct divisors of the same parity that multiply to $n$.
== Proof ==
Thus $n = p q$ where:
Thus for $a$ and $b$ to be integers, both $p$ and $q$ must be:
:distinct, otherwise $p = q$ and so $b = 0$
:either both even or both odd, otherwise both $p + q$ and $p - q$ will be odd, and so neither $\dfrac {p + q} 2$ nor $\dfrac {p - q} 2$ are defined in $\Z$.
Hence the result.
Category:Square Numbers
Category:Difference of Two Squares
Integer as Difference between Two Squares/Formulation 2

== Theorem ==
Any integer can be expressed as the difference of two squares that integer is NOT $n \equiv 2 \pmod 4$
== Proof ==
: Each integer will be in one of the 4 sets of residue classes modulo $4$:
: $n \equiv 0 \pmod 4$
: $n \equiv 1 \pmod 4$
: $n \equiv 2 \pmod 4$
: $n \equiv 3 \pmod 4$
: For $n \equiv 2 \pmod 4$, it is impossible to represent such an integer as the difference of two squares.
Taking the squares of each of the residue classes, we have:
: $0^2 \equiv 0 \pmod 4$
: $1^2 \equiv 1 \pmod 4$
: $2^2 \equiv 0 \pmod 4$
: $3^2 \equiv 1 \pmod 4$
: Therefore, when taking the difference of two squares, $n \equiv 2 \pmod 4$ is never a result.
: $0 - 0 \equiv 0 \pmod 4$
: $1 - 1 \equiv 0 \pmod 4$
: $0 - 1 \equiv 3 \pmod 4$
: $1 - 0 \equiv 1 \pmod 4$
Category:Square Numbers
Category:Difference of Two Squares
Integer as Sum of 2 Cubes in 3 Ways/Mistake

== Source Work ==
:
:The Dictionary
::$4104$
== Mistake ==
:The smallest integer that can be represented as the sum of $2$ cubes in $3$ different ways: $4104 = 16^3 + 2^3 = 15^3 + 9^3 = \paren {-12}^3 + 18^3$.
== Correction ==
There is at least one smaller:
Also note the assumption that it should say positive integer.
There is no indication where got this from.
== Sources ==
* : $4104$
Category:Mistakes/Curious and Interesting Numbers
Integer as Sum of 4 Cubes

== Theorem ==
Let $n \in \Z$ be an integer.
Let $n \not \equiv 4 \pmod 9$ and $n \not \equiv 5 \pmod 9$.
Then it is possible to express $n$ as the sum of no more than $4$ cubes which may be either positive or negative.
== Proof ==
== Examples ==
== Also see ==
* Compare with the Hilbert-Waring theorem for $k = 3$: if the cubes all have to be positive then as many as $9$ may be needed.
== Sources ==
* : $23$
* : $23$
Category:Cube Numbers
Category:Integer as Sum of 4 Cubes
Integer as Sum of 4 Cubes/Examples/23

== Example of Integer as Sum of 4 Cubes ==
:$23 = 8 + 8 + 8 - 1 = 2^3 + 2^3 + 2^3 + \paren {-1}^3$
== Sources ==
* : $23$
* : $23$
Category:Integer as Sum of 4 Cubes
Category:23
Integer as Sum of 5 Non-Zero Squares

== Theorem ==
Let $n \in \Z$ be an integer such that $n > 33$.
Then $n$ can be expressed as the sum of $5$ non-zero squares.
== Proof ==
From Lagrange's Four Square Theorem, every positive integer can be expressed as the sum of $4$ squares, some of which may be zero.
The existence of positive integers which cannot be expressed as the sum of $4$ non-zero squares is noted by the trivial examples $1$, $2$ and $3$.
Thus Lagrange's Four Square Theorem can be expressed in the form:
:$(1): \quad$ Every positive integer can be expressed as the sum of $1$, $2$, $3$ or $4$ non-zero squares.
We note the following from 169 as Sum of up to 155 Squares:
Let $n > 169$.
Then $n$ can be expressed as:
:$n = m + 169$
where $m \ge 1$.
From $(1)$, $m$ can be expressed as the sum of sum of $1$, $2$, $3$ or $4$ non-zero squares.
Thus at least one of the following holds:
:$m = a^2$
:$m = a^2 + b^2$
:$m = a^2 + b^2 + c^2$
:$m = a^2 + b^2 + c^2 + d^2$
Thus one of the following holds:
It remains to be shown that of the positive integers less than $169$, all but the following can be expressed in this way:
:$1, 2, 3, 4, 6, 7, 9, 10, 12, 15, 18, 33$
Note that by Integer as Sum of Three Squares, all integers not of the form:
:$4^n \paren {8 m + 7}$
can be written as a sum of $1$, $2$ or $3$ non-zero squares.
Also note that:
Similar to the above, for $x = y + 18$ where $y \ne 4^n \paren {8 m + 7}$, at least one of the following holds:
the ineligible $0 < y < 151$ are:
:$7, 15, 23, 28, 31, 39, 47, 55, 60, 63, 71, 79, 87, 92, 95, 103, 111, 112, 119, 124, 127, 135, 143$
with corresponding $x$:
:$25, 33, 41, 46, 49, 57, 65, 73, 78, 81, 89, 97, 105, 110, 113, 121, 129, 130, 137, 142, 145, 153, 161$
for $x > 18$.
Similarly, for $45$:
So if we can write $x = y + 45$ where $y \ne 4^n \paren {8 m + 7}$, $x$ can be expressed as a sum of $5$ non-zero squares.
The ineligible $x > 45$ for $0 < y < 124$ are:
:$52, 60, 68, 73, 76, 84, 92, 100, 105, 108, 116, 124, 132, 137, 140, 148, 156, 157, 164$
Comparing both lists, we only need to check:
:$x < 18$ and $x = 25, 33, 41, 73, 105, 137$
And we have:
while for the rest:
:$1, 2, 3, 4 < 5 \times 1^2$
:$5 \times 1^2 < 6, 7 < 2^2 + 4 \times 1^2$
:$2^2 + 4 \times 1^2 < 9, 10 < 2 \times 2^2 + 3 \times 1^2$
$12, 15, 18, 33$ are divisible by $3$.
By Square Modulo 3, $n^2 \equiv 0$ or $1 \pmod 3$.
We must require the $5$ non-zero squares to be equivalent to:
:$0, 0, 1, 1, 1 \pmod 3$
or
:$0, 0, 0, 0, 0 \pmod 3$
The smallest non-zero square divisible by $3$ is $3^2 = 9$.
The sum of the squares must therefore be greater than:
:$3^2 + 3^2 = 18$
hence $12, 15, 18$ cannot be expressed as the sum of $5$ non-zero squares.
Since $6^2 > 33$, we must have $33 = 3^2 + 3^2 + a^2 + b^2 + c^2$.
But $33 - 3^2 - 3^2 = 15$ cannot be expressed as the sum of $3$ non-zero squares, as it is of the form $4^n \paren {8 m + 7}$.
This proves the theorem.
== Sources ==
* 
* : $33$
Category:Sums of Squares
Integer as Sum of 5 Non-Zero Squares/Mistake

== Source Work ==
:
:The Dictionary
::$33$
== Mistake ==
:Any integer greater than $33$ can be written as the sum of $5$ non-zero squares. [ Jackson, Masat, Mitchell, MM v61 41 ]
The reference is incorrect.
It refers to , volume $61$, page $41$.
The actual article itself is on page $41$ of volume $66$, not $61$.
== Sources ==
* : $33$
Category:Mistakes/Curious and Interesting Numbers
Integer as Sum of Polygonal Numbers

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then $n$ is:
:$(1): \quad$ Either triangular or the sum of $2$ or $3$ triangular numbers
:$(2): \quad$ Either square or the sum of $2$, $3$ or $4$ square numbers
:$(3): \quad$ Either pentagonal or the sum of $2$, $3$, $4$ or $5$ pentagonal numbers
:and so on.
That is:
:for all $k \ge 3$, $n$ is the sum of no more than $k$ polygonal numbers of order $k$.
== Proof ==
First some lemmata:
=== Lemma 1 ===
=== Lemma 2 ===
=== Lemma 3 ===
First note that Cauchy's Lemma (Number Theory) gives us:
Now we consider the numbered statements as asserted.
For $(1)$, we have Integer is Sum of Three Triangular Numbers.
For $(2)$, we have Lagrange's Four Square Theorem.
For each $m \ge 3$, we prove that $n$ is the sum of at most $m + 2$ polygonal numbers of order $m + 2$.
We choose to consider polygonal numbers of order $m + 2$, rather than $m$-gonal numbers, because the former have a neater closed form:
:$\map P {m + 2, k} = \dfrac m 2 \paren {k^2 - k} + k$
For $n < 116 m$ the result is shown in Lemma 1.
It remains to investigate the integers $n \ge 116 m$.
Define:
:$I = \openint {\dfrac 2 3 + \sqrt {8 \paren {\dfrac n m} - 8} } {\dfrac 1 2 + \sqrt {6 \paren {\dfrac n m} - 3} }$
By Lemma 2:
:For $\dfrac n m \ge 116$, the length of $I$ is greater than $4$.
Thus $I$ must contain at least $2$ consecutive odd integers.
Let $b_1, b_2$ be those consecutive odd integers.
Because $b_1 + m - 1 = b_2 + m - 3$:
:the set $\set {b + r: b \in \set {b_1, b_2}, r \in \set {0, 1, \dots, m - 2} }$ contains a complete residue system modulo $m$.
Hence one can choose some $b, r$ and write $n \equiv b + r \pmod m$.
Define:
:$a = 2 \paren {\dfrac {n - b - r} m} + b = \paren {1 - \dfrac 2 m} b + 2 \paren {\dfrac {n - r} m}$
Since $m \divides \paren {n - b - r}$ and $b$ is odd:
:$a$ is an odd integer.
Since $b \in I$, from Lemma 3 we have:
:$b^2 < 4 a$
:$3 a < b^2 + 2 b + 4$
Then by Cauchy's Lemma (Number Theory), we can write:
:$a = s^2 + t^2 + u^2 + v^2$
:$b = s + t + u + v$
for nonnegative integers $s, t, u, v$.
Thus:
Since $r \le m - 2$, the above expression comprises at most $m + 2$ polygonal numbers of order $m + 2$.
== Historical Note ==
== Sources ==
* 
* : Chapter $\text {A}.13$: Fermat ()
Category:Polygonal Numbers
Category:Integer as Sum of Polygonal Numbers
Integer as Sum of Polygonal Numbers/Historical Note

== Historical Note on Integer as Sum of Polygonal Numbers ==
=== Fermat's Note ===
As himself put it:
 struggled on and off with the proof for squares without success for nearly $40$ years. finally managed this in $1772$.
 proved the theorem for triangular numbers in $1796$.
Finally, proved the theorem in its completeness in $1815$.
== Sources ==
* : Chapter $\text {A}.13$: Fermat ()
Category:Historical Notes
Integer as Sum of Polygonal Numbers/Historical Note/Fermat's Note

== Historical Note on Integer as Sum of Polygonal Numbers: 's Note in 's ==
:Every positive integer is triangular or the sum of $2$ or $3$ triangular numbers; a square or the sum of $2$, $3$ or $4$ squares; a pentagonal number or the sum of $2$, $3$, $4$ or $5$ pentagonal numbers; and so on to infinity, whether it is a question of hexagonal, heptagonal or any polygonal numbers.
:I cannot give the proof here, for it depends on many abstruse mysteries of numbers; but I intend to devote an entire book to this subject, and to present in this part of number theory astonishing advances beyond previously known boundaries.
No such book was ever written, although it is generally believed that he probably could have done so.
== Sources ==
* : Chapter $\text {A}.13$: Fermat ()
Category:Historical Notes
Integer as Sum of Polygonal Numbers/Lemma 1

== Theorem ==
Let $n, m \in \N_{>0}$ such that $m \ge 3$.
Let $n < 116 m$.
Then $n$ can be expressed as a sum of at most $m + 2$ polygonal numbers of order $m + 2$.
== Proof ==
From Closed Form for Polygonal Numbers:
:$\map P {m + 2, k} = \dfrac m 2 \paren {k^2 - k} + k = m T_{k - 1} + k$
Where $T_{k - 1}$ are triangular numbers.
The first few $\paren {m + 2}$-gonal numbers less than $116 m$ are:
:$0, 1, m + 2, 3 m + 3, 6 m + 4, 10 m + 5, 15 m + 6, 21 m + 7, 28 m + 8, 36 m + 9, 45 m + 10, 55 m + 11, 66 m + 12, 78 m + 13, 91 m + 14, 105 m + 15$
We show the expression of the first few numbers explicitly.
and so on.
Category:Integer as Sum of Polygonal Numbers
Integer as Sum of Polygonal Numbers/Lemma 2

== Theorem ==
Let $n, m \in \R_{>0}$ such that $\dfrac n m \ge 1$.
Define $I$ to be the open real interval:
:$I = \openint {\dfrac 2 3 + \sqrt {8 \paren {\dfrac n m} - 8} } {\dfrac 1 2 + \sqrt {6 \paren {\dfrac n m} - 3} }$
Then:
:For $\dfrac n m \ge 116$, the length of $I$ is greater than $4$.
== Proof ==
We need to show that $\paren {\dfrac 2 3 + \sqrt {8 \paren {\dfrac n m} - 8} } - \paren {\dfrac 1 2 + \sqrt {6 \paren {\dfrac n m} - 3}} > 4$ when $\dfrac n m \ge 116$.
Let $x = \dfrac n m - 1$. 
Then:
To simplify calculations, we consider:
which is true when $x \ge 115$.
Thus this condition is satisfied when $\dfrac n m \ge 116$.
== Sources ==
* 
Category:Integer as Sum of Polygonal Numbers
Integer as Sum of Polygonal Numbers/Lemma 3

== Theorem ==
Let $n, m, r \in \R_{>0}$.
Suppose $\dfrac n m > 1$.
Let $b \in \openint {\dfrac 2 3 + \sqrt {8 \paren {\dfrac n m} - 8} } {\dfrac 1 2 + \sqrt {6 \paren {\dfrac n m} - 3} }$.
Define:
:$a = 2 \paren {\dfrac {n - b - r} m} + b = \paren {1 - \dfrac 2 m} b + 2 \paren {\dfrac {n - r} m}$
Then $a, b$ satisfy:
:$b^2 < 4 a$
:$3 a < b^2 + 2 b + 4$
== Proof ==
=== $b^2 < 4 a$ ===
:$b^2 - 4 a = b^2 - 4 \paren {1 - \dfrac 2 m} b - 8 \paren {\dfrac {n - r} m}$
By the Quadratic Formula, $b^2 - 4 a < 0$ when $b$ is between:
Observing the term in the square root, we have: 
:$2 - \dfrac 4 m - \sqrt {\paren {2 - \dfrac 4 m}^2 + 8 \paren {\dfrac n m} - 8 \paren {\dfrac r m} } < 0$
Since $b > 0$ this is satisfied.
Also we have:
showing that first inequality is satisfied.
=== $3 a < b^2 + 2 b + 4$ ===
:$b^2 + 2 b + 4 - 3 a = b^2 - \paren {1 - \dfrac 6 m} b - 6 \paren {\dfrac {n - r} m} + 4$
By the Quadratic Formula, $b^2 + 2 b + 4 - 3 a > 0$ when $b$ is greater than:
showing that second inequality is satisfied by any $b > \dfrac 1 2 + \sqrt {6 \paren {\dfrac n m} - 3}$.
== Sources ==
* 
Category:Integer as Sum of Polygonal Numbers
Integer as Sum of Seven Positive Cubes/Historical Note

== Historical Note on Integer as Sum of Seven Positive Cubes ==
 knew that some integers required at least $9$ positive cubes to represent them as a sum:
In fact these are the only two integers that need as many as $9$ positive cubes to express them.
All other integers need no more than $8$.
It was shown in $1943$ by that only finitely many numbers do require $8$ positive cubes.
That is, from some point on, $7$ cubes are enough.
It is not known what that point is.
Category:Historical Notes
Integer as Sum of Three Odd Squares

== Theorem ==
Let $r$ be a positive integer.
Then:
:$r \equiv 3 \pmod 8$
:
:$r$ is the sum of $3$ odd squares.
== Proof ==
=== Sufficient Condition ===
From Integer as Sum of Three Squares, every positive integer not of the form $4^n \paren {8 m + 7}$ can be expressed as the sum of three squares.
Hence every positive integer $r$ such that $r \equiv 3 \pmod 8$ can likewise be expressed as the sum of three squares.
From Square Modulo 8, the squares modulo $8$ are $0, 1$ and $4$.
Thus for $r$ to be the sum of three squares, each of those squares needs to be congruent modulo $8$ to $1$.
Thus each square is odd, and $r$ can be expressed in the form $8 n + 3$ as the sum of $3$ odd squares.
=== Necessary Condition ===
Suppose $r$ is the sum of $3$ odd squares.
From Odd Square Modulo 8, each of these odd squares is congruent to $1 \pmod 8$.
Therefore:
:$r \equiv 1 + 1 + 1 \pmod 8 \equiv 3 \pmod 8$
Category:Odd Squares
Category:Number Theory