User:Kip/Sandbox

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

Is an $n^{th}$ root of unity modulo $m$ where:
 * $\displaystyle n=\frac{\phi(m)}{gcd(\phi(m),x)}$

$\phi(m)$ is Euler's Totient function of the modulus and $gcd(\phi(m),x)$ is the greatest common divisor of the totient and the power.

Proof
Let $\alpha=\frac{x}{gcd(\phi(m),x)}\in\Z_{>0}$ be a positive integer.
 * $\displaystyle A^{\alpha \phi(m)}\equiv a^n\equiv 1\,(mod\,m)$

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
 * $m=3$