User:Kip/Sandbox

Theorem
Let $m\in\Z_{>1}$ be a positive integer greater than one.

Let $x\in\Z_{>1}$ be a positive integer greater than one.

Let $A\in\Z_{>0}$ be a positive integer coprime with $m$.

Then:
 * $\displaystyle A^{x}\equiv a \pmod m$

where $a$ is an $n^{th}$ root of unity modulo $m$. The power of the root has the value:
 * $\displaystyle n=\frac{\phi(m)}{\operatorname{gcd} \left\{{\phi(m),x}\right\}}$

where $\phi(m)$ is Euler's Phi function of the modulus and $\operatorname{gcd}$ is the greatest common divisor.

Proof
Let $q=\frac \alpha n$ be a rational number such that:
 * $\displaystyle x=q \phi(m)$

Then
 * $\displaystyle A^x=A^{q \phi(m)}$

Raising to the $n^{th}$ power:
 * $\displaystyle A^{n x}=A^{\alpha \phi(m)}$

From Euler's Theorem:
 * $\displaystyle A^{\alpha \phi(m)}\equiv 1\pmod m$

Therefore, $A^{x}$ is an $n^{th}$ root of unity modulo $m$.

The canonical form of $q$ produces the smallest denominator as:


 * $\displaystyle n=\frac{\phi(m)}{\operatorname{gcd} \left\{{\phi(m),x}\right\}}$

Square Root of Unity Addition Theorem
Let $p\in\Z_{>0}$ be a prime number greater. Let $x,y,z\in\Z_{>1}$ be positive integers greater than one for which:
 * $2gcd(p-1,x)=2gcd(p-1,y)=n gcd(p-1,z)=p-1$

Then
 * $A^x+B^y\ne C^z \pmod m$

Proof

 * $A^{2x}+2A^xB^y+B^{2y}\equiv C^{2z} \pmod m$

Since the binomial sum is:
 * $\displaystyle \sum_{k\mathop =0}^n \binom n k =2^n$
 * $2A^xB^y\equiv -1 \pmod m$
 * $1\pm 2\equiv 0 \pmod m$
 * $mk=m-1$

Or
 * $m=3$

Definition
Let $x,y,z\in\Z_{>0}$ be (strictly) positive integers.

Let $S=\left({A,B,C}\right)\in\Z_{>0}^3$ be the set of all positive integer triples that satisfy:
 * $A^x + B^y = C^z$

Let $p$ be a prime number that is a factor of one of the triplets of every member of $S$.

Then $p$ is a mandatory triple factor.

The factor can be checked by assuming $p$ is not a factor of the triplets and exhaustively search the permutations of the possible roots of unity modulo $p$ and determine when:
 * $A^x + B^y \not \equiv C^z \pmod p$

Case $p=2$
Two always satisfies this condition. If the triplets are not divisible by $p$ then the modulus is always 1. So:
 * $1+1 \not \equiv 1 \pmod 2$

Case $p=3$
Three has two conditions that do not satisfy this condition. These are when the roots of unity modulo 3 are:
 * $1+1 \equiv -1 \pmod 3$
 * $-1-1 \equiv 1 \pmod 3$

Table of Non-Trivial Mandatory Triple Factors

 * $\begin{array}{|c|c|c|l|}

\hline x & y & z & p \\ \hline 1 & 2 & 2 & 3 \\ 1 & 2 & 4 & 3 \\ 1 & 4 & 2 & 3 \\ 1 & 4 & 4 & 3,5 \\ 2 & 2 & 4 & 3,5 \\ 2 & 3 & 2 & 3 \\ 2 & 3 & 4 & 3 \\ 2 & 4 & 2 & 3,5 \\ 2 & 4 & 4 & 3,5,13 \\ 3 & 3 & 3 & 7,13 \\ 3 & 3 & 6 & 7,13 \\ 3 & 4 & 2 & 3 \\ 3 & 4 & 4 & 3,5 \\ 4 & 4 & 2 & 3,5 \\ 4 & 4 & 4 & 3,5,13,17,41 \\ \vdots \\ \hline \end{array}$
 * 2 & 2 & 2 & 3,5 \\
 * Pythagorean Triples