nyuuzyou/wikis huggingface dataset:
Solution to Simultaneous Homogeneous Linear First Order ODEs with Constant Coefficients

== Theorem ==
Consider the system of linear first order ordinary differential equations with constant coefficients:
The general solution to $(1)$ and $(2)$ consists of the linear combinations of the following:
and:
where $A_1 : B_1 = A_2 : B_2 = r$
where $r$ is either of the roots of the quadratic equation:
:$\paren {k + a} \paren {k + d} - b c = 0$
== Proof ==
We look for solutions to $(1)$ and $(2)$ of the form:
We do of course have the Trivial Solution of Homogeneous Linear 1st Order ODE:
:$y = z = 0$
which happens when $A = B = 0$.
So let us investigate solutions where either or both of $A$ and $B$ are non-zero.
Substituting $(3)$ and $(4)$ into $(1)$ and $(2)$ and cancelling $e^{k x}$, we get::
From $(5)$ and $(6)$ we get:
So $A = B = 0$ unless $k$ is a root of the quadratic equation:
:$\paren {k + a} \paren {k + d} - b c = 0$
That is:
:$(8): \quad \begin {vmatrix} k + a & b \\ c & k + d \end {vmatrix} = 0$
where the above notation denotes the determinant.
Assume $(8)$ has distinct roots $k_1$ and $k_2$.
Taking $k = k_1$ and $k = k_2$ in $(7)$, we can obtain ratios $A_1 : B_1$ and $A_2 : B_2$ such that:
and:
are solutions of $(1)$ and $(2)$.
By taking arbitrary linear combinations of these, we obtain the general solution.
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 3$. Equations of higher order and systems of first order equations: $\S 3.2$ First order systems
Category:Systems of Differential Equations
Category:Linear First Order ODEs
Solution to Simultaneous Linear Congruences

== Theorem ==
Let:
be a system of simultaneous linear congruences.
This system has a simultaneous solution :
:$\forall i, j: 1 \le i, j \le r: \gcd \set {n_i, n_j}$ divides $b_j - b_i$.
If a solution exists then it is unique modulo $\lcm \set {n_1, n_2, \ldots, n_r}$.
== Proof ==
We take the case where $r = 2$.
Suppose $x \in \Z$ satisfies both:
That is, $\exists r, s \in \Z$ such that:
Eliminating $x$, we get:
:$b_2 - b_1 = n_1 r - n_2 s$
The is an integer combination of $n_1$ and $n_2$ and so is a multiple of $\gcd \left\{{n_1, n_2}\right\}$.
Thus $\gcd \set {n_1, n_2}$ divides $b_2 - b_1$, so this is a necessary condition for the system to have a solution.
To show sufficiency, we reverse the argument.
Suppose $\exists k \in \Z: b_2 - b_1 = k \gcd \set {n_1, n_2}$.
We know that $\exists u, v \in \Z: \gcd \set {n_1, n_2} = u n_1 + v n_2$ from Bézout's Identity.
Eliminating $\gcd \set {n_1, n_2}$, we have:
:$b_1 + k u n_1 = b_2 - k v n_2$.
Then:
:$b_1 + k u n_1 = b_1 + \paren {k u} n_1 \equiv b_1 \pmod {n_1}$
:$b_1 + k u n_1 = b_2 + \paren {k v} n_2 \equiv b_2 \pmod {n_2}$
So $b_1 + k u n_1$ satisfies both congruences and so simultaneous solutions do exist.
Now to show uniqueness.
Suppose $x_1$ and $x_2$ are both solutions.
That is:
:$x_1 \equiv x_2 \equiv b_1 \pmod {n_1}$
:$x_1 \equiv x_2 \equiv b_2 \pmod {n_2}$
Then from Intersection of Congruence Classes the result follows.
The result for $r > 2$ follows by a tedious induction proof.
Category:Modulo Arithmetic
Solution to Simultaneous Linear Equations

== Theorem ==
Let $\ds \forall i \in \closedint 1 m: \sum _{j \mathop = 1}^n {\alpha_{i j} x_j} = \beta_i$ be a system of simultaneous linear equations
where all of $\alpha_1, \ldots, a_n, x_1, \ldots x_n, \beta_i, \ldots, \beta_m$ are elements of a field $K$.
Then $x = \tuple {x_1, x_2, \ldots, x_n}$ is a solution of this system :
:$\sqbrk \alpha_{m n} \sqbrk x_{n 1} = \sqbrk \beta_{m 1}$
where $\sqbrk a_{m n}$ is an $m \times n$ matrix.
== Proof ==
We can see the truth of this by writing them out in full.
:$\ds \sum_{j \mathop = 1}^n {\alpha_{i j} x_j} = \beta_i$
can be written as:
while $\sqbrk \alpha_{m n} \sqbrk x_{n 1} = \sqbrk \beta_{m 1}$ can be written as:
:$\begin {bmatrix}
\alpha_{1 1} & \alpha_{1 2} & \cdots & \alpha_{1 n} \\
\alpha_{2 1} & \alpha_{2 2} & \cdots & \alpha_{2 n} \\
 \vdots & \vdots & \ddots & \vdots \\
\alpha_{m 1} & \alpha_{m 2} & \cdots & \alpha_{m n}
\end {bmatrix}
\begin {bmatrix}
x_1 \\ x_2 \\ \vdots \\ x_n
\end {bmatrix}
= \begin {bmatrix}
\beta_1 \\ \beta_2 \\ \vdots \\ \beta_m
\end {bmatrix}$
So the question:
:Find a solution to the following system of $m$ simultaneous linear equations in $n$ variables
is equivalent to:
:Given the following element $\mathbf A \in \map {\MM_K} {m, n}$ and $\mathbf b \in \map {\MM_K} {m, 1}$, find the set of all $\mathbf x \in \map {\MM_K} {n, 1}$ such that $\mathbf A \mathbf x = \mathbf b$
where $\map {\MM_K} {m, n}$ is the $m \times n$ matrix space over $S$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 30$. Linear Equations
Category:Simultaneous Linear Equations
Solutions of Diophantine Equation x^4 + y^4 = z^2 + 1 for x = 239

== Theorem ==
Consider the indeterminate Diophantine equation:
:$x^4 + y^4 = z^2 + 1$
When $x = 239$ and $x > y$, there are $3$ solutions:
== Proof ==
== Sources ==
* 
* : $239$
Category:Diophantine Equations
Category:239
Solutions of Diophantine Equation x^4 + y^4 = z^2 + 1 for x = 239/Mistake

== Source Work ==
:
:The Dictionary
::$239$
== Mistake ==
:The 'approximation' to a Fermat equation, $x^4 + y^4 = z^4 + 1$, has $3$ solutions with $x = 239$. The other numbers are $y = 104, z = 58, 136$; $y = 143, z = 60,671$; $y = 208, z = 71, 656$.
== Correction ==
There are $2$ points here:
:$(1): \quad$ The equation in question is $x^4 + y^4 = z^2 + 1$.
:$(2): \quad$ It may not necessarily be the case that there are only $3$ solutions. The cited article claims only that these are the only $3$ solutions where $x > y$.
== Sources ==
* 
* : $239$
Category:Mistakes/Curious and Interesting Numbers
Solutions of Linear 2nd Order ODE have Common Zero iff Linearly Dependent

== Theorem ==
Let $\map {y_1} x$ and $\map {y_2} x$ be particular solutions to the homogeneous linear second order ODE:
:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
on a closed interval $\closedint a b$.
Let $y_1$ and $y_2$ both have a zero for the same value of $x$ in $\closedint a b$.
Then $y_1$ and $y_2$ are constant multiples of each other.
That is, $y_1$ and $y_2$ are linearly dependent.
== Proof ==
Let $\xi \in \closedint a b$ be such that $\map {y_1} \xi = \map {y_2} \xi = 0$.
Consider the Wronskian $\map W {y_1, y_2}$ at $\xi$:
From Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE:
:$\forall x \in \closedint a b: \map W {\map {y_1} \xi, \map {y_2} \xi} = 0$
and so from Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent:
:$y_1$ and $y_2$ are linearly dependent.
== Sources ==
* : $\S 3.14$: Problem $7$
Category:Linear Second Order ODEs
Solutions of Polynomial Congruence

== Theorem ==
Let $\map P x$ be an integral polynomial.
Let $a \equiv b \pmod n$.
Then:
:$\map P a \equiv \map P b \pmod n$
In particular, $a$ is a solution to the polynomial congruence $\map P x \equiv 0 \pmod n$ $b$ is also.
== Proof ==
Let $\map P x = c_m x^m + c_{m - 1} x^{m - 1} + \cdots + c_1 x + c_0$.
Since $a \equiv b \pmod n$, from Congruence of Product and Congruence of Powers, we have:
:$\forall r \in \Z_{>0}: c_r a^r \equiv c_r b^r \pmod n$
From Modulo Addition we then have:
In particular:
:$\map P a \equiv 0 \iff \map P b \equiv 0 \pmod n$
That is, $a$ is a solution to the polynomial congruence $\map P x \equiv 0 \pmod n$ $b$ is also.
Category:Polynomial Congruences
Solutions of Pythagorean Equation

== Theorem ==
=== Primitive Solutions of Pythagorean Equation ===
=== General Solutions of Pythagorean Equation ===
== Sequence ==
== Examples ==
== Historical Note ==
== Sources ==
* 
* : Chapter $\text {A}.9$: Hypatia (A.D. )
* : Chapter $7$: Patterns in Numbers: Diophantus
Category:Diophantine Equations
Category:Pythagorean Triples
Category:Solutions of Pythagorean Equation
Solutions of Pythagorean Equation/General

== Theorem ==
Let $x, y, z$ be a solution to the Pythagorean equation.
Then $x = k x', y = k y', z = k z'$, where:
:$\tuple {x', y', z'}$ is a primitive Pythagorean triple
:$k \in \Z: k \ge 1$
== Proof ==
Let $\tuple {x, y, z}$ be non-primitive solution to the Pythagorean equation.
Let:
:$\exists k \in \Z: k \ge 2, k \divides x, k \divides y$
such that $x \perp y$.
Then we can express $x$ and $y$ as $x = k x', y = k y'$.
Thus:
:$z^2 = k^2 x'^2 + k^2 y'^2 = k^2 z'^2$
for some $z' \in \Z$.
Let:
:$\exists k \in \Z: k \ge 2, k \divides x, k \divides z$
such that $x \perp z$
Then we can express $x$ and $z$ as $x = k x', z = k z'$.
Thus:
:$y^2 = k^2 z'^2 - k^2 x'^2 = k^2 y'^2$
for some $y' \in \Z$.
Similarly for any common divisor of $y$ and $z$.
Thus any common divisor of any pair of $x, y, z$ has to be a common divisor of Integers of the other.
Hence any non-primitive solution to the Pythagorean equation is a constant multiple of some primitive solution.
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.9$: Hypatia (A.D. )
* 
Category:Solutions of Pythagorean Equation
Solutions of Pythagorean Equation/Historical Note

== Historical Note on Solutions of Pythagorean Equation ==
It is clear from the cuneiform tablet that the ancient Babylonians of $2000$ BCE were familiar with this result.
The complete solution of the Pythagorean equation was known to .
The proof was provided by and .
It forms problem $8$ of the second book of his .
It was in the margin of his copy of 's translation of this where made his famous marginal note that led to the hunt for the proof of Fermat's Last Theorem.
== Sources ==
* : Chapter $\text{IV}$: The Prince of Amateurs
* : $13$
* 
* : Pythagorean Triples: $15$
* : $13$
Category:Historical Notes
Solutions of Pythagorean Equation/Primitive

== Theorem ==
The set of all primitive Pythagorean triples is generated by:
:$\tuple {2 m n, m^2 - n^2, m^2 + n^2}$
where:
:$m, n \in \Z_{>0}$ are (strictly) positive integers
:$m \perp n$, that is, $m$ and $n$ are coprime
:$m$ and $n$ are of opposite parity
:$m > n$
== Proof 1 ==
== Proof 2 ==
== Examples ==
== Historical Note ==
== Sources ==
* : $13$
* : $13$
* 
Category:Solutions of Pythagorean Equation
Solutions of Pythagorean Equation/Primitive/Proof 1

== Theorem ==
== Proof ==
First we show that $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ is a Pythagorean triple:
So $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ is indeed a Pythagorean triple.
Now we establish that $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ is primitive:
 to the contrary, that $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ is not primitive.
So there is a prime divisor $p$ of both $2 m n$ and $m^2 - n^2$.
That is:
:$p \in \mathbb P: p \divides \paren {2 m n}, p \divides \paren {m^2 - n^2}$
Then from Prime Divides Power:
:$p \divides \paren {2 m n}^2$ and $p \divides \paren {m^2 - n^2}^2$
Hence from Common Divisor Divides Integer Combination:
:$p \divides \paren {m^2 + n^2}^2$
and from Prime Divides Power again:
:$p \divides \paren {m^2 + n^2}$
So from Common Divisor Divides Integer Combination:
:$p \divides \paren {m^2 + n^2} + \paren {m^2 - n^2} = 2 m^2$
:$p \divides \paren {m^2 + n^2} - \paren {m^2 - n^2} = 2 n^2$
But $p \ne 2$ as, because $m$ and $n$ are of opposite parity, $m^2 - n^2$ must be odd.
So $p \divides n^2$ and $p \divides m^2$ and so from Prime Divides Power, $p \divides n$ and $p \divides m$.
But as we specified that $m \perp n$, this is a contradiction.
Therefore $\paren {2 m n, m^2 - n^2, m^2 + n^2}$ is primitive.
Now we need to show that every primitive Pythagorean triple is of this form:
So, suppose that $\tuple {x, y, z}$ is any primitive Pythagorean triple given in canonical form.
From Parity of Smaller Elements of Primitive Pythagorean Triple, $x$ and $y$ are of opposite parity.
By definition of canonical form $x$ is even and $y$ and $z$ are both odd.
As $y$ and $z$ are both odd, their sum and difference are both even.
Hence we can define:
:$s, t \in Z: s = \dfrac {z + y} 2, t = \dfrac {z - y} 2$.
Note that $s \perp t$ as any common divisor would also divide $s + t = z$ and $s - t = y$, and we know that $z \perp y$ from Elements of Primitive Pythagorean Triple are Pairwise Coprime.
Then from the Pythagorean equation:
:$x^2 = z^2 - y^2 = \paren {z + y} \paren {z - y} = 4 s t$
Hence:
:$\paren {\dfrac x 2}^2 = s t$
As $x$ is even, $\dfrac x 2$ is an integer and so $s t$ is a square.
So each of $s$ and $t$ must be square as they are coprime.
Now, we write $s = m^2$ and $t = n^2$ and substitute back:
:$x^2 = 4 s t = 4 m^2 n^2$ and so $x = 2 m n$
:$y = s - t = m^2 - n^2$
:$z = m^2 + n^2$
Finally, note that:
:$m \perp n$ from $s \perp t$ and Prime Divides Power
:$m$ and $n$ have opposite parity otherwise $y$ and $z$ would be even.
Thus, our primitive Pythagorean triple is of the form $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$.
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.9$: Hypatia (A.D. ): Appendix: A Proof of Diophantus' Theorem on Pythagorean Triples
Category:Solutions of Pythagorean Equation
Solutions of Pythagorean Equation/Primitive/Proof 2

== Theorem ==
== Proof ==
Let $\tuple {A, B, C}$ be a Pythagorean Triple:
:$A^2 + B^2 = C^2$
By the Pythagorean theorem, this equation describes the sides of a right triangle:
:
By the definitions of sine and cosine:
That is,
Next, we invoke Equiangular Triangles are Similar and Proportion is Equivalence Relation to write:
We construct the following restriction of $\tan \dfrac \theta 2$.
From Shape of Tangent Function, $\tan \dfrac \theta 2: \openint 0 \pi \leftrightarrow \openint 0 {+\infty}$ is a bijection.
Restrict $\tan \dfrac \theta 2$ on this interval so that its image is the set of strictly positive rational numbers:
:$\tan \dfrac \theta 2: \tan^{-1} \sqbrk {\Q_{>0} } \cap \openint 0 \pi \leftrightarrow \Q_{>0}$
Then $\ds \tan \frac \theta 2 = \frac p q$ for some $\dfrac p q \in \Q_{>0}$, where $\dfrac p q$ is the canonical form of a rational number.
From the Double Angle Formulas:
Thus these proportions describe the sides of a right triangle:
By the Pythagorean theorem, $\tuple {2 p q, q^2 - p^2, q^2 + p^2}$ is a Pythagorean triple.
That $p, q \in \Z_{>0}$ follows from $\dfrac p q \in \Q_{>0}$.
That $p \perp q$ follows from the assumption that $\dfrac p q$ was written in canonical form.
That every triple is of this form follows from the bijectivity of the tangent function as restricted above.
Recall $q^2 - p^2$ describes the side of a triangle, and so is positive.
Then $q^2 - p^2 > 0$ and therefore $q > p$.
It remains to be proven that:
:this triple is primitive
:$p$ and $q$ are of opposite parity.
== Historical Note ==
Category:Solutions of Pythagorean Equation
Solutions of Pythagorean Equation/Sequence

== Sequence ==
The sequence of solutions of the Pythagorean equation can be tabulated as follows:
:$\begin{array} {r r | r r | r r r | c}
m & n & m^2 & n^2 & 2 m n & m^2 - n^2 & m^2 + n^2 \\
\hline
2 & 1 & 4 & 1 & 4 & 3 & 5 & \text{Primitive} \\
\hline
3 & 1 & 9 & 1 & 6 & 8 & 10 \\
3 & 2 & 9 & 4 & 12 & 5 & 13 & \text{Primitive} \\
\hline
4 & 1 & 16 & 1 & 8 & 15 & 17 & \text{Primitive} \\
4 & 2 & 16 & 4 & 16 & 12 & 20 \\
4 & 3 & 16 & 9 & 24 & 7 & 25 & \text{Primitive} \\
\hline
5 & 1 & 25 & 1 & 10 & 24 & 26 \\
5 & 2 & 25 & 4 & 20 & 21 & 29 & \text{Primitive} \\
5 & 3 & 25 & 9 & 30 & 16 & 34 \\
5 & 4 & 25 & 16 & 40 & 9 & 41 & \text{Primitive} \\
\hline
6 & 1 & 36 & 1 & 12 & 35 & 37 & \text{Primitive} \\
6 & 2 & 36 & 4 & 24 & 32 & 40 \\
6 & 3 & 36 & 9 & 36 & 27 & 45 \\
6 & 4 & 36 & 16 & 48 & 20 & 52 \\
6 & 5 & 36 & 25 & 60 & 11 & 61 & \text{Primitive} \\
\hline
7 & 1 & 49 & 1 & 14 & 48 & 50 \\
7 & 2 & 49 & 4 & 28 & 45 & 53 & \text{Primitive} \\
7 & 3 & 49 & 9 & 42 & 40 & 58 \\
7 & 4 & 49 & 16 & 56 & 33 & 65 & \text{Primitive} \\
7 & 5 & 49 & 25 & 70 & 24 & 74 \\
7 & 6 & 49 & 36 & 84 & 13 & 85 & \text{Primitive} \\
\hline
\end{array}$
Category:Pythagorean Triples
Category:Solutions of Pythagorean Equation
Solutions of Ramanujan-Nagell Equation

== Theorem ==
Integer solutions to the Ramanujan-Nagell equation:
:$x^2 + 7 = 2^n$
exist for only $5$ values of $n$:
:$3, 4, 5, 7, 15$
The corresponding values of $x$ are:
:$1, 3, 5, 11, 181$
== Proof ==
By direct implementation:
== Also see ==
* Five Ramanujan-Nagell Numbers
== Sources ==
* : $15$
::Beware of the terminological mistake: he omits to point out that the equation is Diophantine.
Category:Ramanujan-Nagell Equation
Category:15
Solutions of Ramanujan-Nagell Equation/Mistake

== Source Work ==
:
:The Dictionary
::$15$
== Mistake ==
:The equation $x^2 + 7 = 2^n$ has solutions for only $5$ values of $n$: $3, 4, 5, 7$ and $15$.
== Correction ==
It needs to be pointed out that integer solutions only are under consideration here.
The equation in question is presented as a Diophantine equation.
== Sources ==
* : $15$
Category:Mistakes/Curious and Interesting Numbers
Solutions of cos x equals cos a

== Theorem ==
Let $\alpha \in \closedint {-1} 1$ be fixed.
Let:
:$(1): \quad \cos x = \cos \alpha$
The solution set of $(1)$ is:
:$\set {x \in \R: \forall n \in \Z: x = 2 n \pi \pm \alpha}$
== Proof ==
== Examples ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(35)$
Category:Solutions of cos x equals cos a
Category:Cosine Function
Solutions of cos x equals cos a/Examples/cos 2 x equals sin 3 x

== Example of Use of Solutions of $\cos x = \sin a$ ==
The equation
:$\cos 2 x = \sin 3 x$
has the general solution:
:$x = \paren {4 n + 1} \dfrac \pi {10}$
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Solution of equations: Example $1$.
Category:Solutions of cos x equals cos a
Solutions of cos x equals cos a/Proof

== Theorem ==
== Proof ==
From Cosine of Supplementary Angle:
:$\map \cos {\pi - x} = -\cos x$
and so from Real Cosine Function is Periodic:
== Sources ==
* : $\text V$. Trigonometry: General solution of $\cos x = \cos \alpha$
Category:Solutions of cos x equals cos a
Solutions of sin x equals sin a

== Theorem ==
Let $\alpha \in \closedint {-1} 1$ be fixed.
Let:
:$(1): \quad \sin x = \sin \alpha$
The solution set of $(1)$ is:
:$\set {x \in \R: \forall n \in \Z: x = n \pi + \paren {-1}^n \alpha}$
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(34)$
Category:Solutions of sin x equals sin a
Category:Sine Function
Solutions of sin x equals sin a/Proof

== Theorem ==
== Proof ==
From Sine of Supplementary Angle:
:$\map \sin {\pi - x} = \sin x$
and so from Real Sine Function is Periodic:
== Sources ==
* : $\text V$. Trigonometry: General solution of $\sin x = \sin \alpha$
Category:Solutions of sin x equals sin a
Solutions of tan x equals tan a

== Theorem ==
Let $\alpha \in \R$ be fixed.
Let:
:$(1): \quad \tan x = \tan \alpha$
The solution set of $(1)$ is:
:$\set {x \in \R: \forall n \in \Z: x = n \pi + \alpha}$
== Proof ==
== Examples ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(36)$
Category:Solutions of tan x equals tan a
Category:Tangent Function
Solutions of tan x equals tan a/Examples/2 sec squared x equals 5 tan x

== Examples of Use of Solutions of $\tan x = \tan a$ ==
The equation
:$2 \sec^2 x = 5 \tan x$
has the general solution:
:$\set {n \pi + \arctan \dfrac 1 2: n \in \Z} \cup \set {n \pi + \arctan 2: n \in \Z}$
where $\arctan$ denotes the (real) arctangent function.
== Proof ==
Hence the result.
== Sources ==
* : $\text V$. Trigonometry: Solution of equations: Example $2$.
Category:Solutions of tan x equals tan a
Solutions to Approximate Fermat Equation x^3 = y^3 + z^3 Plus or Minus 1

== Theorem ==
The approximate Fermat equation:
:$x^3 = y^3 + z^3 \pm 1$
has the solutions:
== Proof ==
Performing the arithmetic:
== Proof that there are infinitely many solutions ==
$(1 - 9 t^3)^3 + (9 t^4)^3 + (3 t - 9 t^4)^3=1$
holds for all $t$.
== Sources ==
* : $729$
Category:Approximate Fermat Equations
Solutions to Approximate Fermat Equation x^3 = y^3 + z^3 Plus or Minus 1/Mistake

== Source Work ==
:
:The Dictionary
::$729$
== Mistake ==
Mistake $1$:
:$9^3$ and the $2$nd smallest cube to be the sum of $3$ cubes: $9^3 = 1^3 + 6^3 + 8^3$. This makes $7^3$ the smallest solution to the approximate Fermat equation, $x^3 = y^3 + z^3 + 1$.
Mistake $2$:
:The next such solution is $104^3 = 64^3 + 94^3 + 1$.
The first mistake is that it is not $7^3$ but $9^3$ which is the solution to that equation.
The second mistake is that $104^3 = 64^3 + 94^3 - 1$.
== Sources ==
* : $729$
Category:Mistakes/Curious and Interesting Numbers
Solutions to Diophantine Equation 16x^2+32x+20 = y^2+y

== Theorem ==
The indeterminate Diophantine equation:
:$16x^2 + 32x + 20 = y^2 + y$
has exactly $4$ solutions:
:$\tuple {0, 4}, \tuple {-2, 4}, \tuple {0, -5}, \tuple {-2, -5}$
== Proof ==
$17$ is prime and therefore the solution of only two sets of integer products:
This leaves us with four systems of equations with four solutions:
Hence the solution:
:$\tuple {0, 4}$
Hence the solution:
:$\tuple {-2, 4}$
Hence the solution:
:$\tuple {0, -5}$
Hence the solution:
:$\tuple {-2, -5}$
== Also see ==
* Diophantine Equation
== Sources ==
* 
Category:Diophantine Equations
Solutions to Diophantine Equation x (x + 1) = y (y + 5) (y + 10) (y + 15)

== Theorem ==
The Diophantine equation
: $n = x \paren {x + 1} = y \paren {y + 5} \paren {y + 10} \paren {y + 15}$
has exactly $2$ solutions in $\N \setminus \set 0$:
== Proof ==
First, we observe that:
We note here that the square root of the product $x \paren {x + 1}$ has a fractional part which is less than one half.
Therefore, the fractional part of the square root of the product of the four integers on the must also be less than one half.
Next, we observe that:
Let $a = \paren {y^2 + 15 y + 25}$
We know that $\sqrt {a^2 - 25^2 } \lt a$
We can narrow the search for solutions tremendously as follows:
We need $\sqrt {a^2 - 25^2 } \lt a - \dfrac 1 2$ so that the fractional part of $\sqrt {a^2 - 25^2 }$ might be less than one half.
Therefore:
Therefore:
We now only need to review $18$ cases which are shown below.
:
== Historical Note ==
== Sources ==
* : $1056$
Category:Diophantine Equations
Solutions to Diophantine Equation x (x + 1) = y (y + 5) (y + 10) (y + 15)/Historical Note

== Historical Note on Solutions to Diophantine Equation $x \paren {x + 1} = y \paren {y + 5} \paren {y + 10} \paren {y + 15}$ ==
According to in his of $1997$, this result appeared in on page $194$ of volume $7$.
However, this contributor has tried to identify the article in question, and it appears to have nothing to do with the theorem discussed here.
== Sources ==
* : $1056$
Category:Diophantine Equations
Solutions to p^2 Divides 10^p - 10

== Theorem ==
The known prime numbers $p$ which satisfy the equation:
:$p^2 \divides \paren {10^p - 10}$
where $\divides$ denotes divisibility, are:
:$3, 487, 56 \, 598 \, 313$
== Proof ==
== Also see ==
* Definition:Wieferich Prime
== Sources ==
* 
* : $487$
Category:Prime Numbers
Category:10
Solutions to p^2 Divides 10^p - 10/Mistake

== Source Work ==
:
:The Dictionary
::$487$
== Mistake ==
:One of the only $3$ primes, less than $2^{32}$ such that $p^2$ divides $10 p - 10$. The others are $3$ and $56,598,313$.
== Correction ==
That should read:
:... such that $p^2$ divides $10^p - 10$.
Otherwise it is just not true:
etc.
== Sources ==
* : $487$
Category:Mistakes/Curious and Interesting Numbers
Solutions to x^3 + y^3 + z^3 = 6xyz

== Theorem ==
The Diophantine equation:
:$x^3 + y^3 + z^3 = 6 x y z$
has exactly two kinds of solutions in the integers:
== Proof ==
== Sources ==
* 
* : $6$
Category:Cubic Equations
Category:Diophantine Equations
Solutions to x^3 + y^3 + z^3 = 6xyz/Mistake

== Source Work ==
:
:The Dictionary
:: $6$
== Mistake ==
:The equation $x^3 + y^3 + z^3 = 6 x y z$ has the unique solution $x = 1$, $y = 2$, $z = 3$.
== Correction ==
The correct answer is as given in Solutions to $x^3 + y^3 + z^3 = 6 x y z$
== Sources ==
* : $6$
Category:Mistakes/Curious and Interesting Numbers